Worked examples · Sheet 1

Energy balances of heat engines and refrigerators

Theory used: Heat engine Efficiency Refrigerator Coefficient of performance of refrigerators and heat pumps

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

In each cycle a heat engine absorbs $3.0 kJ$ from the hot reservoir and releases $2.1 kJ$ to the cold reservoir, and it performs $20$ cycles per second. Compute the work produced in one cycle, the efficiency and the power of the engine.

Q_{in} = 3.0 kJ, Q_{out} = 2.1 kJ, ν = 20 s^{- 1}

L = Q_{in} - Q_{out}, η = \frac{L}{Q _{in}}, P = ν L

L = 3.0 kJ - 2.1 kJ = 0.90 kJ

η = \frac{0.90 kJ}{3.0 kJ} = 0.30

P = 20 s^{- 1} \cdot 900 J = 1.8 \cdot 1 0^{4} W = 18 kW

The engine goes through a cycle, therefore its internal energy at the end of each cycle equals the initial one and, by the first law, the work produced equals the net heat absorbed, that is, the difference between the heat absorbed and the heat released. The efficiency is the ratio of the work to the heat absorbed, therefore $30%$ of the heat taken from the hot reservoir becomes work. Finally, the engine performs $20$ cycles in one second, therefore in one second it produces a work equal to $20$ times that of one cycle, and the power is $18 kW$ .

Exercise 2

A car engine has efficiency $η = 0.25$ and delivers a mechanical power $P = 40 kW$ . The heat comes from the combustion of gasoline, which releases $4.5 \cdot 1 0^{7} J$ for each kilogram burned. Compute the thermal power absorbed, the thermal power released to the surroundings and the mass of gasoline burned in one hour of operation.

η = 0.25, P = 40 kW, H = 4.5 \cdot 1 0^{7} J/kg, Δ t = 3600 s

P_{in} = \frac{P}{η}, P_{out} = P_{in} - P, m = \frac{P _{in} Δ t}{H}

P_{in} = \frac{40 kW}{0.25} = 160 kW, P_{out} = 160 kW - 40 kW = 120 kW

m = \frac{1.6 \cdot 1 0 ^{5} W \cdot 3600 s}{4.5 \cdot 1 0 ^{7} J/kg} = \frac{5.76 \cdot 1 0 ^{8} J}{4.5 \cdot 1 0 ^{7} J/kg} \approx 13 kg

The efficiency also applies to powers, since in the same time interval the work produced and the heat absorbed are in the same ratio as in one cycle. Therefore the thermal power absorbed is the mechanical power divided by the efficiency. The part of the heat absorbed that does not become work is released to the surroundings, through the radiator and the exhaust gases, and amounts to $120 kW$ , three times the useful power. In one hour the engine absorbs $5.76 \cdot 1 0^{8} J$ ; since each kilogram of gasoline supplies $4.5 \cdot 1 0^{7}$ of them, the mass burned is about $13 kg$ . The order of magnitude is realistic for a car that maintains a high power for a long time.

Exercise 3

A refrigerator has coefficient of performance $ε = 4.0$ and draws an electrical power of $150 W$ from the grid, entirely converted into work on the refrigerant fluid. Compute the heat extracted from the interior and the heat released to the kitchen in $10$ minutes, and the time needed to turn into ice $0.50 kg$ of water that is already at $0^{\circ} C$ (latent heat of fusion $λ_{f} = 3.33 \cdot 1 0^{5} J/kg$ ).

ε = 4.0, P = 150 W, Δ t = 600 s

ε = \frac{Q _{2}}{∣ L ∣}, ∣ Q_{1} ∣ = Q_{2} + ∣ L ∣, m λ_{f} = εP t

∣ L ∣ = P Δ t = 150 W \cdot 600 s = 9.0 \cdot 1 0^{4} J

Q_{2} = ε ∣ L ∣ = 4.0 \cdot 9.0 \cdot 1 0^{4} J = 3.6 \cdot 1 0^{5} J

∣ Q_{1} ∣ = 3.6 \cdot 1 0^{5} J + 0.90 \cdot 1 0^{5} J = 4.5 \cdot 1 0^{5} J

t = \frac{m λ _{f}}{εP} = \frac{0.50 kg \cdot 3.33 \cdot 1 0 ^{5} J/kg}{4.0 \cdot 150 W} \approx 2.8 \cdot 1 0^{2} s \approx 4.6 min

The work received by the fluid in $10$ minutes is the electrical energy absorbed in that time. By the definition of the coefficient of performance, the heat extracted from the interior is $ε$ times the work spent, that is, $3.6 \cdot 1 0^{5} J$ . In a cycle the internal energy of the fluid does not change, therefore the heat released to the kitchen is the sum of the heat extracted from the interior and of the work received, $4.5 \cdot 1 0^{5} J$ ; for this reason a running refrigerator warms the room it is in. To freeze the water at $0^{\circ} C$ the heat $m λ_{f}$ must be extracted from it; the refrigerator extracts heat from the interior at the rate $εP = 600 W$ , therefore the time required is the ratio of the two quantities, about 4.6 minutes.

Exercise 4

A house loses a thermal power of $12 kW$ to the outside, and it is kept at constant temperature with a heat pump that has coefficient of performance $ε_{hp} = 3.5$ . Compute the electrical power drawn by the pump and the thermal power taken from the outside air, and compare the consumption with that of an electric heater that heats the house by converting all the electrical energy into heat.

P_{1} = 12 kW, ε_{hp} = 3.5

ε_{hp} = \frac{∣ Q _{1} ∣}{∣ L ∣} = \frac{P _{1}}{P _{el}}, P_{2} = P_{1} - P_{el}

P_{el} = \frac{P _{1}}{ε _{hp}} = \frac{12 kW}{3.5} \approx 3.4 kW

P_{2} = 12 kW - 3.43 kW \approx 8.6 kW

\frac{P _{heater}}{P _{el}} = \frac{12 kW}{3.43 kW} = ε_{hp} = 3.5

The temperature of the house stays constant, therefore every second the pump must supply to it the heat that the house loses, that is, the power $P_{1} = 12 kW$ . For a heat pump the useful effect is the heat released to the hot reservoir, therefore the electrical power is $P_{1}$ divided by the coefficient of performance. The energy balance of the cycle says that the heat released to the house is the sum of the heat taken from the outside and of the work received; therefore the outside air supplies $8.6 kW$ . To release $12 kW$ , the electric heater draws $12 kW$ from the grid, therefore it consumes 3.5 times the electrical energy of the heat pump. The result is consistent with the first law, since most of the heat released to the house comes from the outside air.