Work done by a gas and the heat engine
Why does a thermal power plant get less than half of the energy of its fuel as electricity, and where does the rest of the energy go?
Prior knowledge
- The first law $\Delta U = Q + W$, where $W$ is the work done on the system by the surroundings, and conservation of energychapter Internal energy, heat and conservation of energy
- Isobaric change of state: at constant pressure the volume of a gas grows when its temperature riseschapter Changes of state of a gas and Boyle's law
- Work is force times distance, and the area under an $(s, F)$ graphchapter Work and energy
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
In the cylinder of a museum locomotive, hot steam pushes a piston, and the piston turns the wheels of the locomotive. A large condensing power plant does basically the same thing: steam expands in a turbine and turns a generator. The plant is supplied with the energy of fuel at a rate of $1000\,\text{MW}$, but only about $400\,\text{MW}$ goes to the electricity grid. Energy is still conserved, as the first law says.
Could better technology turn all the energy of the fuel into electricity, and can the part that does not become electricity still be put to use?
Optional Review from earlier lessons
Predict
Make your predictions before investigating 2 questions
Prediction 1: work done by a gas
Prediction 2: all the heat into work?
Investigate
In the simulation a gas is heated in a cylinder whose piston is pressed down by two weights; the pressure of the gas is $200\,\text{kPa}$.
Simulation
How to use
Choose the heat $Q$ and press Heat: the piston rises, and the work done by the gas fills in the $(V, p)$ graph. Then press Return the piston. With the choice at the same pressure the weights push the piston back; with the choice cool first the gas first cools to the pressure $p_2$ (slider). Speed changes only the pace, not the results. The orange area is the work $W_\text{g}$ done by the gas, and the blue area is the work done on the gas; in the bars the same colour means the same energy.
Assumptions of the model
The readings are values of the model: the cylinder contains a monatomic ideal gas. The piston is frictionless and weightless, and there is no air above it, so the pressure is caused by the weights: $200\,\text{kPa}$ with two weights, and the lower weight alone gives the pressure $p_2$. The hot plate is always hotter than the gas and the cold water colder. No heat escapes elsewhere. The initial volume is $V_1 = 1.00\,\text{l}$.
How the calculation works
The change in internal energy is $\Delta U = \tfrac{3}{2}nR\Delta T = \tfrac{3}{2}\Delta(pV)$. At constant pressure $p_1$ the work done by the gas is $W_\text{g} = p_1\Delta V$ and $Q = \Delta U + W_\text{g} = \tfrac{5}{2}p_1\Delta V$, so $\Delta V = \dfrac{2Q}{5p_1}$. In the option cool first the upper weight is moved up onto a shelf and a new one is taken from below; when the piston is locked, the volume does not change. The net work of the cycle $W = (p_1 - p_2)\Delta V$ is the energy used to lift the weight, and $Q_1 - Q_2 = W$. The efficiency is $\eta = \dfrac{W}{Q_1}$.
Experiment 1: heating at constant pressure
Test prediction 1. The pressure stays at $200\,\text{kPa}$, because the weights do not change. Change only the heat $Q$, for example $500\,\text{J}$, $750\,\text{J}$ and $1000\,\text{J}$. Press Heat each time and record the measurement only when the piston has stopped. Record at least the row $Q = 1000\,\text{J}$.
Initial state: heat $Q$ = $500\,\text{J}$, piston return = at the same pressure, pressure $p_2$ = $100\,\text{kPa}$. Kept constant: piston return. Varied: heat $Q$.
| heat $Q$ | $\Delta V$ $(\text{l})$ | $\Delta U$ $(\text{J})$ |
|---|
From the change in volume on the row $Q = 1000\,\text{J}$, calculate the work done by the gas $W_\text{g} = p\,\Delta V$ when $p = 200\,\text{kPa}$. Give the answer in joules.
Hint 1: a guiding question
The pressure stays constant, so the work is the area of the orange rectangle in the $(V, p)$ graph: height $p$, width $\Delta V$.
Hint 2: what is the connection
Convert the units: $200\,\text{kPa} = 200\,000\,\text{Pa}$ and $1\,\text{l} = 0.001\,\text{m}^3$.
Hint 3: an intermediate step
On the row
Show the solution
Given
the measurement row with $Q = 1000\,\text{J}$ and constant pressure, $p = 200\,\text{kPa}$; from the table the change in volume $\Delta V = 2.00\,\textsf{l}$
Find
the work done by the gas $W_\text{g}$ in joules
Principle
The pressure stays constant because the weights do not change, so the force of the gas on the piston is the same throughout the expansion. Then the work is the pressure times the change in volume, that is, the area of the rectangle in the $(V, p)$ graph.
Formula
$W_\text{g} = p\,\Delta V$
Conversions
$200\,\text{kPa} = 200\,000\,\text{Pa}$ and $1\,\textsf{l} = 10^{-3}\,\text{m}^3$, so $2.00\,\textsf{l} = 0.00200\,\text{m}^3$. Then $\text{Pa} \cdot \text{m}^3 = \text{J}$.
Substitution
Answer
$400\,\text{J}$
Three significant figures. The limiting value is the change in volume $2.00\,\textsf{l}$ (three figures); the pressure $200\,\text{kPa}$ is a setting of the simulation that is not measured, so it does not limit the precision.
Reasonableness
Estimate:
The work is smaller than the whole quantity of heat $Q = 1000\,\text{J}$, as it should be, because part of the heat stays as internal energy: on the row
For the monatomic gas of the model, $\tfrac{2}{5}$ of the heat always goes into work at constant pressure.
Experiment 2: two ways to return the piston
Test prediction 2. The heat $Q = 500\,\text{J}$ and the pressure $p_2 = 100\,\text{kPa}$ are locked; change only the way the piston returns. Press Heat, then Return the piston, and record the measurement only when the cycle is complete. The efficiency $\eta$ is shown only then, and it is used only to compare ways A and B.
Way A. The weights press the piston down at the same pressure.
Way B.
- The piston is locked at the top, and the upper weight is moved onto a shelf.
- Cold water cools the gas. The volume does not change, so the pressure drops to $p_2$.
- The piston descends at the pressure $p_2$.
- At the bottom a new weight is added, and a hot plate heats the gas back to the initial state.
$Q_1$ is all the heat from the hot plate, $Q_2$ all the heat to the cold water and $W$ the net work of the cycle.
Initial state: heat $Q$ = $500\,\text{J}$, pressure $p_2$ = $100\,\text{kPa}$, piston return = at the same pressure. Kept constant: heat $Q$, pressure $p_2$. Varied: piston return.
| piston return | $Q_1$ $(\text{J})$ | $Q_2$ $(\text{J})$ | $W$ $(\text{J})$ | $\eta$ $(\%)$ |
|---|
Conclusion: where does the heat go?
Use your own tables. Write briefly: (1) how the heat $Q$ of experiment 1 was divided, (2) which return method gave net work and why, and (3) whether a machine can turn all the heat it receives from the hot plate into work.
Model answer
In experiment 1 only part of the heat became work: for example, on the row $Q = 1000\,\text{J}$ the volume grew by $2.0\,\text{l}$, so the gas did the work $p\,\Delta V = 400\,\text{J}$, and the remaining $600\,\text{J}$ stayed as internal energy ($\Delta U$). In experiment 2, method A returned the piston at the same pressure: the net work was $W = 0$, and all the heat went to the cold water. Method B cooled the gas first, so the piston came down at a lower pressure and the net work was $100\,\text{J}$; even then $Q_2 = 550\,\text{J}$ of the heat $Q_1 = 650\,\text{J}$ was transferred to the cold water, so $\eta = 100/650 \approx 15\,\%$. So a machine that works repeatedly cannot turn all the heat into work, even though energy is conserved: $Q_1 = W + Q_2$. The numbers are values of the simulation model; the model is frictionless, so the efficiency of a real machine is lower.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
When a gas expands and pushes a piston, it does work. The gas pushes the piston with a force $F = pA$, where $A$ is the area of the piston. If the piston moves a distance $s$ and the pressure stays the same, the work done by the gas is force times distance, $pAs$. The product $As$ is the change in the volume of the gas $\Delta V$, so the work done by the gas is $W_\text{g} = p\,\Delta V$. In the ($V$, $p$) graph it is the area between the graph and the $V$ axis, just like the work in an ($s$, $F$) graph. When the gas is compressed, the surroundings do work on the gas, and $W_\text{g}$ is negative. A gas that does work gives up energy, so in this lesson the first law is written in the form $\Delta U = Q - W_\text{g}$.
A heat engine repeats its motion
A heat engine turns heat into work continuously, so it has to repeat its motion. The piston has to be brought back down. If it is pressed down at the same pressure, the return takes all the work done by the gas. The return takes less work if the gas is first cooled to a lower pressure, but then the gas gives up heat to a colder environment. In a steam power plant, water boils in the boiler, the hot steam expands in the turbine and turns the generator, and the steam coming out of the turbine condenses back into water in the condenser, which is cooled by seawater. The working substance goes round the same path again and again, so at the end of a cycle its internal energy is the same as at the start. Energy is conserved: the heat $Q_1$ from the hot source equals the work $W$ done by the engine plus the heat $Q_2$ given up to the colder environment.
Experience shows that a heat engine cannot turn all the heat it receives into work; part of it must always be given up to a colder environment. This is the content of the second law of thermodynamics. The larger the temperature difference between the hot source and the cold environment compared with the temperature of the hot source, the larger the share of the heat the engine can turn into work. Cyclic processes are not calculated in this course; the simulation only shows why the heat $Q_2$ cannot be avoided.
Efficiency
Efficiency tells how large a share of the energy supplied to a machine is put to use, so its value depends on what is counted as useful. In the efficiency of a power plant's electricity production, only the electricity is useful. A Finnish condensing power plant that produces only electricity gets about $40\,\%$ of the energy of its fuel as electricity, and a nuclear power plant about $35\,\%$. Most of the rest is transferred to the sea through the condenser, and in a plant that burns fuel part also leaves into the air with the flue gases. The efficiency of a car engine is about $30\,\%$, and the rest of the energy leaves with the exhaust gases and the radiator.
In combined production of electricity and heat, the heat $Q_2$ of the condenser is recovered: it heats the water of the district heating network, which carries it to the houses of the city. So the heat engine still gives up heat to something colder, but this heat is put to use. In the overall efficiency, the electricity and the district heating together count as useful, and it is 85–$90\,\%$. The efficiency of electricity production therefore does not rise; it is usually lower than when only electricity is produced. This answers the opening question: less than half becomes electricity, but the rest does not disappear; it is transferred as heat to the environment or to the district heating network.
Refrigerator and heat pump
A refrigerator and a heat pump work the opposite way to a heat engine: they use electrical energy to transfer heat from a cold place to a warm one. A refrigerator transfers heat from inside the cabinet to the room, and an air-source heat pump from the outdoor air to the inside. The heat given out indoors is the sum of the heat taken from outside and the electricity used, so a pump can heat a house with, for example, three times as much heating power as the electric power it uses. The ratio of the heating power to the electric power is called the coefficient of performance. The coefficient of performance is not an efficiency: it can be greater than 1, because part of the heat that comes indoors is transferred from the outdoor air and does not come from the electricity. The pump does not create energy; it transfers it.
Work done by a gas and efficiency
When the pressure of a gas stays constant, the work done by the gas is
$$W_\text{g} = p\,\Delta V$$
where $W_\text{g}$ is the work done by the gas (J), $p$ the pressure of the gas (Pa) and $\Delta V$ the change in the volume of the gas (m³). The work is positive when the gas expands and negative when the gas is compressed. The change in the internal energy of the gas is
$$\Delta U = Q - W_\text{g}$$
where $Q$ is the quantity of heat transferred to the gas (J). This is the first law $\Delta U = Q + W$ presented in the chapter Internal energy, heat and conservation of energy, where $W$ is the work done on the gas by the surroundings: $W = -W_\text{g}$.
During one cycle of a heat engine, the work done and the efficiency are
$$W = Q_1 - Q_2$$
$$\eta = \frac{W}{Q_1}$$
where $Q_1$ is the heat transferred to the engine from the hot source (J), $Q_2$ the heat given up to the colder environment (J), $W$ the work done by the engine during the cycle (J) and $\eta$ the efficiency, which is often given as a percentage. Efficiency can be calculated in the same way from powers, for example the electric power divided by the fuel power.
A heat pump gives out indoors a quantity of heat that is the sum of the heat taken from the outdoor air and the electrical energy used by the pump:
$$Q_\text{indoors} = Q_\text{outdoors} + E_\text{el}$$
The coefficient of performance tells how many times as much heat the pump gives out indoors as the electrical energy it uses:
$$\varepsilon = \frac{Q_\text{indoors}}{E_\text{el}}$$
where $Q_\text{indoors}$ is the heat given out indoors (J), $Q_\text{outdoors}$ the heat taken from the outdoor air (J), $E_\text{el}$ the electrical energy used by the pump (J) and $\varepsilon$ the coefficient of performance. The coefficient of performance can be calculated in the same way from powers.
Common misconception
Many people think that better technology could make a heat engine almost perfect, so that all the heat would turn into work. Reducing friction and other losses helps, but part of the heat must always be given up to a colder environment, so the efficiency of a heat engine stays clearly below $100\,\%$. The frictionless piston of the simulation shows the reason: for the piston to be brought down with less work than it took to raise it, the gas has to be cooled, and in the cooling heat is transferred to the cold water. The limit applies to a machine that turns heat into work, not to all devices: for example, in an electric heater practically all the electrical energy increases the internal energy of the resistor and is transferred as heat to the room.
Another mistake is to think of a heat pump as a perpetual motion machine, because it gives out more heat than it uses electricity. The extra heat comes from the outdoor air, and energy is not created out of nothing. A third mistake is to think that a pressurised gas does work just because its pressure is high. Work is done only when the volume changes and the piston moves.
Summary
- A gas that expands at constant pressure does the work $W_\text{g} = p\,\Delta V$, which is the area under the ($V$, $p$) graph. The internal energy changes by $\Delta U = Q - W_\text{g}$.
- A piston machine can repeat its work only if the gas is cooled before the piston is returned; then heat is transferred to something colder.
- A heat engine takes heat $Q_1$ from a hot source, does work $W$ and gives up heat $Q_2$ to a colder environment: $W = Q_1 - Q_2$.
- Efficiency is $\eta = W/Q_1$. A heat engine cannot turn all the heat into work.
- The efficiency of electricity production of a condensing power plant is about $40\,\%$. In combined production of electricity and heat the heat of the condenser is used as district heating, so the overall efficiency, in which electricity and heat both count as useful, is 85–$90\,\%$.
- A refrigerator and a heat pump use work to transfer heat from cold to warm. They do not create energy. The coefficient of performance of a heat pump, $\varepsilon = Q_\text{indoors}/E_\text{el}$, is not an efficiency, and it can be greater than 1, because part of the heat comes from the outdoor air.
Optional Go back to your predictions
Compare your predictions with the values of the model in the experiments and with the formulas $W_\text{g} = p\,\Delta V$, $W = Q_1 - Q_2$ and $\eta = W/Q_1$. Then return to the opening question: where does the part of the fuel's energy that does not become electricity go, why can even better technology not turn all of it into electricity, and how can it still be put to use?
Would you change your prediction?
Apply
Worked example
Problem
Wood chip energy is fed into the boiler of a biopower plant at a power of $500\,\text{MW}$. When the plant produces only electricity, the electric power is $190\,\text{MW}$. In winter the same plant works as a combined heat and power plant: it produces $150\,\text{MW}$ of electricity and $290\,\text{MW}$ of district heating. What is the efficiency in each case, and how much power is transferred to the environment as waste heat?
Step 1/7
Given and find. Fuel power $P_1 = 500\,\text{MW}$. When only electricity is produced, the electric power is $P_\text{e} = 190\,\text{MW}$; in combined production $P_\text{e} = 150\,\text{MW}$ and the district heating power is $P_\text{dh} = 290\,\text{MW}$. The efficiencies $\eta$ and the power of the waste heat $P_2$ are asked for in both cases.
Step 2/7
Principle. Efficiency is the useful power divided by the power supplied; it is calculated from powers just like from energies, because both are energy in the same time. Energy is conserved, so the fuel power is divided into the useful power and the waste heat. In electricity production alone only the electricity is useful; in combined production both the electricity and the district heating are.
Step 3/7
Rearranged formula. Solve for the quantities asked for, with symbols before numbers:
$$\begin{aligned} \eta &= \frac{P_\text{useful}}{P_1} \\ P_2 &= P_1 - P_\text{useful} \end{aligned}$$
For electricity alone $P_\text{useful} = P_\text{e}$, and for combined production $P_\text{useful} = P_\text{e} + P_\text{dh}$.
Step 4/7
Unit conversions. All powers are given in megawatts, so no conversions are needed: in the efficiency the unit cancels, and the waste heat comes out in megawatts. A decimal number is changed to a percentage by multiplying by one hundred.
Step 5/7
Substitution. For electricity alone and for combined production separately:
$$\begin{aligned} \eta_1 &= \frac{P_\text{e}}{P_1} \\ &= \frac{190\,\text{MW}}{500\,\text{MW}} \\ &= 0.380 \\ P_{2,1} &= P_1 - P_\text{e} \\ &= 500\,\text{MW} - 190\,\text{MW} \\ &= 310\,\text{MW} \\[6pt] \eta_2 &= \frac{P_\text{e} + P_\text{dh}}{P_1} \\ &= \frac{150\,\text{MW} + 290\,\text{MW}}{500\,\text{MW}} \\ &= 0.880 \\ P_{2,2} &= P_1 - (P_\text{e} + P_\text{dh}) \\ &= 500\,\text{MW} - 440\,\text{MW} \\ &= 60\,\text{MW} \end{aligned}$$
Step 6/7
Result. With electricity production alone the efficiency is $\eta_1 = 0.380 = 38.0\,\%$ and $310\,\text{MW}$ of waste heat is transferred to the environment, most of it through the condenser to the sea. In combined production the overall efficiency is $\eta_2 = 0.880 = 88.0\,\%$ and there is only $60\,\text{MW}$ of waste heat. The powers are given in whole megawatts, so the differences $310\,\text{MW}$ and $60\,\text{MW}$ are known to the nearest megawatt. The efficiencies are quotients, so they have as many figures as the least precise power: $190\,\text{MW}$, $150\,\text{MW}$, $290\,\text{MW}$ and $500\,\text{MW}$ all have three figures, so the efficiencies are given with three figures.
Step 7/7
Reasonableness. Energy is conserved: $190\,\text{MW} + 310\,\text{MW} = 500\,\text{MW}$ and $150\,\text{MW} + 290\,\text{MW} + 60\,\text{MW} = 500\,\text{MW}$. The efficiency of electricity production stays clearly below $100\,\%$, because a heat engine must give up part of the heat to something colder. Even the overall efficiency is not $100\,\%$, because part of the heat leaves with the flue gases and other losses. Less electricity is obtained in combined production because the steam is condensed at a higher temperature so that its heat is suitable for district heating. The energy of the fuel is still put to much better use.
Optional Find the mistake
Design the energy production of a city
A city needs $300\,\text{MW}$ of electricity and $500\,\text{MW}$ of district heating. Compare three options by how much fuel power each one needs. Assumptions:
- Condensing power plant: electricity is $40\,\%$ of the fuel power.
- Electric heating: $1\,\text{MW}$ of electricity gives $1\,\text{MW}$ of heat.
- Combined heat and power plant: electricity is $45\,\%$ and district heat $40\,\%$ of the fuel power.
- Heat pump: coefficient of performance 3.0.
The options:
- Condensing power plant and electric heating. The condensing power plant produces all the electricity, including that used for heating.
- Combined heat and power. The combined heat and power plant produces all the district heating, and at the same time more electricity is produced than the city needs.
- Combined heat and power and a heat pump. Part of the plant's electricity is used in a heat pump, which takes heat from purified waste water. The plant is sized so that the city gets exactly the electricity and heat it needs.
The assumption for the combined heat and power plant corresponds to a plant that burns natural gas; in a plant that burns wood the share of electricity is smaller, as in the Worked example.
First calculate the fuel power of options 1 and 2. How much fuel power does option 3 need? Give the answer in megawatts.
Hint 1: a guiding question
Denote the fuel power of the combined heat and power plant by $P$ and the electric power of the heat pump by $P_\text{pump}$. What two conditions must the electricity and heat of the city meet?
Hint 2: what is the connection
The city is left with electricity $0.45P - P_\text{pump} = 300\,\text{MW}$, and heat comes from the plant and the pump: $0.40P + 3P_\text{pump} = 500\,\text{MW}$.
Hint 3: an intermediate step
From the first condition $P_\text{pump} = 0.45P - 300\,\text{MW}$. Substitute it into the second condition: $0.40P + 1.35P - 900\,\text{MW} = 500\,\text{MW}$.
Show the solution
Given
the electricity need of the city $P_\text{e} = 300\,\text{MW}$ and the district heating need $P_\text{h} = 500\,\text{MW}$; the efficiency of the condensing power plant $0.40$, the combined heat and power plant gives electricity $0.45P$ and heat $0.40P$ from the fuel power $P$, the coefficient of performance of the heat pump $3.0$; option 3 is examined
Find
the fuel power $P$ of option 3 in megawatts (for comparison, the fuel power of options 1 and 2)
Principle
In options 1 and 2 a single plant meets the demand, and the fuel power is the useful power divided by the efficiency. In option 3 part of the plant's electricity goes to the heat pump, so the balances of electricity and heat must be written separately. Energy is conserved: the pump gives out heat equal to three times its electric power, and the difference comes from the waste water.
Formula
Option 1:
Option 2: $P = \dfrac{P_\text{h}}{0.40}$. In option 3 the conditions are $0.45P - P_\text{pump} = P_\text{e}$ and $0.40P + 3P_\text{pump} = P_\text{h}$. Eliminating $P_\text{pump}$:
$$P = \frac{P_\text{h} + 3P_\text{e}}{0.40 + 3 \cdot 0.45}$$
Conversions
All powers are in megawatts and the efficiencies are decimal numbers, so no conversions are needed.
Substitution
Option 1:
Option 2:
and the electricity produced is $0.45 \cdot 1250\,\text{MW} = 562.5\,\text{MW}$, which is $262.5\,\text{MW}$ more than needed. Option 3:
Answer
$800\,\text{MW}$
Two significant figures. The limiting values are the efficiencies $0.40$ and $0.45$ (two figures); the needs of the city $300\,\text{MW}$ and $500\,\text{MW}$ and the coefficient of performance $3.0$ are design values of the problem that are read as having at least two figures ($3.0 \cdot 10^2$, $5.0 \cdot 10^2$), so the result is $8.0 \cdot 10^2\,\text{MW}$.
Reasonableness
Check the balance again: the electric power of the pump is $0.45 \cdot 800\,\text{MW} - 300\,\text{MW} = 60\,\text{MW}$ and it gives out heat $3 \cdot 60\,\text{MW} = 180\,\text{MW}$, of which $120\,\text{MW}$ comes from the waste water. The plant gives heat $0.40 \cdot 800\,\text{MW} = 320\,\text{MW}$, and together $320\,\text{MW} + 180\,\text{MW} = 500\,\text{MW}$, as needed. The losses of the plant, $0.15 \cdot 800\,\text{MW} = 120\,\text{MW}$, are made up by the $120\,\text{MW}$ taken from the waste water. Option 3 needs only $40\,\%$ of the fuel of option 1.
Optional Apply to a new situation
A large heat pump that produces district heating takes heat from purified waste water and gives it to the district heating network. The electric power of the pump is $9.00\,\text{MW}$ and its coefficient of performance is 3.50.
How much heat power does the pump take from the waste water? Give the answer in megawatts.
Hint 1: a guiding question
From which two sources does the heat given to the district heating network come?
Hint 2: what is the connection
The coefficient of performance is $\varepsilon = P_\text{indoors}/P_\text{pump}$, and energy is conserved: $P_\text{indoors} = P_\text{sewage} + P_\text{pump}$.
Hint 3: an intermediate step
The power given to the network is
Subtract the electric power from it.
Show the solution
Given
the electric power of the heat pump $P_\text{pump} = 9.00\,\text{MW}$, coefficient of performance $\varepsilon = 3.50$
Find
the heat power taken from the waste water $P_\text{sewage}$ in megawatts
Principle
The coefficient of performance tells how many times as much heat power the pump gives to the network as its electric power. Energy is conserved: the power given to the network is the electric power of the pump and the power taken from the waste water together, because the pump only transfers heat.
Formula
$P_\text{indoors} = \varepsilon P_\text{pump}$ and $P_\text{indoors} = P_\text{sewage} + P_\text{pump}$, so
$$\begin{aligned} &P_\text{sewage} \\ &= P_\text{indoors} - P_\text{pump} \\ &= (\varepsilon - 1) P_\text{pump} \end{aligned}$$
Conversions
The power is already in megawatts and the coefficient of performance has no unit, so no conversions are needed.
Substitution
Answer
$22.5\,\text{MW}$
Three significant figures. The limiting values are the electric power $9.00\,\text{MW}$ and the difference $\varepsilon - 1 = 3.50 - 1 = 2.50$ (three figures, because the integer $1$ is exact); both are given with three figures.
Reasonableness
The power going to the network is $3.50 \cdot 9.00\,\text{MW} = 31.5\,\text{MW}$, and $22.5\,\text{MW} + 9.00\,\text{MW} = 31.5\,\text{MW}$ agrees. The power from the waste water is greater than the electric power, as it should be when the coefficient of performance is above 2. The pump does not create energy but transfers it.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The pressure of a gas in a cylinder stays at $200\,\text{kPa}$ while its volume grows by $0.50\,\textsf{l}$. How much work does the gas do while the piston moves?
Answer
$W_\text{g}$$= p\,\Delta V$$= 200\,000\,\text{Pa} \cdot 0.000\,50\,\text{m}^3$$= 100\,\text{J}$ - The energy of fuel is supplied to a car engine at a power of $60\,\text{kW}$, and the efficiency of the engine is $30\,\%$. What is the mechanical power, and where does the rest of the energy go?
Answer
The mechanical power is $0.30 \cdot 60\,\text{kW} = 18\,\text{kW}$. The remaining $42\,\text{kW}$ is transferred as heat to the surroundings through the exhaust gases and the radiator.
1. Concept
2. Heat pump
3. Graph and calculation
The $(V, p)$ graph of a gas is a horizontal line segment at the height $p = 150\,\text{kPa}$, starting at $V = 2.0\,\text{l}$ and ending at $V = 5.0\,\text{l}$.
How much work $W_\text{g}$ does the gas do? Give the answer in joules.
Hint 1: a guiding question
What does the area between the line segment and the $V$ axis mean, and what shape is it?
Hint 2: what is the connection
The area of the rectangle is $W_\text{g} = p\,\Delta V$, where $\Delta V = V_2 - V_1$.
Hint 3: an intermediate step
Show the solution
Given
the $(V, p)$ graph is a horizontal line segment with $p = 150\,\text{kPa}$, initial volume $V_1 = 2.0\,\textsf{l}$ and final volume $V_2 = 5.0\,\textsf{l}$
Find
the work done by the gas $W_\text{g}$ in joules
Principle
The work done by a gas is the area between the $(V, p)$ graph and the $V$ axis. The pressure stays constant because the line segment is horizontal, so the region is a rectangle and the work is the pressure times the change in volume.
Reading the graph
The height of the rectangle is the height of the line segment, $p = 150\,\text{kPa}$, and the width is the distance between the start and end points of the segment,
The values of the start and end points are not multiplied together; their difference is the width.
Formula
$W_\text{g} = p\,\Delta V$, where $\Delta V = V_2 - V_1$
Conversions
Then $\text{Pa} \cdot \text{m}^3 = \text{J}$.
Substitution
Answer
$450\,\text{J}$
Two significant figures. The limiting value is the change in volume $\Delta V = 5.0\,\textsf{l} - 2.0\,\textsf{l} = 3.0\,\textsf{l}$ (the difference of the readings has one decimal and two figures); the pressure $150\,\text{kPa}$ has at least two figures.
Reasonableness
The work is positive because the volume grows. Estimate:
If the final volume $5.0\,\textsf{l}$ had been used, the result would be $750\,\text{J}$, which would correspond to the whole rectangle from zero volume. During the expansion work is done only between $2.0$ and $5.0\,\textsf{l}$.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
4. Explain and transfer
Your friend suggests: “Let’s build such a good power plant that it needs no condenser at all. Then all the energy of the fuel becomes electricity.” Explain why this is not possible, and say how the heat of the condenser can still be put to use.
Model answer
A heat engine turns heat into work only when part of the heat is given up to a colder environment; this is the second law of thermodynamics. The piston of the simulation showed the reason: for the piston to be brought back with less work than it took to raise it, the gas has to be cooled. In a power plant the condenser does the same: without it the steam would not condense back into water, and the cycle could not continue. Energy is conserved, $Q_1 = W + Q_2$, so the part $W$ that becomes electricity is always smaller than the heat $Q_1$ obtained from the fuel. Reducing friction raises the efficiency somewhat, but the heat $Q_2$ that has to be given up cannot be removed. The heat of the condenser can, however, be led to the district heating network instead of being released into the sea. A combined heat and power plant of this kind gets 85–$90\,\%$ of the energy of the fuel as useful electricity and heat together. This is the overall efficiency: the heat $Q_2$ is still given up, but it is put to use, and no more electricity is obtained.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Work done by a gas
- An expanding gas does work on the surroundings as its volume grows.
- Heat engine
- Turns part of the energy received from a hot system into mechanical work.
- Hot reservoir
- The system at higher temperature from which the engine receives energy as heat.
- Cold reservoir
- The system at lower temperature to which the engine gives up part of the energy it received.
- Waste heat
- Energy passed to the colder surroundings that cannot be turned into useful work.
- Efficiency
- The fraction of the energy received that is turned into useful work.
- Cycle
- A system goes through changes of state and finally returns to its original state.
- Refrigerator
- Uses work to move energy from a colder space to warmer surroundings.
- Heat pump
- Uses external work to move energy from cooler surroundings to the space being heated.
- Compressor
- Compresses the refrigerant, so its pressure and usually its temperature rise.
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 13-1Core
Steam flows from the boiler into the cylinder of a steam locomotive on a heritage railway and pushes the piston. While the piston moves, the pressure of the steam stays at $480.0\,\text{kPa}$, and the volume of the steam in the cylinder grows by $0.0250\,\text{m}^3$.
How much work does the steam do while the piston moves? Give the answer in kilojoules.
Answer
$12.0\,\text{kJ}$
Show the solution
Given
the steam pressure stays constant, $p = 480.0\,\text{kPa}$, change in volume $\Delta V = 0.0250\,\text{m}^3$
Find
the work $W_\text{g}$ done by the steam on the piston in kilojoules
Principle
The pressure stays constant, so the force $F = pA$ that the steam exerts on the piston is the same throughout the motion, and the work is force times distance. Since the change in volume is the area of the piston times the distance, $\Delta V = A s$, the work is the pressure times the change in volume.
Formula
$W_\text{g} = F s = pA s$, so $W_\text{g} = p\,\Delta V$
Conversions
$480.0\,\text{kPa} = 480\,000\,\text{Pa}$. The volume is already in cubic metres, so $\text{Pa} \cdot \text{m}^3 = \text{J}$ without further conversion. Finally $1\,\text{kJ} = 1000\,\text{J}$.
Substitution
$W_\text{g}$$= 480\,000\,\text{Pa} \cdot 0.0250\,\text{m}^3$$= 12\,000\,\text{J}$$= 12.0\,\text{kJ}$Answer
$12.0\,\text{kJ}$
Three significant figures. Both values, the pressure $480.0\,\text{kPa}$ and the change in volume $0.0250\,\text{m}^3$, are given with at least three figures.
Reasonableness
The unit is correct:
$\text{Pa} \cdot \text{m}^3$$= \tfrac{\text{N}}{\text{m}^2} \cdot \text{m}^3$$= \text{N} \cdot \text{m}$$= \text{J}$Estimate: $500\,\text{kPa} \cdot 0.025\,\text{m}^3 \approx 12\,\text{kJ}$. The work is positive because the steam expands and pushes the piston. It would be enough to lift a load of about 1.2 tonnes by a metre, and the pistons of a locomotive do this much work many times a second.
- Exercise 13-2Core
Water is thrown on the stones of a sauna stove. When one gram of water evaporates, the resulting steam at $100\,{}^\circ\text{C}$ takes up $1.70\,\textsf{l}$ more space than the water did. The evaporation takes place at atmospheric pressure, which is $101\,\text{kPa}$.
How much work does the steam do as it pushes the surrounding air out of the way?
Answer
$172\,\text{J}$
Show the solution
Given
atmospheric pressure $p = 101\,\text{kPa}$, additional volume taken by the steam $\Delta V = 1.70\,\textsf{l}$
Find
the work $W_\text{g}$ done by the steam against the surrounding air in joules
Principle
The steam pushes the air out of the way, and the atmospheric pressure stays approximately constant throughout the evaporation. At constant pressure the work done by a gas is the pressure times the change in volume.
Formula
$W_\text{g} = p\,\Delta V$
Conversions
$101\,\text{kPa} = 101\,000\,\text{Pa}$ and $1\,\textsf{l} = 10^{-3}\,\text{m}^3$, so $1.70\,\textsf{l} = 0.00170\,\text{m}^3$. Then $\text{Pa} \cdot \text{m}^3 = \text{J}$.
Substitution
$W_\text{g}$$= 101\,000\,\text{Pa} \cdot 0.00170\,\text{m}^3$$= 171.7\,\text{J}$$\approx 172\,\text{J}$Answer
$172\,\text{J}$
Three significant figures, like the values $101\,\text{kPa}$ and $1.70\,\textsf{l}$.
Reasonableness
Estimate: $100\,\text{kPa} \cdot 0.0017\,\text{m}^3 = 170\,\text{J}$. Evaporating a gram takes about $2.3\,\text{kJ}$ of heat, so the share of the work is $\tfrac{172\,\text{J}}{2300\,\text{J}} \approx 7.5\,\%$, less than a tenth. The rest of the heat stays as internal energy of the steam. The puff of steam from the sauna stones is exactly this expansion.
- Exercise 13-3
The reactor of the third nuclear power plant unit at Olkiluoto produces steam with a thermal power of $4.30 \cdot 10^{3}\,\text{MW}$, and the electric power of the unit is $1.60 \cdot 10^{3}\,\text{MW}$.
What is the efficiency of the unit? Give the answer as a percentage.
Answer
$37.2\,\%$
Show the solution
Given
thermal power of the reactor $P_\text{heat} = 4.30 \cdot 10^3\,\text{MW}$, electric power $P_\text{el} = 1.60 \cdot 10^3\,\text{MW}$
Find
the efficiency $\eta$ as a percentage
Principle
A heat engine turns part of the heat it receives into work, that is, electricity. Efficiency is the share of the energy supplied that is put to use, and it can also be calculated from powers, because both powers are energy in the same time.
Formula
$\eta$$= \dfrac{P_\text{el}}{P_\text{heat}}$Conversions
Both powers are in megawatts, so the unit cancels and no conversion is needed. A decimal number is changed to a percentage by multiplying by one hundred.
Substitution
$\eta$$= \dfrac{1.60 \cdot 10^3\,\text{MW}}{4.30 \cdot 10^3\,\text{MW}}$$= 0.3721$$= 37.21\,\%$Answer
$37.2\,\%$
Three significant figures. Both powers ($1.60 \cdot 10^3\,\text{MW}$ and $4.30 \cdot 10^3\,\text{MW}$) are given with three figures.
Reasonableness
The efficiency is below $100\,\%$, as it should be for a heat engine, because part of the heat has to be given up to something colder, and the typical value for a nuclear power plant is about $35\,\%$. Energy balance: the seawater takes $4300\,\text{MW} - 1600\,\text{MW} = 2700\,\text{MW}$ from the condenser, more than the unit produces as electricity. Estimate:
$\tfrac{1600}{4300}$$\approx \tfrac{1}{2.7}$$\approx 0.37$ - Exercise 13-4
The gas in the cylinder of a small gas engine is heated, and the gas pushes a piston forward. The graph shows the pressure $p$ of the gas as a function of the volume $V$ during the expansion.
How much work does the gas do as it expands? Read the values you need from the graph. Give the answer in joules.
Answer
$W_\text{g} = 750\,\text{J}$
Show the solution
Given
read from the graph, the pressure stays constant at $p = 250\,\text{kPa}$, and the volume grows from $V_1 = 1.0\,\textsf{l}$ to $V_2 = 4.0\,\textsf{l}$
Find
the work $W_\text{g}$ done by the gas in joules
Principle
The pressure is the same throughout the expansion, so the work of the gas is the product of the pressure and the change in volume. The work is positive, because the gas expands and pushes the piston.
Reading the graph
The segment is horizontal and runs along the grid line $p = 250\,\text{kPa}$. Its ends are at grid crossings, $V_1 = 1.0\,\textsf{l}$ and $V_2 = 4.0\,\textsf{l}$, so
$\Delta V$$= V_2 - V_1$$= 3.0\,\textsf{l}$Formula
$W_\text{g} = p\,\Delta V$
Conversions
$p$$= 250\,\text{kPa}$$= 250\,000\,\text{Pa}$$\Delta V$$= 3.0\,\textsf{l}$$= 0.0030\,\text{m}^3$because $1\,\textsf{l} = 10^{-3}\,\text{m}^3$.
Substitution
$W_\text{g}$$= 250\,000\,\text{Pa} \cdot 0.0030\,\text{m}^3$$= 750\,\text{J}$Answer
$W_\text{g} = 750\,\text{J}$
Two significant figures, because the volumes $1.0\,\textsf{l}$ and $4.0\,\textsf{l}$ are read to two figures. The values are read from the grid lines, so the accepted range is $735$–$765\,\text{J}$.
Reasonableness
The units match: $1\,\text{Pa} \cdot 1\,\text{m}^3 = 1\,\text{J}$. The result equals the area of the rectangle in the graph: a height of $250\,\text{kPa}$ and a width of $3.0\,\textsf{l}$ give $250 \cdot 3.0 = 750\,\text{kPa}\cdot\textsf{l}$, and $1\,\text{kPa}\cdot\textsf{l} = 1\,\text{J}$. The work is positive in an expansion, as it should be.
- Exercise 13-5
A generator at a cottage produces $2.60\,\text{kWh}$ of electricity in an evening. In the same time the energy of the petrol burnt in the engine of the generator is $12.0\,\text{kWh}$.
What is the efficiency of the generator? Give the answer as a percentage.
Answer
$21.7\,\%$
Show the solution
Given
electrical energy $W = 2.60\,\text{kWh}$, energy of the petrol $Q_1 = 12.0\,\text{kWh}$
Find
the efficiency $\eta$ as a percentage
Principle
A generator is a heat engine: only part of the energy of the petrol becomes electricity, and efficiency tells this share. Both energies are given in the same unit.
Formula
$\eta = \dfrac{W}{Q_1}$
Conversions
Both energies are in kilowatt-hours, so the unit cancels and there is no need to convert to joules. A decimal number is changed to a percentage by multiplying by one hundred.
Substitution
$\eta$$= \dfrac{2.60\,\text{kWh}}{12.0\,\text{kWh}}$$= 0.2167$$= 21.67\,\%$Answer
$21.7\,\%$
Three significant figures. Both energies ($2.60\,\text{kWh}$ and $12.0\,\text{kWh}$) are given with three figures.
Reasonableness
The efficiency is below $100\,\%$, as it should be for a heat engine. A small engine falls short of the roughly $30\,\%$ of a car engine. Energy balance: the remaining $12.0\,\text{kWh} - 2.60\,\text{kWh} = 9.4\,\text{kWh}$ is transferred as heat to the surroundings, and $\tfrac{9.4}{12.0} \approx 78\,\%$ completes the efficiency to one hundred per cent.
- Exercise 13-6
A gas expands in a cylinder at constant pressure, and in the $(V, p)$ coordinate system the expansion appears as a horizontal line segment.
Which of the following correctly tells how the work done by the gas can be seen in the coordinate system?
Answer
The work is the area of the rectangle between the segment and the volume axis, because the height is the pressure and the width is the change in volume.
Show the solution
Find
how the work done by a gas expanding at constant pressure can be seen in the $(V, p)$ coordinate system.
Principle
At constant pressure the work done by the gas is $W_\text{g} = p\,\Delta V$. In the coordinate system the pressure is the height of the horizontal segment and the change in volume $\Delta V$ is its length, that is, its width, so the product is the area of the rectangle between the segment and the volume axis.
Model answer
The work is the area of the rectangle under the segment, because its height is the pressure and its width is the change in volume. The height alone or the length alone is not the work; to get the work you need both the pressure and the change in volume. The work is not zero either, even though the pressure does not change: the piston moves when the volume grows, and the force does work. The unit $\text{Pa} \cdot \text{m}^3$ is the joule.
- Exercise 13-7
A heat pump gives out more heat indoors than the electrical energy it uses. Why is this not in conflict with the conservation of energy?
Answer
The pump transfers heat from the outdoor air to the inside, so the energy given out indoors is the sum of the heat taken from outdoors and the electricity.
Show the solution
Find
why the heat that a heat pump gives out indoors can be greater than the electrical energy it uses without breaking the conservation of energy.
Principle
A heat pump uses electricity to transfer heat from the colder outdoor air to the warmer indoor air. Energy is conserved, so the heat given out indoors is the sum of the heat taken from outdoors and the electrical energy used by the pump: $Q_\text{indoors} = Q_\text{outdoors} + E_\text{el}$.
Model answer
Part of the heat given out indoors has been taken from the outdoor air, so the energy going indoors is the sum of the heat from the outdoor air and the electrical energy. The pump does not produce energy but transfers it, and the outdoor air cools a little. The coefficient of performance $\varepsilon = Q_\text{indoors}/E_\text{el}$ can therefore be over 1, although the efficiency cannot exceed $100\,\%$. Electricity and heat are both energy, so they can be compared with each other.
- Exercise 13-8
An air-to-water heat pump uses an electric power of $1.20\,\text{kW}$ and gives a thermal power of $3.90\,\text{kW}$ to the heating circuit.
What is the coefficient of performance of the pump?
Answer
$\varepsilon = 3.25$
Show the solution
Given
electric power of the pump $P_\text{el} = 1.20\,\text{kW}$, thermal power given out to the heating circuit $P_\text{indoors} = 3.90\,\text{kW}$
Find
the coefficient of performance $\varepsilon$
Principle
The coefficient of performance tells how many times as much heat the pump gives out indoors as the electrical energy it uses. It can also be calculated from the powers, because both powers are energy in the same time.
Formula
$\varepsilon$$= \dfrac{Q_\text{indoors}}{E_\text{el}}$$= \dfrac{P_\text{indoors}}{P_\text{el}}$Conversions
Both powers are in kilowatts, so the unit cancels and no conversion is needed. The coefficient of performance has no unit.
Substitution
$\varepsilon$$= \dfrac{3.90\,\text{kW}}{1.20\,\text{kW}}$$= 3.25$Answer
$\varepsilon = 3.25$
Three significant figures, because both powers are given with three figures.
Reasonableness
The coefficient of performance is over 1, as it can be for a heat pump. The power taken from the outdoor air is $3.90\,\text{kW} - 1.20\,\text{kW} = 2.70\,\text{kW}$, that is, $2.25$ times the electric power. Estimate:
$\tfrac{3.9}{1.2}$$\approx \tfrac{4}{1.2}$$\approx 3.3$
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 13-9Core
A condensing power plant that burns wood chips and peat produces $3\,830\,\text{MWh}$ of electricity in a day. In the same time the energy of the fuel burnt in the boiler is $33.6\,\text{TJ}$ ($1\,\text{TJ} = 10^{12}\,\text{J}$).
What is the efficiency of the plant? Give the answer as a percentage.
Answer
$41.0\,\%$
Show the solution
Given
electrical energy $W = 3830\,\text{MWh}$, energy of the fuel $Q_1 = 33.6\,\text{TJ} = 33.6 \cdot 10^{12}\,\text{J}$, in a day
Find
the efficiency $\eta$ as a percentage
Principle
A condensing power plant is a heat engine, and its efficiency is the share of the energy of the fuel that has become electricity. To calculate the share, both energies must be expressed in the same unit.
Formula
$\eta = \dfrac{W}{Q_1}$
Conversions
$1\,\text{MWh}$$= 10^6\,\text{W} \cdot 3600\,\text{s}$$= 3.6 \cdot 10^9\,\text{J}$so
$W$$= 3830 \cdot 3.6 \cdot 10^9\,\text{J}$$= 1.3788 \cdot 10^{13}\,\text{J}$Also $33.6\,\text{TJ} = 3.36 \cdot 10^{13}\,\text{J}$.
Substitution
$\eta$$= \dfrac{1.3788 \cdot 10^{13}\,\text{J}}{3.36 \cdot 10^{13}\,\text{J}}$$= 0.4104$$= 41.04\,\%$Answer
$41.0\,\%$
Three significant figures. Both values ($3830\,\text{MWh}$ and $33.6\,\text{TJ}$) are given with at least three figures, and the factor $3.6 \cdot 10^9\,\text{J/MWh}$ adds no uncertainty because it is a definition.
Reasonableness
The efficiency of a condensing power plant is typically about $40\,\%$, so the result is believable. Estimate:
$\tfrac{3.8 \cdot 3.6}{33.6}$$\approx \tfrac{13.7}{33.6}$$\approx 0.41$The average electric power is
$\dfrac{3830\,\text{MWh}}{24\,\text{h}}$$\approx 160\,\text{MW}$and the fuel power $\tfrac{160}{0.41} \approx 390\,\text{MW}$.
- Exercise 13-10Core
In the boiler of an old steamship that sails on Lake Saimaa, $120\,\text{kg}$ of birch logs are burnt in an hour. The heat of combustion of air-dry birch is $15\,\text{MJ/kg}$, that is, $15\,\text{MJ}$ of energy is released when a kilogram burns. The steam engine of the ship does work at an average power of $45\,\text{kW}$.
What is the overall efficiency of the boiler and the steam engine? Give the answer as a percentage.
Answer
$9.0\,\%$
Show the solution
Given
mass of firewood $m = 120\,\text{kg}$ per hour, heat of combustion $H = 15\,\tfrac{\text{MJ}}{\text{kg}}$, power of the steam engine $P = 45\,\text{kW}$, time $t = 1\,\text{h}$
Find
the overall efficiency $\eta$ of the boiler and the steam engine as a percentage
Principle
The whole plant is a heat engine: the burning of the firewood releases heat $Q_1$ and the engine does work $W$. Efficiency is the share of this heat that becomes work, and both are calculated for the same hour.
Formula
$Q_1 = mH$ and $W = Pt$, so
$\eta$$= \dfrac{W}{Q_1}$$= \dfrac{Pt}{mH}$Conversions
$45\,\text{kW} = 45\,000\,\text{W}$,
$1\,\text{h}$$= 3600\,\text{s}$$15\,\tfrac{\text{MJ}}{\text{kg}}$$= 15 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}$so that both energies are in joules.
Substitution
$\eta$$= \dfrac{45\,000\,\text{W} \cdot 3600\,\text{s}}{120\,\text{kg} \cdot 15 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}}$$= \dfrac{1.62 \cdot 10^8\,\text{J}}{1.8 \cdot 10^9\,\text{J}}$$= 0.090$$= 9.0\,\%$Answer
$9.0\,\%$
Two significant figures. The limiting values are the power $45\,\text{kW}$ and the heat of combustion $15\,\tfrac{\text{MJ}}{\text{kg}}$ (both two figures); the mass $120\,\text{kg}$ has at least two figures and does not limit the precision, and $3600\,\text{s/h}$ is a definition.
Reasonableness
The efficiency of old steam engines is 5–$10\,\%$, so the result fits. The fuel power is $\tfrac{1.8 \cdot 10^9\,\text{J}}{3600\,\text{s}} = 500\,\text{kW}$, of which only $45\,\text{kW}$ becomes work and the remaining $455\,\text{kW}$ leaves through the chimney and with the exhaust steam. That is why steamships were replaced by diesel engines.
- Exercise 13-11
In a physics lesson there is air in a cylinder closed by a lightly moving piston. The air is heated, and a quantity of heat of $700.0\,\text{J}$ is transferred to it. The air expands so that its pressure stays at $125\,\text{kPa}$ and its volume grows by $1.60\,\textsf{l}$.
By how much does the internal energy of the air change? Use the first law $\Delta U = Q - W_\text{g}$, where $W_\text{g}$ is the work done by the air. Give the answer with its sign.
Answer
$+500\,\text{J}$
Show the solution
Given
heat transferred to the air $Q = 700.0\,\text{J}$, the pressure stays $p = 125\,\text{kPa}$, increase in volume $\Delta V = 1.60\,\textsf{l}$
Find
the change in internal energy $\Delta U$ in joules with its sign
Principle
Energy is conserved, so the heat supplied to the air is divided into a change in internal energy and the work done by the air: the first law $\Delta U = Q - W_\text{g}$. The piston moves lightly, so the pressure stays constant and the work is $p\,\Delta V$.
Formula
$W_\text{g} = p\,\Delta V$, so
$\Delta U$$= Q - W_\text{g}$$= Q - p\,\Delta V$Conversions
$125\,\text{kPa} = 125\,000\,\text{Pa}$ and $1.60\,\textsf{l} = 0.00160\,\text{m}^3$, so $\text{Pa} \cdot \text{m}^3 = \text{J}$.
Substitution
$\Delta U$$= 700.0\,\text{J} - 125\,000\,\text{Pa} \cdot 0.00160\,\text{m}^3$$= 700.0\,\text{J} - 200\,\text{J}$$= 500\,\text{J}$Answer
$+500\,\text{J}$
That is, the internal energy grows, three significant figures. The limiting value is the work $W_\text{g} = 200\,\text{J}$, which has three figures and is known to the nearest joule, because the pressure $125\,\text{kPa}$ and the change in volume $1.60\,\textsf{l}$ are given with three figures. In the difference $700.0 - 200$ the less precise number decides the decimals, so the result is known to the nearest joule.
Reasonableness
Of the heat, $200\,\text{J}$ is used in pushing the piston and the remaining $500\,\text{J}$ raises the temperature of the air, so $\Delta U > 0$ and $\Delta U < Q$. If the work were added to the heat, the result would be $900\,\text{J}$, more than the heat that entered the air, although the air gives up energy as it pushes the piston. So the sign is correct.
- Exercise 13-12
The diesel engine of an archipelago ferry turns the propeller with a mechanical power of $480\,\text{kW}$. The efficiency of the engine is $40.0\,\%$.
How much heat power is transferred to the surroundings with the exhaust gases and the cooling water?
Answer
$720\,\text{kW}$
Show the solution
Given
mechanical power $P_\text{mech} = 480\,\text{kW}$, efficiency $\eta = 40.0\,\% = 0.400$
Find
the power of the waste heat $P_\text{waste}$ in kilowatts
Principle
The engine is a heat engine, and energy is conserved: the fuel power is divided into mechanical power and waste heat. Efficiency links the mechanical power to the fuel power, so the fuel power is calculated first.
Formula
$\eta$$= \dfrac{P_\text{mech}}{P_\text{in}}$so
$P_\text{in}$$= \dfrac{P_\text{mech}}{\eta}$$P_\text{waste} = P_\text{in} - P_\text{mech}$
Conversions
Efficiency $40.0\,\% = \tfrac{40.0}{100} = 0.400$. The powers are already in kilowatts, so no other conversion is needed.
Substitution
$P_\text{in}$$= \dfrac{480\,\text{kW}}{0.400}$$= 1200\,\text{kW}$so
$P_\text{waste}$$= 1200\,\text{kW} - 480\,\text{kW}$$= 720\,\text{kW}$Answer
$720\,\text{kW}$
Two significant figures. The limiting value is the mechanical power $480\,\text{kW}$, whose first two figures are certain; the efficiency $40.0\,\%$ has three figures and does not limit the precision.
Reasonableness
$60\,\%$ is wasted and $40\,\%$ is put to use, so there is $\tfrac{60}{40} = 1.5$ times as much waste heat as mechanical power: $1.5 \cdot 480\,\text{kW} = 720\,\text{kW}$. The fuel power is greater than the mechanical power, as it should be. Part of the waste heat can be used to heat the passenger cabin.
- Exercise 13-13
Example data: in a test run of a diesel generator, the mechanical power $P$ of the engine and the fuel consumption $q_m$ in kilograms per hour were measured at four different loads. The heat of combustion of diesel fuel is $43\,\text{MJ/kg}$.
At which load is the efficiency of the engine greatest? Give this greatest efficiency as a percentage as the answer.
Power and fuel consumption of a diesel engine (example data) $P\,(\text{kW})$ $q_m\,(\text{kg/h})$ 20.0 7.9 40.0 11.6 60.0 16.0 80.0 22.5 Answer
$31.4\,\%$
Show the solution
Given
heat of combustion of the fuel $H = 43\,\text{MJ/kg}$. Power and consumption: $20.0\,\text{kW}$ and $7.9\,\text{kg/h}$; $40.0\,\text{kW}$ and $11.6\,\text{kg/h}$; $60.0\,\text{kW}$ and $16.0\,\text{kg/h}$; $80.0\,\text{kW}$ and $22.5\,\text{kg/h}$
Find
the greatest efficiency $\eta$ of the engine as a percentage
Principle
The engine is a heat engine, and its efficiency is the share of the power released in the burning of the fuel that becomes mechanical power. The fuel power is the consumption times the heat of combustion. The efficiency is calculated separately for each load, and the greatest is chosen.
Formula
$P_1$$= q_m H$$\eta$$= \dfrac{P}{P_1}$$= \dfrac{P}{q_m H}$Conversions
$43\,\text{MJ/kg} = 43 \cdot 10^6\,\text{J/kg}$ and $1\,\text{h} = 3600\,\text{s}$, so for example
$16.0\,\text{kg/h}$$= \tfrac{16.0}{3600}\,\text{kg/s}$$= 4.444 \cdot 10^{-3}\,\text{kg/s}$The power $60.0\,\text{kW} = 60\,000\,\text{W}$.
Substitution
At the load $60.0\,\text{kW}$:
$$\begin{aligned} P_1 &= 4.444 \cdot 10^{-3}\,\tfrac{\text{kg}}{\text{s}} \cdot 43 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}} \\ &= 191\,100\,\text{W} \\ \eta &= \frac{60\,000\,\text{W}}{191\,100\,\text{W}} \\ &= 0.314 \end{aligned}$$
For the other loads in the same way: $20.0\,\text{kW}$ gives $21.2\,\%$, $40.0\,\text{kW}$ gives $28.9\,\%$ and $80.0\,\text{kW}$ gives $29.8\,\%$.
Answer
The greatest efficiency is $31.4\,\%$ at the load $60.0\,\text{kW}$, three significant figures. The heat of combustion $43\,\text{MJ/kg}$ is given with two figures, so the result is really known to two figures ($31\,\%$); the third figure is given so that the loads can be compared with each other and the rounding does not take the answer outside the accepted interval.
Reasonableness
The efficiency is below $100\,\%$ and in the range of $20$–$40\,\%$ typical of diesel engines. It grows with the load and decreases again at the greatest load, so the maximum is in the middle. Estimate:
$\tfrac{60}{16 \cdot 43 / 3.6}$$= \tfrac{60}{191}$$\approx 0.31$ - Exercise 13-14
A student claims: “When the friction and other losses of a heat engine are removed, the engine can turn all the heat it receives into work, because energy is conserved.” What is wrong in the claim?
Answer
The engine repeats its work only by cooling the gas, so part of the heat must always be given out to the colder surroundings.
Show the solution
Find
what is wrong in the student's claim about the efficiency of a heat engine.
Principle
A heat engine takes heat $Q_1$ from a hot source, does work $W$ and gives out heat $Q_2$ to a colder environment: $W = Q_1 - Q_2$. A piston engine can repeat its work only if the gas is cooled before the piston is returned, and in the cooling heat is transferred to the colder surroundings. So $Q_2 > 0$ and $\eta = W/Q_1 < 100\,\%$.
Model answer
The claim is wrong, because the engine can repeat its work only by cooling the gas, and therefore part of the heat must always be given out to the colder surroundings. Energy is indeed conserved, but that does not mean that all the heat could become work: part of the heat is transferred to the surroundings in the cooling. Removing friction improves the efficiency, but the limit does not go away. The size of the engine or the heat of combustion of the fuel does not change this limit.
- Exercise 13-15
Example data: the operation of the air-source heat pump of a school was followed for four days. For each day the electrical energy $E_\text{el}$ used by the pump and the heat given out indoors $Q_\text{indoors}$ were measured.
How much heat did the pump take from the outdoor air in total during these four days? Give the answer in kilowatt-hours.
Measurements of the heat pump (example data) Day $T_\text{outdoor}\,({}^\circ\text{C})$ $E_\text{el}\,(\text{kWh})$ $Q_\text{indoors}\,(\text{kWh})$ 1 2.0 2.1 6.4 2 −3.0 3.0 8.1 3 −9.0 4.2 10.0 4 −5.0 3.6 9.3 Answer
$20.9\,\text{kWh}$
Show the solution
Given
electrical energies of the four days in the table $E_\text{el}$: $2.1$, $3.0$, $4.2$ and $3.6\,\text{kWh}$, and the heats given out indoors $Q_\text{indoors}$: $6.4$, $8.1$, $10.0$ and $9.3\,\text{kWh}$
Find
the heat $Q_\text{outdoors}$ taken from the outdoor air in total in the four days
Principle
A heat pump does not produce energy. The heat given out indoors is the sum of the heat taken from the outdoor air and the electrical energy used by the pump, and this holds also for sums over several days.
Formula
$Q_\text{outdoors} = Q_\text{indoors} - E_\text{el}$
Conversions
All the energies are given in kilowatt-hours, so no conversions are needed.
Substitution
The sums are
$E_\text{el}$$= 2.1 + 3.0 + 4.2 + 3.6$$= 12.9\,\text{kWh}$$Q_\text{indoors}$$= 6.4 + 8.1 + 10.0 + 9.3$$= 33.8\,\text{kWh}$So
$$\begin{aligned} Q_\text{outdoors} &= 33.8\,\text{kWh} - 12.9\,\text{kWh} \\ &= 20.9\,\text{kWh} \end{aligned}$$
Answer
$20.9\,\text{kWh}$
Three significant figures, because the energies are given to a tenth and the difference is known to that precision.
Reasonableness
The heat taken from outdoors is less than the heat given out indoors, as it should be. Energy balance: $20.9\,\text{kWh} + 12.9\,\text{kWh} = 33.8\,\text{kWh}$. The outdoor temperatures $T_\text{outdoor}$ are not part of the calculation, because the energy balance does not need them. The coefficient of performance $Q_\text{indoors}/E_\text{el}$ of each day is over 1 ($3.05$; $2.70$; $2.38$; $2.58$) and decreases as the frost gets harder, which fits the fact that it is harder to transfer heat from cold air.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 13-16
A city combined heat and power plant burns wood chips at a power of $320\,\text{MW}$ and at the same time produces $96\,\text{MW}$ of electricity and $180\,\text{MW}$ of district heating.
If the plant did not exist, the same electric power would be produced in a condensing power plant with an efficiency of $40.0\,\%$, and the same district heating in a heating plant whose boiler heats the water of the district heating network with an efficiency of $90.0\,\%$.
How much more fuel power would such separate production need? Justify why the heating plant reaches a much higher efficiency than the condensing power plant.
Answer
$120\,\text{MW}$
Show the solution
Given
in combined production fuel power $P_\text{CHP} = 320\,\text{MW}$, electric power $P_\text{e} = 96\,\text{MW}$, district heating power $P_\text{dh} = 180\,\text{MW}$; in separate production the efficiency of the condensing plant $\eta_\text{cond} = 0.400$ and of the heating plant $\eta_\text{h} = 0.900$
Find
the additional fuel power $\Delta P$ needed by separate production in megawatts, and a justification for the difference in efficiencies
Principle
The same electric and heating powers are produced in two separate plants, and the fuel power of each is obtained from the efficiency: $P_\text{in} = P_\text{useful}/\eta$. The comparison is fair because the useful powers are the same in both cases.
Formula
$P_\text{cond}$$= \dfrac{P_\text{e}}{\eta_\text{cond}}$$P_\text{hp}$$= \dfrac{P_\text{dh}}{\eta_\text{h}}$so
$\Delta P$$= \dfrac{P_\text{e}}{\eta_\text{cond}} + \dfrac{P_\text{dh}}{\eta_\text{h}} - P_\text{CHP}$Conversions
The efficiencies are $40.0\,\% = 0.400$ and $90.0\,\% = 0.900$. All powers are in megawatts, so no other conversion is needed.
Substitution
$\Delta P$$= \dfrac{96\,\text{MW}}{0.400} + \dfrac{180\,\text{MW}}{0.900} - 320\,\text{MW}$$= 240\,\text{MW} + 200\,\text{MW} - 320\,\text{MW}$$= 120\,\text{MW}$Answer
$120\,\text{MW}$
Two significant figures. The limiting value is the electric power $96\,\text{MW}$ (two figures): it determines the fuel power of the condensing plant, $240\,\text{MW}$, to the nearest ten megawatts, and the difference $440\,\text{MW} - 320\,\text{MW}$ keeps this precision. The efficiencies $40.0\,\%$ and $90.0\,\%$ have three figures and do not limit the precision.
Reasonableness
The overall efficiency is $\dfrac{96 + 180}{320} \approx 86\,\%$ in combined production and $\dfrac{276}{440} \approx 63\,\%$ in separate production, so separate production needs more fuel. A heating plant does not turn heat into work, so its losses are only in the flue gases. A condensing power plant turns heat into work, and then part of the heat must always be given up to a colder environment, usually the sea.
- Exercise 13-17
Heating a detached house takes $16.0\,\text{MWh}$ of heat a year. The house now has direct electric heating, in which electric resistors turn practically all the electrical energy into internal energy, which is transferred as heat to the room.
The owner is considering a heat pump that would take heat from the outdoor air and take care of all the heating. The coefficient of performance of the pump is 2.50 as a yearly average.
How many megawatt-hours of electricity would the pump save in a year? How much heat would it take from the outdoor air during the year? Use conservation of energy to justify why the pump is not a perpetual motion machine even though it gives the house more heat than the electricity it uses.
Answer
$9.60\,\text{MWh}$
Show the solution
Given
yearly heat demand $Q = 16.0\,\text{MWh}$, coefficient of performance of the heat pump $\varepsilon = 2.50$, current heating is direct electric heating
Find
the electrical energy saved $\Delta E$ and the heat taken from the outdoor air $Q_\text{outdoors}$ in megawatt-hours, and a justification for why the pump is not a perpetual motion machine
Principle
The coefficient of performance tells how many times as much heat the pump gives out as the electricity it uses. In direct electric heating the resistors turn practically all the electrical energy into internal energy, which is transferred as heat to the room, so it needs as much electricity as heat. Energy is conserved: the heat entering the house is the pump's electricity and the heat taken from the outdoor air together.
Formula
$\varepsilon = \dfrac{Q}{E_\text{el}}$, so $E_\text{el} = \dfrac{Q}{\varepsilon}$. The saving is $\Delta E = Q - E_\text{el}$ and from conservation of energy $Q_\text{outdoors} = Q - E_\text{el}$, so
$\Delta E$$= Q_\text{outdoors}$$= Q - \dfrac{Q}{\varepsilon}$Conversions
All energies are in megawatt-hours and the coefficient of performance has no unit, so no conversions are needed.
Substitution
$E_\text{el}$$= \dfrac{16.0\,\text{MWh}}{2.50}$$= 6.40\,\text{MWh}$so
$\Delta E$$= Q_\text{outdoors}$$= 16.0\,\text{MWh} - 6.40\,\text{MWh}$$= 9.60\,\text{MWh}$or directly
$\Delta E$$= Q\left(1 - \dfrac{1}{\varepsilon}\right)$$= 16.0\,\text{MWh} \cdot 0.600$$= 9.60\,\text{MWh}$Answer
$9.60\,\text{MWh}$
Both as electricity saved and as heat taken from the outdoor air, three significant figures. The electricity of the pump, $6.40\,\text{MWh}$, is calculated from the heat demand $16.0\,\text{MWh}$ and is not a separate measurement, so the rule for differences is not used. The result is the product $Q\left(1 - \tfrac{1}{\varepsilon}\right)$, and its figures are limited by the heat demand $16.0\,\text{MWh}$ and the coefficient of performance $2.50$ (both three figures).
Reasonableness
The electricity saved is exactly the heat brought in from the outdoor air, and the pump uses only $\tfrac{1}{2.50} = 40\,\%$ of the electricity. Checking the balance: the heat entering the house is $9.60\,\text{MWh} + 6.40\,\text{MWh} = 16.0\,\text{MWh}$. The pump is not a perpetual motion machine, because it does not create energy but uses electricity to transfer heat from cold to warm.
- Exercise 13-18
A petrol car uses $6.00\,\textsf{l}$ of petrol over $100\,\text{km}$ when it drives on a main road at a steady speed of $80.0\,\text{km/h}$. The density of petrol is $0.750\,\text{kg/l}$ and its heat of combustion is $43.0\,\text{MJ/kg}$. The efficiency of the engine is $28.0\,\%$.
What is the average mechanical power of the engine? Give the answer in kilowatts. Where does the rest of the energy of the fuel go, and why does an electric car, whose motor has an efficiency of about $90\,\%$, need much less energy for the same distance?
Answer
$12.0\,\text{kW}$
Show the solution
Given
volume of fuel $V = 6.00\,\textsf{l}$ over a distance $s = 100\,\text{km}$ (the reference distance for consumption), speed $v = 80.0\,\tfrac{\text{km}}{\text{h}}$; density of petrol $\rho = 0.750\,\tfrac{\text{kg}}{\textsf{l}}$ and heat of combustion $H = 43.0\,\tfrac{\text{MJ}}{\text{kg}}$; efficiency $\eta = 28.0\,\% = 0.280$
Find
the average mechanical power $P$ of the engine in kilowatts, and justifications for where the energy goes and why an electric car needs less
Principle
The engine is a heat engine: the energy $Q_1$ released when petrol burns is divided into work $W = \eta Q_1$ and waste heat. Power is work divided by the time taken for the distance, and the speed is steady, so the power is constant.
Formula
$Q_1 = \rho V H$ and $t = \dfrac{s}{v}$, so
$P$$= \dfrac{W}{t}$$= \dfrac{\eta \rho V H}{t}$Conversions
$t$$= \dfrac{100\,\text{km}}{80.0\,\text{km/h}}$$= 1.25\,\text{h}$$= 4500\,\text{s}$$43.0\,\tfrac{\text{MJ}}{\text{kg}}$$= 43.0 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}$Finally $1\,\text{kW} = 1000\,\text{W}$.
Substitution
The work obtained from the energy of the fuel is
$W$$= 0.280 \cdot 0.750\,\tfrac{\text{kg}}{\textsf{l}} \cdot 6.00\,\textsf{l} \cdot 43.0 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}$$= 5.418 \cdot 10^7\,\text{J}$so
$$\begin{aligned} P &= \dfrac{5.418 \cdot 10^7\,\text{J}}{4500\,\text{s}} \\ &= 12\,040\,\text{W} \\ &= 12.04\,\text{kW} \end{aligned}$$
Answer
$12.0\,\text{kW}$
Three significant figures. All the values ($6.00\,\textsf{l}$, $0.750\,\text{kg/l}$, $43.0\,\text{MJ/kg}$, $28.0\,\%$ and $80.0\,\text{km/h}$) are given with three figures; the distance $100\,\text{km}$ is the reference distance for consumption and does not limit the precision.
Reasonableness
The fuel power is $\tfrac{1.935 \cdot 10^8\,\text{J}}{4500\,\text{s}} = 43\,\text{kW}$, so $31\,\text{kW}$ leaves as waste heat through the exhaust gases and the radiator. The petrol used is $54\,\text{kWh}$ of energy, and the work is $28\,\%$ of that, $15\,\text{kWh}$. An electric car needs about $\tfrac{15\,\text{kWh}}{0.90} \approx 17\,\text{kWh}$ from the battery to do the same work, because its motor turns electricity into work with an efficiency of $90\,\%$ and no heat has to be turned into work.
- Exercise 13-19
Example data: the heating of a detached house is being planned with a heat pump. The table shows the monthly heat demand $Q$ of the house in the winter season and the coefficient of performance $\varepsilon$ of the pump at the average outdoor temperature of each month. The heat demand is covered entirely by the pump.
a) Calculate the electrical energy $E_\text{el}$ that the pump uses in each month, and the total electrical energy of all the months.
b) Calculate the coefficient of performance of the season $\varepsilon_\text{season}$, which is the ratio of the heat given out during the whole season to the electrical energy of the whole season.
c) Explain why the coefficient of performance of the season must not be calculated as the ordinary average of the monthly coefficients of performance, and calculate how much electricity the pump saves compared with direct electric heating, where all the heat comes from electricity.
Write the result of part b) in the answer box.
Heat demand of the house and coefficient of performance of the heat pump (example data) Month $T_\text{outdoor}\,({}^\circ\text{C})$ $Q\,(\text{MWh})$ $\varepsilon$ November 1.0 1.8 3.4 December −4.0 2.6 2.9 January −12.0 4.0 2.4 February −7.0 3.1 2.6 March −1.0 2.2 3.1 Answer
$\varepsilon_\text{season} = 2.74$
Show the solution
Given
monthly heat demands $Q$ and coefficients of performance $\varepsilon$ from the table: November $1.8\,\text{MWh}$ and $3.4$; December $2.6\,\text{MWh}$ and $2.9$; January $4.0\,\text{MWh}$ and $2.4$; February $3.1\,\text{MWh}$ and $2.6$; March $2.2\,\text{MWh}$ and $3.1$
Find
a) the monthly electrical energies and their sum, b) the coefficient of performance of the season $\varepsilon_\text{season}$ (for the answer box), c) why the average is not used, and the saving of electricity compared with direct electric heating
Principle
The coefficient of performance is $\varepsilon = Q_\text{indoors}/E_\text{el}$, so the electrical energy of each month is $E_\text{el} = Q/\varepsilon$. The coefficient of performance of the whole season is defined in the same way from the energies of the whole season: the heat given out during the season divided by the electrical energy of the season. The electricity use of direct electric heating is equal to the heat demand.
Formula
$E_\text{el}$$= \dfrac{Q}{\varepsilon}$$\varepsilon_\text{season}$$= \dfrac{\sum Q}{\sum E_\text{el}}$Conversions
All the energies are in megawatt-hours, so no conversions are needed. The coefficient of performance has no unit.
Substitution
a) Month by month:
$$\begin{aligned} E_\text{el,Nov} &= \frac{1.8\,\text{MWh}}{3.4} = 0.529\,\text{MWh} \\ E_\text{el,Dec} &= \frac{2.6\,\text{MWh}}{2.9} = 0.897\,\text{MWh} \\ E_\text{el,Jan} &= \frac{4.0\,\text{MWh}}{2.4} = 1.667\,\text{MWh} \\ E_\text{el,Feb} &= \frac{3.1\,\text{MWh}}{2.6} = 1.192\,\text{MWh} \\ E_\text{el,Mar} &= \frac{2.2\,\text{MWh}}{3.1} = 0.710\,\text{MWh} \end{aligned}$$
The sum of the electrical energies is $4.995\,\text{MWh}$ and the sum of the heat demands $1.8 + 2.6 + 4.0 + 3.1 + 2.2 = 13.7\,\text{MWh}$.
b) The coefficient of performance of the season:
$$\begin{aligned} \varepsilon_\text{season} &= \frac{13.7\,\text{MWh}}{4.995\,\text{MWh}} \\ &= 2.74 \end{aligned}$$
c) Direct electric heating uses $13.7\,\text{MWh}$ of electricity, so the saving is $13.7\,\text{MWh} - 4.995\,\text{MWh} = 8.7\,\text{MWh}$.
Answer
b) $\varepsilon_\text{season} = 2.74$, three significant figures. The initial values are given with two figures, so the result is really known to two figures ($2.7$), and the third figure is given so that the rounding does not move the answer outside the accepted interval. This number goes in the answer box.
Reasonableness
The coefficient of performance of the season, $2.74$, lies between the smallest ($2.4$) and the greatest ($3.4$) monthly coefficient of performance. The ordinary average of the monthly coefficients of performance would be: the sum of the coefficients is $14.4$, so the average is $\tfrac{14.4}{5} = 2.88$, about $5\,\%$ too high. Justification: the average would give every month the same weight, although in the coldest month the most heat is needed and the coefficient of performance is the smallest. Then most electricity is used at a small coefficient of performance, so the coefficient of performance of the whole season must be smaller than the average. Energy balance: $4.995\,\text{MWh} + 8.7\,\text{MWh} = 13.7\,\text{MWh}$, so the heat taken from the outdoor air covers about $64\,\%$ of the heat demand. The coefficient of performance is the smallest in the coldest month, because it is harder to transfer heat from the cold outdoor air to the warm inside when the temperature difference grows.
- Exercise 13-20
Example data: in a vertical cylinder there is air closed by a lightly moving piston. The cross-sectional area of the cylinder is $A = 200\,\text{cm}^2$, and according to the measurements the pressure of the air stays constant at $p = 125\,\text{kPa}$. The air is heated with an electric resistor whose power is $260\,\text{W}$. Assume that all the energy given out by the resistor is transferred to the air as heat. The table gives the height $h$ of the piston, measured from the bottom of the cylinder, during the heating.
a) Calculate the heat $Q$ transferred to the air in four seconds and the work $W_\text{g}$ done by the air, using the first and last values of the table.
b) Calculate the change in the internal energy of the air $\Delta U$.
c) Calculate what share of the heat turned into work, and justify why a single expansion like this is not yet a heat engine.
Write the result of part b) in kilojoules in the answer box.
Height of the piston during the heating (example data) $t\,(\text{s})$ $h\,(\text{cm})$ 0.0 10.0 1.0 13.1 2.0 16.0 3.0 19.2 4.0 22.0 Answer
$\Delta U = 0.74\,\text{kJ}$
Show the solution
Given
cross-sectional area $A = 200\,\text{cm}^2$, constant pressure $p = 125\,\text{kPa}$, power of the resistor $P = 260\,\text{W}$, time $t = 4.0\,\text{s}$, height of the piston $h_1 = 10.0\,\text{cm}$ at the start and $h_2 = 22.0\,\text{cm}$ at the end
Find
a) the heat $Q$ and the work $W_\text{g}$ done by the gas, b) the change in internal energy $\Delta U$ in kilojoules (for the answer box), c) the share that became work, and a justification of why a single expansion is not a heat engine
Principle
The energy of the resistor is transferred to the air as heat, $Q = Pt$. The pressure is constant, so the work done by the air is $W_\text{g} = p\,\Delta V$, and the change in volume is obtained from the displacement of the piston: $\Delta V = A\,\Delta h$. Energy is conserved, so the first law is $\Delta U = Q - W_\text{g}$.
Reading the graph
From the table the height of the piston grows from $h_1 = 10.0\,\text{cm}$ to $h_2 = 22.0\,\text{cm}$, so $\Delta h = 12.0\,\text{cm}$. The height grows by about the same amount every second ($3.0\,\text{cm/s}$). The pressure is given as constant, and the steady rise fits heating with a constant power.
Formula
$Q = Pt$, $\Delta V = A\,\Delta h$, $W_\text{g} = p\,\Delta V$ and $\Delta U = Q - W_\text{g}$
Conversions
$A$$= 200\,\text{cm}^2$$= 200 \cdot 10^{-4}\,\text{m}^2$$= 0.0200\,\text{m}^2$$\Delta h$$= 12.0\,\text{cm}$$= 0.120\,\text{m}$$p$$= 125\,\text{kPa}$$= 125\,000\,\text{Pa}$Substitution
a)
$$\begin{aligned} Q &= 260\,\text{W} \cdot 4.0\,\text{s} \\ &= 1040\,\text{J} \\ \Delta V &= 0.0200\,\text{m}^2 \cdot 0.120\,\text{m} \\ &= 2.40 \cdot 10^{-3}\,\text{m}^3 \\ W_\text{g} &= 125\,000\,\text{Pa} \cdot 2.40 \cdot 10^{-3}\,\text{m}^3 \\ &= 300\,\text{J} \end{aligned}$$
b)
$$\begin{aligned} \Delta U &= 1040\,\text{J} - 300\,\text{J} \\ &= 740\,\text{J} \\ &= 0.74\,\text{kJ} \end{aligned}$$
c) The share that became work is
$\tfrac{W_\text{g}}{Q}$$= \tfrac{300\,\text{J}}{1040\,\text{J}}$$= 0.288$$\approx 29\,\%$Answer
b) $\Delta U = 0.74\,\text{kJ}$, two significant figures, because the power $260\,\text{W}$ and the time $4.0\,\text{s}$ are given with two figures. This number goes in the answer box.
Justification c)
About $29\,\%$ of the heat became work, and the remaining about $71\,\%$ stayed as the internal energy of the air, so the air warmed up. This fits air: for a diatomic gas at constant pressure the share of the work is about $\tfrac{2}{7} \approx 29\,\%$ of the heat received. A single expansion is not a heat engine, because an engine repeats its work cycle after cycle. To bring the piston back to the start, the air has to be cooled and compressed, and in the cooling heat is transferred to the colder surroundings. The work share $29\,\%$ of one expansion is therefore not the efficiency of a cycle: that is $\eta = W/Q_1$, where the work is $W = Q_1 - Q_2$.
Reasonableness
The change in internal energy is positive and smaller than the heat, $0 < 740\,\text{J} < 1040\,\text{J}$, as it should be when a gas expands and receives heat. Energy balance: $740\,\text{J} + 300\,\text{J} = 1040\,\text{J}$. The speed of the piston, $3.0\,\text{cm/s}$, is slow, so the motion of a lightly moving piston is calm.
Photographs
- Close-up of red steam locomotive wheel. Masood Aslami. source, Pexels License