Energy production and the efficiency of power plants

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 14-1Core

   The turbine of a hydropower plant converts $92.0\,\%$ of the kinetic energy of the water into mechanical energy, and the generator converts $98.0\,\%$ of the energy of the turbine shaft into electricity.

   What is the efficiency of the whole plant? Give your answer as a percentage.

   %Check
   Answer

   $90.2\,\%$

   Show the solution

   Given

   efficiency of the turbine $\eta_1 = 92.0\,\% = 0.920$, efficiency of the generator $\eta_2 = 98.0\,\% = 0.980$

   Find

   the efficiency $\eta$ of the whole plant as a percentage

   Principle

   The plant is a chain in which the output power of the turbine is the input power of the generator. Each step lets through only part of the energy coming into it, so the efficiency of the chain is the product of the efficiencies of the steps.

   Formula

   $P_1 = \eta_1 P_0$ and $P_2 = \eta_2 P_1$, so

   $\eta$$= \dfrac{P_2}{P_0}$

   $= \eta_1 \eta_2$

   Conversions

   The efficiencies are fractions of a whole: $92.0\,\% = 0.920$ and $98.0\,\% = 0.980$. At the end the result is multiplied by a hundred to give a percentage.

   Substitution

   $\eta$$= 0.920 \cdot 0.980$

   $= 0.9016$

   $= 90.16\,\%$

   Answer

   $90.2\,\%$

   Three significant figures like the given efficiencies $92.0\,\%$ and $98.0\,\%$.

   Reasonableness

   The product is smaller than either efficiency, as it should be when the factors are less than 1. Estimate:

   $0.92 \cdot 0.98$$\approx 0.92 - 0.02 \cdot 0.92$

   $\approx 0.90$

   A hydropower plant is not a heat engine, so its efficiency can be over $90\,\%$.

2. Exercise 14-2Core

   The power of the kinetic energy of the air flow swept by the rotor of a wind turbine is $2.50\,\text{MW}$, and the turbine delivers a power of $1.05\,\text{MW}$ to the electricity grid.

   What is the efficiency of the turbine? Give your answer as a percentage.

   %Check
   Answer

   $42.0\,\%$

   Show the solution

   Given

   power of the kinetic energy of the air flow $P_\text{wind} = 2.50\,\text{MW}$ and electric power $P_\text{elec} = 1.05\,\text{MW}$

   Find

   the efficiency $\eta$ of the turbine as a percentage

   Principle

   Efficiency is the share of the useful power in the power coming into the plant. Here the input power is the power of the kinetic energy of the air flowing through the rotor and the useful power is the power going to the electricity grid; the rest of the energy stays in the motion of the air or turns into heat.

   Formula

   $\eta$$= \dfrac{P_\text{elec}}{P_\text{wind}}$

   Conversions

   Both powers are in megawatts, so the units cancel. The result is multiplied by a hundred to give a percentage.

   Substitution

   $\eta$$= \dfrac{1.05\,\text{MW}}{2.50\,\text{MW}}$

   $= 0.420$

   $= 42.0\,\%$

   Answer

   $42.0\,\%$

   Three significant figures like the powers $2.50\,\text{MW}$ and $1.05\,\text{MW}$.

   Reasonableness

   Estimate: $1 / 2.5 = 0.4$, which matches the result. The efficiency of a modern wind turbine is at best about 40–$45\,\%$. You cannot take all the energy from the wind, because the air has to keep moving after the rotor too. Part of the energy taken by the rotor turns into internal energy in friction and electrical losses, and this is transferred to the surroundings as heat.

3. Exercise 14-3

   The area of a solar panel installed on a roof is $20.0\,\text{m}^2$, and sunlight hits it with a power of $1.00\,\text{kW}$ for every square metre. The efficiency of the panel is 0.210.

   How much electric power does the panel produce? Give your answer in kilowatts.

   kWCheck
   Answer

   $4.20\,\text{kW}$

   Show the solution

   Given

   area $A = 20.0\,\text{m}^2$, radiation power per unit area $I = 1.00\,\tfrac{\text{kW}}{\text{m}^2}$, efficiency $\eta = 0.210$

   Find

   the electric power $P$ of the panel in kilowatts

   Principle

   The power of the radiation hitting the panel is the intensity times the area, and the efficiency tells what fraction of it turns into electricity. The rest turns into heat or is reflected, so the electric power is the $\eta$ fraction of the input power.

   Formula

   $P = \eta P_\text{in} = \eta I A$

   Conversions

   The units go along with the numbers: $\tfrac{\text{kW}}{\text{m}^2} \cdot \text{m}^2 = \text{kW}$, so the result comes directly in kilowatts and no conversions are needed.

   Substitution

   $P$$= 0.210 \cdot 1.00\,\tfrac{\text{kW}}{\text{m}^2} \cdot 20.0\,\text{m}^2$

   $= 0.210 \cdot 20.0\,\text{kW}$

   $= 4.20\,\text{kW}$

   Answer

   $4.20\,\text{kW}$

   Three significant figures, because the area $20.0\,\text{m}^2$, the radiation power $1.00\,\tfrac{\text{kW}}{\text{m}^2}$ and the efficiency $0.210$ all have three figures.

   Reasonableness

   Radiation of $20.0\,\text{kW}$ hits the panel, and about a fifth of it ($0.210 \approx 0.2$) is $\approx 4\,\text{kW}$. The rest, $20.0\,\text{kW} - 4.20\,\text{kW} = 15.8\,\text{kW}$, heats the panel or is reflected away. On the roof of a detached house $4\,\text{kW}$ is a usual peak power in clear weather.

4. Exercise 14-4

   The figure shows the power of a condensing power plant at different steps of the energy chain: $P_0$ is the fuel power, $P_1$ the power after the boiler, $P_2$ the power after the steam cycle and $P_3$ the electric power.

   In which step does the power decrease the most, and how large is the power transferred to the surroundings as heat in this step? Read the values from the figure. Give your answer in megawatts.

   MWCheck
   Answer

   $200\,\text{MW}$

   Show the solution

   Given

   read from the bars of the figure: $P_0 = 400\,\text{MW}$, $P_1 = 350\,\text{MW}$, $P_2 = 150\,\text{MW}$ and $P_3 = 140\,\text{MW}$

   Find

   the step in which the power decreases the most, and the power transferred to the surroundings as heat in that step

   Principle

   The power plant runs steadily and energy is conserved, so when the power decreases in a step, the difference is transferred to the surroundings as heat. The power coming out of a step is smaller than the power going in, and the difference is the waste heat produced in it.

   Reading the graph

   The tops of the bars $P_0$, $P_1$ and $P_2$ are on the gridlines, and they give the powers $400$, $350$ and $150\,\text{MW}$. The electric power $P_3$ is a little below the line $150\,\text{MW}$, about $140\,\text{MW}$; it is not needed for the answer. The decrease of power step by step is $P_0 - P_1 = 50\,\text{MW}$ in the boiler, $P_1 - P_2 = 200\,\text{MW}$ in the steam cycle and $P_2 - P_3 = 10\,\text{MW}$ in the generator and transformer. The biggest drop is between the bars $P_1$ and $P_2$, that is in the steam cycle.

   Formula

   $P_\text{h} = P_\text{in} - P_\text{out}$, here $P_\text{h} = P_1 - P_2$

   Conversions

   Not needed: all powers are in megawatts.

   Substitution

   $P_\text{h}$$= 350\,\text{MW} - 150\,\text{MW}$

   $= 200\,\text{MW}$

   Answer

   The power decreases the most in the steam cycle, where $200\,\text{MW}$ is transferred to the surroundings as heat. The values are read from the gridlines as whole megawatts, so the result to three significant figures is $200\,\text{MW}$.

   Reasonableness

   The losses of the steps add up to

   $50 + 200 + 10$$= 260\,\text{MW}$

   $= P_0 - P_3$

   $= 400\,\text{MW} - 140\,\text{MW}$

   so the balance works out. The efficiency of the steam cycle is $150 / 350 \approx 0.43$, which suits a steam cycle, and its loss is over $75\,\%$ of the waste heat of the whole plant.

5. Exercise 14-5

   The fuel power of a combined heat and power plant is $250\,\text{MW}$. The plant produces $75.0\,\text{MW}$ of electricity and $150\,\text{MW}$ of district heating.

   What is the overall efficiency of the plant? Give your answer as a percentage.

   %Check
   Answer

   $90.0\,\%$

   Show the solution

   Given

   fuel power $P_0 = 250\,\text{MW}$, electric power $P_\text{s} = 75.0\,\text{MW}$ and district heating power $P_\text{k} = 150\,\text{MW}$

   Find

   the overall efficiency $\eta_\text{tot}$ as a percentage

   Principle

   In a combined heat and power plant both electricity and district heating are useful, so both are counted in the useful power. Only the rest is wasted, and the overall efficiency is the share of the useful power in the fuel power.

   Formula

   $\eta_\text{tot}$$= \dfrac{P_\text{s} + P_\text{k}}{P_0}$

   Conversions

   All powers are in megawatts and the units cancel. The result is multiplied by a hundred to give a percentage.

   Substitution

   $\eta_\text{tot}$$= \dfrac{75.0\,\text{MW} + 150\,\text{MW}}{250\,\text{MW}}$

   $= \dfrac{225\,\text{MW}}{250\,\text{MW}}$

   $= 0.900$

   $= 90.0\,\%$

   Answer

   $90.0\,\%$

   Three significant figures, because all powers are given to the nearest megawatt and the sum $75.0 + 150 = 225\,\text{MW}$ keeps the same precision; so the quotient $225/250$ has three figures.

   Reasonableness

   The result is in the usual range of 85–$90\,\%$ for combined production. The power of waste heat is $250\,\text{MW} - 225\,\text{MW} = 25\,\text{MW}$, and $225 + 25 = 250$, so the balance works out. The share of electricity is only

   $\dfrac{75.0\,\text{MW}}{250\,\text{MW}}$$= 30\,\%$

   so the electrical efficiency is lower than in a condensing power plant.

6. Exercise 14-6

   A condensing power plant and a combined heat and power plant burn the same amount of the same fuel. The overall efficiency of the combined heat and power plant is still clearly higher.

   What explains this?

   Choose an option.

   • More energy is released when the fuel burns in combined production, so there is enough for both electricity and heat.
   • In combined production the turbine converts a larger share of the heat into electricity, so less waste heat is left.
   • The heat of the condenser is used as district heating, so a larger share of the energy is put to use.
   • The district heating network adds energy as the water circulates in the pipes, so more comes out than goes in.

   Answer
   I have done my solution.
   Answer

   The heat of the condenser is used as district heating, so a larger share of the energy is put to use.

   Show the solution

   Find

   why the overall efficiency of a combined heat and power plant is higher than that of a condensing power plant when the fuel is the same

   Principle

   Efficiency tells what share of the energy of the fuel is put to use. The energy released from the fuel is the same in both, so the difference comes from what is counted as useful: in a condensing power plant only electricity, in combined production electricity and district heating, $\eta_\text{tot} = \dfrac{P_\text{s} + P_\text{k}}{P_0}$.

   Model answer

   In combined production the heat of the condenser is led into the district heating network instead of the sea, so a larger share of the energy of the fuel is put to use. No more energy is released from the fuel, because the heat of combustion is the same. The turbine does not convert a larger share of the heat into electricity either: usually less electricity is obtained than in a condensing power plant. The district heating network does not produce extra energy, since the electricity and heat together do not exceed the fuel power. Waste heat is still produced, but less of it.

7. Exercise 14-7

   The thermal power of the reactor of a nuclear power plant is $1500\,\text{MW}$, and the plant delivers electricity to the grid at a power of $510\,\text{MW}$.

   What is the efficiency of the plant's electricity production? Give your answer as a percentage.

   %Check
   Answer

   $34.0\,\%$

   Show the solution

   Given

   thermal power of the reactor $P_0 = 1500\,\text{MW}$ and electric power $P_\text{s} = 510\,\text{MW}$

   Find

   the efficiency $\eta$ of the electricity production as a percentage

   Principle

   Efficiency is the share of the useful power in the power supplied to the plant. The useful power here is only electricity, and the rest of the heat of the reactor is transferred to the surroundings as waste heat.

   Formula

   $\eta = \dfrac{P_\text{s}}{P_0}$

   Conversions

   Both powers are in megawatts, so the units cancel. The result is multiplied by one hundred to get a percentage.

   Substitution

   $\eta$$= \dfrac{510\,\text{MW}}{1500\,\text{MW}}$

   $= 0.340$

   $= 34.0\,\%$

   Answer

   $34.0\,\%$

   Three significant figures like the electric power $510\,\text{MW}$.

   Reasonableness

   Estimate: $1500 : 3 = 500$, and $510\,\text{MW}$ is slightly more, so the efficiency is slightly over a third. The waste heat is $1500\,\text{MW} - 510\,\text{MW} = 990\,\text{MW}$, about two thirds. The electrical efficiency of a thermal power plant is typically 30–$40\,\%$, and the result is in that range.

8. Exercise 14-8

   The fuel power of a power plant is $500\,\text{MW}$, and the plant runs at a steady power for a day.

   How much fuel energy enters the plant in a day? Give your answer in gigawatt-hours.

   GWhCheck
   Answer

   $12.0\,\text{GWh}$

   Show the solution

   Given

   fuel power $P_0 = 500\,\text{MW}$ and running time $t = 1\,\text{day} = 24\,\text{h}$

   Find

   the energy of the fuel $E$ in gigawatt-hours

   Principle

   Power is energy per unit of time. When the power is constant, the energy is the power times the time.

   Formula

   $P = \dfrac{E}{t}$, so $E = P_0\,t$

   Conversions

   $1\,\text{day} = 24\,\text{h}$. The power is converted into gigawatts: $500\,\text{MW} = 0.500\,\text{GW}$. Then the result comes directly in gigawatt-hours.

   Substitution

   $E$$= 0.500\,\text{GW} \cdot 24\,\text{h}$

   $= 12.0\,\text{GWh}$

   Answer

   $12.0\,\text{GWh}$

   Three significant figures. The power $500\,\text{MW}$ is given with three figures, and a day is exactly $24\,\text{h}$.

   Reasonableness

   Estimate:

   $500\,\text{MW} \cdot 24\,\text{h}$$\approx 500 \cdot 24\,\text{MWh}$

   $= 12\,000\,\text{MWh}$

   which is $12\,\text{GWh}$. One hour would give $0.5\,\text{GWh}$, and a day has 24 hours, so the number is of the right order of magnitude. The same energy is $12.0 \cdot 3.6 \cdot 10^{12}\,\text{J} \approx 4.3 \cdot 10^{13}\,\text{J}$.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 14-9Core

   The fuel power of a peat-burning condensing power plant is $300\,\text{MW}$. The efficiency of the boiler is 0.880, of the steam cycle (turbine and condenser) 0.420 and of the generator 0.980.

   How large is the electric power of the plant? Give your answer in megawatts.

   MWCheck
   Answer

   $109\,\text{MW}$

   Show the solution

   Given

   fuel power $P_0 = 300\,\text{MW}$, efficiency of the boiler $\eta_1 = 0.880$, of the steam cycle $\eta_2 = 0.420$ and of the generator $\eta_3 = 0.980$

   Find

   the electric power $P_3$ of the plant in megawatts

   Principle

   The steps are in series, and the output power of each is the input power of the next. Each step keeps only the $\eta$ fraction of its power, so the electric power is the fuel power multiplied by all three efficiencies.

   Formula

   $P_1 = \eta_1 P_0$, $P_2 = \eta_2 P_1$ and $P_3 = \eta_3 P_2$, so $P_3 = \eta_1 \eta_2 \eta_3 P_0$

   Conversions

   The efficiencies are dimensionless and the power is already in megawatts, so no conversions are needed.

   Substitution

   $P_3$$= 0.880 \cdot 0.420 \cdot 0.980 \cdot 300\,\text{MW}$

   $= 0.362208 \cdot 300\,\text{MW}$

   $= 108.6624\,\text{MW}$

   Answer

   $109\,\text{MW}$

   Three significant figures, because the efficiencies $0.880$, $0.420$ and $0.980$ have three figures and the fuel power $300\,\text{MW}$ is given to the nearest megawatt.

   Reasonableness

   Estimate: $0.9 \cdot 0.4 \cdot 1.0 \cdot 300\,\text{MW} \approx 110\,\text{MW}$. The efficiency of the whole chain is $0.362$ or $36\,\%$, less than the weakest step ($0.420$) would give alone. The losses of the boiler and the steam cycle together take $1 - 0.880 \cdot 0.420 = 0.630$, that is almost two thirds (about $63\,\%$) of the energy of the fuel.

2. Exercise 14-10Core

   The efficiency of the whole energy chain of a condensing power plant is 0.380. The efficiency of the boiler is 0.900 and of the generator 0.980.

   What is the efficiency of the steam cycle, that is of the turbine and the condenser? Give your answer as a percentage.

   %Check
   Answer

   $43.1\,\%$

   Show the solution

   Given

   efficiency of the whole chain $\eta = 0.380$, of the boiler $\eta_1 = 0.900$ and of the generator $\eta_3 = 0.980$

   Find

   the efficiency $\eta_2$ of the steam cycle as a percentage

   Principle

   The chain is the product of three steps in series, so the efficiency of the whole chain is the product of the efficiencies of the steps. When two steps and the whole are known, the missing step is found by division.

   Formula

   $\eta = \eta_1 \eta_2 \eta_3$, so $\eta_2 = \dfrac{\eta}{\eta_1 \eta_3}$

   Conversions

   The efficiencies are dimensionless numbers, so no unit conversions are needed. At the end the result is multiplied by a hundred to give a percentage.

   Substitution

   $\eta_2$$= \dfrac{0.380}{0.900 \cdot 0.980}$

   $= \dfrac{0.380}{0.882}$

   $= 0.43084$

   $= 43.08\,\%$

   Answer

   $43.1\,\%$

   Three significant figures, because all three efficiencies, $0.380$, $0.900$ and $0.980$, have three figures.

   Reasonableness

   Check: $0.900 \cdot 0.4308 \cdot 0.980 = 0.380$. The efficiency of the steam cycle is greater than that of the whole chain, $0.380$, as it should be, because the other steps only reduce it. It is a little greater than the usual efficiency of a whole condensing power plant (about $40\,\%$, typically $38$–$42\,\%$), because the losses of the boiler and the generator are not included in it, and it is the weakest link of the chain.

3. Exercise 14-11

   A condensing power plant burning wood chips produces electricity at an average power of $150\,\text{MW}$, and the efficiency of the whole plant is 0.360. The heat of combustion of wood chips is $9.00\,\text{MJ/kg}$.

   How many tonnes of chips does the plant burn in an hour? Give your answer in tonnes.

   tCheck
   Answer

   $167\,\text{t}$

   Show the solution

   Given

   electric power $P_\text{s} = 150\,\text{MW}$, efficiency $\eta = 0.360$, heat of combustion $H = 9.00\,\tfrac{\text{MJ}}{\text{kg}}$, time $t = 1\,\text{h}$

   Find

   the mass $m$ of wood chips burned in an hour, in tonnes

   Principle

   Energy is conserved: the chemical energy of the fuel is either electricity or waste heat. The efficiency connects the electric power to the fuel power, and the fuel power is the heat of combustion times the mass flow, because the heat of combustion tells the energy per unit mass.

   Formula

   $P_0 = \dfrac{P_\text{s}}{\eta}$ and $P_0 = \dfrac{H m}{t}$, so $m = \dfrac{P_\text{s}\, t}{\eta H}$

   Conversions

   $150\,\text{MW}$$= 150 \cdot 10^6\,\text{W}$

   $9.00\,\tfrac{\text{MJ}}{\text{kg}}$$= 9.00 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}$

   $1\,\text{h} = 3600\,\text{s}$. At the end $1\,\text{t} = 1000\,\text{kg}$.

   Substitution

   $m$$= \dfrac{150 \cdot 10^6\,\text{W} \cdot 3600\,\text{s}}{0.360 \cdot 9.00 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}}$

   $= 1.66667 \cdot 10^5\,\text{kg}$

   $= 166.67\,\text{t}$

   Answer

   $167\,\text{t}$

   Three significant figures, because the efficiency $0.360$ and the heat of combustion $9.00\,\tfrac{\text{MJ}}{\text{kg}}$ have three figures and the electric power $150\,\text{MW}$ is given to the nearest megawatt; the hour is exact ($3600\,\text{s}$) and does not limit the precision.

   Reasonableness

   The fuel power is

   $\dfrac{150\,\text{MW}}{0.360}$$= 417\,\text{MW}$

   and the mass flow

   $\dfrac{417 \cdot 10^6\,\text{W}}{9.00 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}}$

   $= 46\,\tfrac{\text{kg}}{\text{s}}$

   In a day that is $46 \cdot 86\,400\,\text{kg} \approx 4000\,\text{t}$, which requires continuous lorry or train transport and suits a large bio power plant.

4. Exercise 14-12

   Heat is transferred to the condenser of a coastal condensing power plant at a power of $270\,\text{MW}$. The sea water cooling the condenser warms up by $9.00\,\text{K}$. The specific heat capacity of water is $4190\,\text{J/(kg}\cdot\text{K)}$.

   How many tonnes of water flow through the condenser per second? Give your answer in tonnes per second.

   t/sCheck
   Answer

   $7.16\,\tfrac{\text{t}}{\text{s}}$

   Show the solution

   Given

   heat power transferred to the condenser $P_\text{L} = 270\,\text{MW}$, rise in the temperature of the water $\Delta T = 9.00\,\text{K}$ and the given specific heat capacity $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$

   Find

   the water flow $q_m = m/t$ through the condenser in tonnes per second

   Principle

   The heat released by the condenser is transferred entirely to the cooling water when there are no other losses. The water only warms up and does not vaporise, so $Q = c m \Delta T$ applies to the quantity of heat; dividing by time gives the power.

   Formula

   $P_\text{L}$$= \dfrac{Q}{t}$

   $= c\, q_m \Delta T$

   so

   $q_m$$= \dfrac{P_\text{L}}{c\, \Delta T}$

   Conversions

   $270\,\text{MW} = 270 \cdot 10^6\,\text{W}$ and $1\,\text{W} = 1\,\tfrac{\text{J}}{\text{s}}$. The change in temperature $9.00\,\text{K}$ is the same in kelvins and in degrees Celsius. At the end $1\,\text{t} = 1000\,\text{kg}$.

   Substitution

   $q_m$$= \dfrac{270 \cdot 10^6\,\text{W}}{4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 9.00\,\text{K}}$

   $= 7159.9\,\tfrac{\text{kg}}{\text{s}}$

   $= 7.1599\,\tfrac{\text{t}}{\text{s}}$

   Answer

   $7.16\,\tfrac{\text{t}}{\text{s}}$

   Three significant figures, because the change in temperature $9.00\,\text{K}$ and the specific heat capacity $4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ have three figures and the power $270\,\text{MW}$ is given to the nearest megawatt.

   Reasonableness

   Check: $4190 \cdot 7160 \cdot 9.00\,\text{W} = 2.70 \cdot 10^8\,\text{W}$. The water flow corresponds to about $7\,\tfrac{\text{m}^3}{\text{s}}$, which is the discharge of a small river. A condensing power plant needs a large amount of cooling water on the shore, and that is why they are built on the coast or by a river.

5. Exercise 14-13

   A combined heat and power plant producing district heating and electricity records every year the energy bound in the fuel and the electricity and district heating it produces (table, example data). The year 2023 was cold.

   a) Calculate the electrical efficiency $\eta_\text{s}$ for each year.

   b) Calculate the overall efficiency $\eta_\text{tot}$ for each year.

   c) Explain why the electrical efficiency decreased but the overall efficiency increased.

   In part b), enter the overall efficiency of 2023 in the answer box as a percentage.

   Annual energies of a combined heat and power plant (example data)

   Year	$E_\text{fuel}\,(\text{GWh})$	$E_\text{electricity}\,(\text{GWh})$	$E_\text{district heating}\,(\text{GWh})$
   2021	2400	840	1190
   2022	2500	800	1410
   2023	2450	735	1470

   %Check
   Answer

   $90.0\,\%$

   Show the solution

   Given

   from the table, in 2023 the energy of the fuel $E_0 = 2450\,\text{GWh}$, electricity $E_\text{s} = 735\,\text{GWh}$ and district heating $E_\text{k} = 1470\,\text{GWh}$; the values for 2021 and 2022 are in the table

   Find

   the overall efficiency $\eta_\text{tot}$ of 2023 as a percentage

   Principle

   In a combined heat and power plant both electricity and district heating are put to use, so both are counted in the overall efficiency. The ratio is the same for energies as for powers, because the plant runs for the same time.

   Formula

   $\eta_\text{s}$$= \dfrac{E_\text{s}}{E_0}$

   $\eta_\text{tot}$$= \dfrac{E_\text{s} + E_\text{k}}{E_0}$

   Conversions

   All energies are in gigawatt-hours, so the units cancel and no conversions are needed. The results are multiplied by one hundred to get percentages.

   Substitution

   In 2023 the sum of electricity and district heating is $735\,\text{GWh} + 1470\,\text{GWh} = 2205\,\text{GWh}$, so

   $\eta_\text{tot}$$= \dfrac{2205\,\text{GWh}}{2450\,\text{GWh}}$

   $= 0.900$

   $= 90.0\,\%$

   The electrical efficiencies are

   $\eta_\text{s}$$= \tfrac{840}{2400}$

   $= 35.0\,\%$

   (2021), $\tfrac{800}{2500} = 32.0\,\%$ (2022) and $\tfrac{735}{2450} = 30.0\,\%$ (2023). The overall efficiencies are $\tfrac{2030}{2400} = 84.6\,\%$ (2021) and $\tfrac{2210}{2500} = 88.4\,\%$ (2022).

   Answer

   $90.0\,\%$

   Three significant figures like the energies in the table.

   Reasonableness

   The overall efficiency is at most $100\,\%$ and greater than the electrical efficiency, as it should be. In 2023 the waste heat is $2450\,\text{GWh} - 2205\,\text{GWh} = 245\,\text{GWh}$, about a tenth. In a cold year more district heating is needed, so a larger share of the heat of the condenser is put to use and less electricity is obtained: the electrical efficiency decreased from $35.0\,\%$ to $30.0\,\%$, but the overall efficiency increased from $84.6\,\%$ to $90.0\,\%$. Electricity and heat are still not equally valuable, even though they are added together.

6. Exercise 14-14

   The table gives the fuel or reactor power $P_0$ and the electric power $P_\text{s}$ of three power plants (example data). All the plants are condensing power plants, and the rest of the energy is transferred to the surroundings as waste heat.

   a) Calculate the efficiency of each plant.

   b) Calculate the waste heat power $P_\text{h}$ of each plant.

   c) Does the plant with the lowest efficiency also have the largest waste heat power? Justify your answer.

   In part b), enter the largest waste heat power in the answer box in megawatts.

   Powers of three condensing power plants (example data)

   Plant	$P_0\,(\text{MW})$	$P_\text{s}\,(\text{MW})$
   Nuclear power plant	2400	840
   Coal power plant	1200	432
   Bio power plant	350	112

   MWCheck
   Answer

   $1560\,\text{MW}$

   Show the solution

   Given

   from the table, the nuclear power plant $P_0 = 2400\,\text{MW}$ and $P_\text{s} = 840\,\text{MW}$, the coal power plant $1200\,\text{MW}$ and $432\,\text{MW}$, the bio power plant $350\,\text{MW}$ and $112\,\text{MW}$

   Find

   the largest waste heat power $P_\text{h}$ in megawatts

   Principle

   In a condensing power plant only electricity is put to use, and energy is conserved, so the fuel or reactor power is either electricity or waste heat: $P_0 = P_\text{s} + P_\text{h}$. Efficiency is a ratio, but the waste heat power also depends on the size of the plant.

   Formula

   $\eta = \dfrac{P_\text{s}}{P_0}$ and $P_\text{h} = P_0 - P_\text{s}$

   Conversions

   All powers are in megawatts, so no conversions are needed.

   Substitution

   Nuclear power plant:

   $\eta$$= \tfrac{840\,\text{MW}}{2400\,\text{MW}}$

   $= 0.350$

   $P_\text{h}$$= 2400\,\text{MW} - 840\,\text{MW}$

   $= 1560\,\text{MW}$

   Coal power plant:

   $\eta$$= \tfrac{432}{1200}$

   $= 0.360$

   $P_\text{h}$$= 1200\,\text{MW} - 432\,\text{MW}$

   $= 768\,\text{MW}$

   Bio power plant:

   $\eta$$= \tfrac{112}{350}$

   $= 0.320$

   $P_\text{h}$$= 350\,\text{MW} - 112\,\text{MW}$

   $= 238\,\text{MW}$

   Answer

   $1560\,\text{MW}$

   (nuclear power plant), three significant figures like the electric powers.

   Reasonableness

   For each plant $P_\text{s} + P_\text{h}$ equals $P_0$: for example $840 + 1560 = 2400$. The lowest efficiency belongs to the bio power plant ($32.0\,\%$), but it has the smallest waste heat power ($238\,\text{MW}$), because it is small. The largest waste heat power belongs to the nuclear power plant, even though its efficiency ($35.0\,\%$) is almost as high as that of the coal power plant ($36.0\,\%$). So efficiency and waste heat power do not come in the same order.

7. Exercise 14-15

   The efficiency of the boiler of a condensing power plant is 0.920, of the steam cycle 0.410 and of the generator 0.970.

   Student's solution: “The efficiencies are added together, because all the steps affect the result: $0.920 + 0.410 + 0.970 = 2.30$. So the efficiency of the plant is $230\,\%$.”

   a) Explain why the student's solution is wrong.

   b) Calculate the correct efficiency of the plant.

   In part b), enter the result in the answer box as a percentage.

   %Check
   Answer

   $36.6\,\%$

   Show the solution

   Given

   $\eta_1 = 0.920$ (boiler), $\eta_2 = 0.410$ (steam cycle) and $\eta_3 = 0.970$ (generator); the student's result $230\,\%$

   Find

   the correct efficiency $\eta$ of the plant as a percentage and an explanation of the error

   Principle

   The steps are in a row, and each step lets through only its efficiency's share of the power coming into it. When $P_1 = \eta_1 P_0$, $P_2 = \eta_2 P_1$ and $P_3 = \eta_3 P_2$, the efficiency of the chain is $\eta = \dfrac{P_3}{P_0} = \eta_1\,\eta_2\,\eta_3$, that is the efficiencies are multiplied.

   Formula

   $\eta = \eta_1\,\eta_2\,\eta_3$

   Conversions

   The efficiencies are dimensionless numbers, so no conversions are needed. At the end the result is multiplied by one hundred to get a percentage.

   Substitution

   $\eta$$= 0.920 \cdot 0.410 \cdot 0.970$

   $= 0.3659$

   $= 36.6\,\%$

   Answer

   $36.6\,\%$

   Three significant figures like the given efficiencies.

   Reasonableness

   The student's error: a sum of efficiencies of $230\,\%$ would be over $100\,\%$, which is impossible, because a plant cannot produce more electricity than there is energy in the fuel. Each step reduces the power that is left, so the efficiencies are multiplied, not added. The product $0.366$ is smaller than the weakest step $0.410$, as it should be when all the factors are below 1. Estimate: $0.9 \cdot 0.4 \cdot 1.0 = 0.36$. The waste heat is $100\,\% - 36.6\,\% = 63.4\,\%$ of the fuel power.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 14-16

   The fuel power of a combined heat and power plant is $600\,\text{MW}$.

   When the plant produces only electricity, the electric power is $240\,\text{MW}$. When all the heat of the condenser is led into the district heating network, the electric power is $180\,\text{MW}$ and the district heating power $348\,\text{MW}$. Between these modes the electric power decreases linearly as the district heating power increases.

   The district heating network now needs $200\,\text{MW}$ of heat. How large is the electric power of the plant? Give your answer in megawatts.

   Also calculate the overall efficiency of the plant and justify why the combined efficiency of electricity and heat still does not tell which way of production is more worthwhile.

   MWCheck
   Answer

   $P_\text{s} = 206\,\text{MW}$

   Show the solution

   Given

   $P_0 = 600\,\text{MW}$. Electricity only: $P_\text{s,0} = 240\,\text{MW}$. Full combined production: $P_\text{s,full} = 180\,\text{MW}$ and $P_\text{k,full} = 348\,\text{MW}$. Now the district heating power is $P_\text{k} = 200\,\text{MW}$.

   Find

   the electric power $P_\text{s}$ in megawatts and the overall efficiency $\eta_\text{tot}$

   Principle

   According to the problem the electric power decreases linearly as the district heating power increases, so the two extreme modes define a straight line. District heating is taken in an intermediate mode, so the electric power is read from that line. The overall efficiency is the combined power of electricity and district heating divided by the fuel power.

   Formula

   $P_\text{s} = P_\text{s,0} - k P_\text{k}$, where

   $k$$= \dfrac{P_\text{s,0} - P_\text{s,full}}{P_\text{k,full}}$

   and

   $\eta_\text{tot}$$= \dfrac{P_\text{s} + P_\text{k}}{P_0}$

   Conversions

   All powers are in megawatts and $k$ is a dimensionless number, so no conversions are needed.

   Substitution

   $k$$= \dfrac{240\,\text{MW} - 180\,\text{MW}}{348\,\text{MW}}$

   $= 0.17241$

   $P_\text{s}$$= 240\,\text{MW} - 0.17241 \cdot 200\,\text{MW}$

   $= 205.517\,\text{MW}$

   Then

   $\eta_\text{tot}$$= \dfrac{205.517\,\text{MW} + 200\,\text{MW}}{600\,\text{MW}}$

   $= 0.67586$

   $= 67.6\,\%$

   Answer

   $P_\text{s} = 206\,\text{MW}$

   And $\eta_\text{tot} = 67.6\,\%$, three significant figures, that is megawatt precision, because all powers are given to the nearest megawatt: the difference $240 - 180 = 60\,\text{MW}$ has that precision, and the subtracted $0.172 \cdot 200 \approx 34\,\text{MW}$ is to the nearest unit.

   Reasonableness

   The results are between the extreme modes: the electric power between $180$ and $240\,\text{MW}$ and the efficiency between $40\,\%$ and $88\,\%$. Limiting case: when $P_\text{k} = 348\,\text{MW}$, the formula gives $240 - 60 = 180\,\text{MW}$. For every megawatt of district heating $\tfrac{60}{348} = 0.17\,\text{MW}$ of electricity is lost. The combined efficiency does not tell which is more worthwhile, because electricity can be used for anything, but district heat only for heating; profitability depends on need and prices.

2. Exercise 14-17

   A wind farm produces $36.0\,\text{GWh}$ of electricity in a year ($1\,\text{GWh} = 3.60 \cdot 10^{12}\,\text{J}$).

   If the same electricity were produced with hard coal in a condensing power plant with an efficiency of 0.380 and where the heat of combustion of coal is $25.0\,\text{MJ/kg}$, how many kilotonnes of coal ($1\,\text{kt} = 10^{6}\,\text{kg}$) would be burned in a year? Also explain why the efficiency of a wind farm cannot be compared with the efficiency of a coal power plant as directly as those of two condensing power plants.

   ktCheck
   Answer

   $13.6\,\text{kt}$

   Show the solution

   Given

   electrical energy $E = 36.0\,\text{GWh}$, efficiency $\eta = 0.380$ and heat of combustion of coal $H = 25.0\,\tfrac{\text{MJ}}{\text{kg}}$; given $1\,\text{GWh} = 3.60 \cdot 10^{12}\,\text{J}$

   Find

   the mass $m$ of coal burned in a year, in kilotonnes

   Principle

   Energy is conserved: the chemical energy $Q_1$ of the coal turns into electricity with efficiency $\eta$ and the rest into waste heat. The denominator of the efficiency is the bought fuel, and the energy of the fuel is the heat of combustion times the mass.

   Formula

   $E = \eta Q_1$ and $Q_1 = H m$, so $m = \dfrac{E}{\eta H}$

   Conversions

   $E$$= 36.0 \cdot 3.60 \cdot 10^{12}\,\text{J}$

   $= 1.296 \cdot 10^{14}\,\text{J}$

   $H = 25.0 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}$. At the end $1\,\text{kt} = 10^6\,\text{kg}$.

   Substitution

   $m$$= \dfrac{1.296 \cdot 10^{14}\,\text{J}}{0.380 \cdot 25.0 \cdot 10^6\,\tfrac{\text{J}}{\text{kg}}}$

   $= 1.3642 \cdot 10^7\,\text{kg}$

   $= 13.642\,\text{kt}$

   Answer

   $13.6\,\text{kt}$

   Three significant figures, because the electrical energy $36.0\,\text{GWh}$, the efficiency $0.380$, the heat of combustion $25.0\,\tfrac{\text{MJ}}{\text{kg}}$ and the conversion factor $3.60 \cdot 10^{12}\,\text{J}/\text{GWh}$ all have three figures.

   Reasonableness

   Estimate:

   $\tfrac{1.3 \cdot 10^{14}\,\text{J}}{0.4 \cdot 2.5 \cdot 10^7\,\tfrac{\text{J}}{\text{kg}}}$

   $\approx 1.3 \cdot 10^7\,\text{kg}$

   The amount corresponds to about 270 wagonloads of coal, when a wagon holds $50\,\text{t}$. The efficiencies cannot be compared directly: in a coal power plant the denominator is bought fuel, which gets used up: about $38\,\%$ of its energy turns into electricity and the rest into waste heat, and burning produces emissions, but the energy of the wind is not a store that gets used up. The efficiency of a wind farm only tells what fraction of the kinetic energy of the air flowing through the rotors is turned into electricity.

3. Exercise 14-18

   Heating a house takes $100\,\text{kWh}$ of heat.

   The electricity is produced in a condensing power plant with an efficiency of 0.400, and $5.0\,\%$ of it is lost in power lines and transformers, so the house's electricity meter receives the fraction $0.950 \cdot 0.400 = 0.380$ of the energy of the fuel. The house has an air-source heat pump with a coefficient of performance of 3.00.

   How much fuel energy does heating with the pump require at the power plant? Give your answer in kilowatt-hours.

   Compare the result directly with electric heating and an oil boiler (efficiency 0.900) and conclude which way saves the most fuel.

   kWhCheck
   Answer

   $87.7\,\text{kWh}$

   Show the solution

   Given

   heating need $Q = 100\,\text{kWh}$, coefficient of performance $\varepsilon = 3.00$, efficiency of the power plant $\eta_\text{plant} = 0.400$ and share left after transmission $\eta_\text{transm} = 0.950$

   Find

   the energy of the fuel $E_\text{fuel}$ at the power plant in kilowatt-hours

   Principle

   A heat pump moves heat from outside to inside and uses electricity for it, so the coefficient of performance tells the ratio of heat to electricity. The electricity is produced from fuel in a chain where each step lets through only part of the energy, so the fuel is found by dividing the electricity by the efficiency of the chain.

   Formula

   $\varepsilon = \dfrac{Q}{E_\text{s}}$, so $E_\text{s} = \dfrac{Q}{\varepsilon}$, and

   $E_\text{fuel}$$= \dfrac{E_\text{s}}{\eta_\text{transm}\, \eta_\text{plant}}$

   thus

   $E_\text{fuel}$$= \dfrac{Q}{\eta_\text{transm}\, \eta_\text{plant}\, \varepsilon}$

   Conversions

   All energies are in kilowatt-hours and the factors are dimensionless, so no conversions are needed.

   Substitution

   $E_\text{fuel}$$= \dfrac{100\,\text{kWh}}{0.950 \cdot 0.400 \cdot 3.00}$

   $= \dfrac{100\,\text{kWh}}{1.14}$

   $= 87.719\,\text{kWh}$

   Answer

   $87.7\,\text{kWh}$

   Three significant figures, because the factors $0.950$, $0.400$ and $3.00$ have three figures and the heating need $100\,\text{kWh}$ is given to the nearest kilowatt-hour.

   Reasonableness

   Direct electric heating would need

   $\dfrac{100\,\text{kWh}}{0.380}$$= 263\,\text{kWh}$

   of fuel and an oil boiler

   $\dfrac{100\,\text{kWh}}{0.900}$$= 111\,\text{kWh}$

   The heat pump saves the most fuel, because in addition to the electricity used, heat transferred from the outside air comes into the house, so electricity is needed for only a third of the heat. Direct electric heating is the worst, because over $60\,\%$ of the energy of the fuel has already been wasted in producing the electricity.

4. Exercise 14-19

   A city needs electricity of $100\,\text{MW}$ and district heating of $200\,\text{MW}$. Option A: the electricity is produced in a condensing power plant with an efficiency of 0.400 and the heat in a heating boiler with an efficiency of 0.900. Option B: the electricity and heat are produced in a combined heat and power plant with an electrical efficiency of 0.320 and a district heating power equal to 0.560 times the fuel power. The heat of the plant is not enough for the whole district heating need, so the missing part is produced in a similar heating boiler as in option A. All the numbers are example data.

   a) Calculate the total fuel power of option A.

   b) Calculate the total fuel power of option B.

   c) By how much is the fuel power smaller in option B than in option A? Justify where the saving comes from.

   In part c), enter the difference of the fuel powers in the answer box in megawatts.

   MWCheck
   Answer

   $132\,\text{MW}$

   Show the solution

   Given

   electricity need $P_\text{s} = 100\,\text{MW}$ and district heating need $P_\text{k} = 200\,\text{MW}$. A: efficiency of the condensing power plant $\eta_\text{s,A} = 0.400$ and of the heating boiler $\eta_\text{boiler} = 0.900$. B: electrical efficiency of the combined heat and power plant $\eta_\text{s,B} = 0.320$ and district heating power $0.560\,P_{0,\text{B}}$; the missing heat is produced in a similar heating boiler ($\eta_\text{boiler} = 0.900$)

   Find

   the difference of the fuel powers $P_{0,\text{A}} - P_{0,\text{B}}$ in megawatts

   Principle

   The fuel power is found by dividing the useful power by the efficiency. In option A electricity and heat are produced separately, so the fuel powers are added. In option B the plant is sized according to the electricity need, and the district heating it produces is subtracted from the heat need; the heat that is left is produced in a boiler.

   Formula

   A:

   $P_{0,\text{A}}$$= \dfrac{P_\text{s}}{\eta_\text{s,A}} + \dfrac{P_\text{k}}{\eta_\text{boiler}}$

   B:

   $P_{0,\text{CHP}}$$= \dfrac{P_\text{s}}{\eta_\text{s,B}}$

   $P_\text{k,CHP}$$= 0.560\,P_{0,\text{CHP}}$

   $P_{0,\text{B}}$$= P_{0,\text{CHP}} + \dfrac{P_\text{k} - P_\text{k,CHP}}{\eta_\text{boiler}}$

   The difference is $P_{0,\text{A}} - P_{0,\text{B}}$.

   Conversions

   All powers are in megawatts and the efficiencies are dimensionless numbers, so no conversions are needed.

   Substitution

   A:

   $P_{0,\text{A}}$$= \dfrac{100\,\text{MW}}{0.400} + \dfrac{200\,\text{MW}}{0.900}$

   $= 250\,\text{MW} + 222.22\,\text{MW}$

   $= 472.22\,\text{MW}$

   B:

   $P_{0,\text{CHP}}$$= \dfrac{100\,\text{MW}}{0.320}$

   $= 312.5\,\text{MW}$

   $P_\text{k,CHP}$$= 0.560 \cdot 312.5\,\text{MW}$

   $= 175.0\,\text{MW}$

   The boiler has to produce $200\,\text{MW} - 175.0\,\text{MW} = 25.0\,\text{MW}$, so the fuel power of the boiler is

   $\dfrac{25.0\,\text{MW}}{0.900}$$= 27.78\,\text{MW}$

   $P_{0,\text{B}}$$= 312.5\,\text{MW} + 27.78\,\text{MW}$

   $= 340.28\,\text{MW}$

   The difference is $472.22\,\text{MW} - 340.28\,\text{MW} = 131.94\,\text{MW}$.

   Answer

   $132\,\text{MW}$

   Three significant figures like the given efficiencies. Option B needs about $28\,\%$ less fuel.

   Reasonableness

   The overall efficiency of combined production is $0.320 + 0.560 = 0.880$, and $0.880 \cdot 312.5\,\text{MW} = 275\,\text{MW}$ is the sum of electricity and heat, $100\,\text{MW} + 175\,\text{MW}$. The overall efficiency of option A is only

   $\dfrac{300\,\text{MW}}{472.22\,\text{MW}}$$= 0.635$

   The saving comes from the fact that in combined production the heat of the condenser, which in a condensing power plant would go to the sea, heats the city.

5. Exercise 14-20

   The thermal power of the reactor of a nuclear power plant is $2700\,\text{MW}$ and the electric power $900\,\text{MW}$. The flow of sea water cooling the condenser is a constant $38.0\,\text{t/s}$. The temperatures of the water entering and leaving the condenser have been measured in four months (table, example data). The specific heat capacity of water is $4190\,\text{J/(kg}\cdot\text{K)}$.

   a) Calculate the rise in the temperature of the water $\Delta T$ in each month and its average $\Delta T_\text{avg}$.

   b) Calculate the heat power $P_\text{L}$ transferred to the condenser, using the average $\Delta T_\text{avg}$.

   c) Compare the result with the waste heat power of the plant $P_\text{h} = P_0 - P_\text{s}$. Where does the rest of the waste heat go?

   d) Why can the heat of the condenser that warms the sea water no longer be used to produce electricity, and what effect does the heat have on the nearby water in July?

   In part b), enter the result in the answer box in megawatts.

   Temperatures of the condenser water (example data)

   Month	$T_\text{in}\,({}^\circ\text{C})$	$T_\text{out}\,({}^\circ\text{C})$
   January	1.2	12.1
   April	3.5	14.2
   July	17.8	28.4
   September	12.3	23.2

   MWCheck
   Answer

   $1720\,\text{MW}$

   Show the solution

   Given

   $P_0 = 2700\,\text{MW}$, $P_\text{s} = 900\,\text{MW}$, water flow $q_m = 38.0\,\text{t/s}$ and $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$; from the table the temperatures in and out in January $1.2\,{}^\circ\text{C}$ and $12.1\,{}^\circ\text{C}$, in April $3.5\,{}^\circ\text{C}$ and $14.2\,{}^\circ\text{C}$, in July $17.8\,{}^\circ\text{C}$ and $28.4\,{}^\circ\text{C}$ and in September $12.3\,{}^\circ\text{C}$ and $23.2\,{}^\circ\text{C}$

   Find

   the heat power $P_\text{L}$ transferred to the condenser in megawatts

   Principle

   The heat of the condenser warms the cooling water, and the quantity of heat is $Q = cm\,\Delta T$ as long as the water stays liquid. Power is the quantity of heat per unit of time, so $P_\text{L} = c\,q_m\,\Delta T$. The rise in temperature varies slightly from month to month, so the average is used.

   Reading the graph

   From the table the rise $\Delta T = T_\text{out} - T_\text{in}$ is calculated for each month: January $10.9\,\text{K}$, April $10.7\,\text{K}$, July $10.6\,\text{K}$ and September $10.9\,\text{K}$. A temperature difference is the same on the Celsius and Kelvin scales.

   Formula

   $\Delta T_\text{avg}$$= \dfrac{\Sigma\,\Delta T}{4}$

   $P_\text{L} = c\,q_m\,\Delta T_\text{avg}$

   Conversions

   $q_m$$= 38.0\,\text{t/s}$

   $= 38\,000\,\text{kg/s}$

   The result is converted from watts to megawatts, $1\,\text{MW} = 10^6\,\text{W}$.

   Substitution

   The sum of the rises is $10.9 + 10.7 + 10.6 + 10.9 = 43.1\,\text{K}$, so $\Delta T_\text{avg} = 10.775\,\text{K}$. Then

   $P_\text{L}$$= 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 38\,000\,\tfrac{\text{kg}}{\text{s}} \cdot 10.775\,\text{K}$

   $= 1.7156 \cdot 10^{9}\,\text{W}$

   $= 1716\,\text{MW}$

   Answer

   $1720\,\text{MW}$

   Three significant figures like the water flow and the rise in temperature.

   Reasonableness

   Estimate: $4000 \cdot 4 \cdot 10^4 \cdot 11\,\text{W} \approx 1.8 \cdot 10^9\,\text{W}$. The waste heat power is

   $P_\text{h}$$= 2700\,\text{MW} - 900\,\text{MW}$

   $= 1800\,\text{MW}$

   and the condenser's share of it is $\tfrac{1716}{1800} = 95\,\%$; the rest, about $84\,\text{MW}$, goes into the air, for example as losses of the generator and the transformer. Less heat is transferred to the sea water than the waste heat, as it should be. After the condenser the water is only about $11\,\text{K}$ warmer than the surroundings, so it can no longer be used to produce electricity, but the energy has not disappeared: it warms the sea. In July the water leaves at a temperature of $28.4\,{}^\circ\text{C}$, so the sea is warmer in the discharge area than elsewhere, and conditions for aquatic life may worsen.