Question

Two reversible heat engines operate in series between a source at 527 celsius and a sink at 17 celsius deg. If the engines have equal efficiencies and the first rejects 400 kJ to the second, calculate:
(1) the temperature at which heat is supplied to the second engine
(2) the heat taken from the source
(3) the work done by each engine

Assume that the each engine operates on the carnot cycle.

Answer 1

Answer on Question #56620-Engineering-Mechanical Engineering

Two reversible heat engines operate in series between a source at 527 celsius and a sink at 17 celsius deg. If the engines have equal efficiencies and the first rejects 400 kJ to the second, calculate:

(1) the temperature at which heat is supplied to the second engine

(2) the heat taken from the source

(3) the work done by each engine

Solution

The figure below shows two engines where the heat sink of engine 1 is the heat source of engine 2. In fact, the two reversible heat engines taken together are fully equivalent to one single reversible heat engine that accepts heat transfer $Q_{H1}$ at temperature $T_{H1}$ and rejects heat transfer $Q_{L2}$ at temperature $T_{L2}$ while providing a net work output of $W_{1} + W_{2}$.

The thermal efficiency of a Carnot cycle engine is given by

$$E_{th} = 1 - \frac{Q_{L}}{Q_{H}} = 1 - \frac{T_{L}}{T_{H}}$$

where $T_{H}$ and $T_{L}$ are the absolute temperatures of the heat source and heat sink respectively and where $Q_{H}$ and $Q_{L}$ are the amounts of heat transfer to and from the engine. The equivalence between the heat transfer ratio $\frac{Q_{L}}{Q_{H}}$ and the ratio of absolute temperatures $\frac{T_{L}}{T_{H}}$ comes from the way the absolute temperature scale is defined. It always applies for a reversible heat engine operating between two fixed-temperature reservoirs.

Hence

$$\frac{Q_{L}}{Q_{H}} = \frac{T_{L}}{T_{H}} \quad (1)$$

Coming back to your problem, the two reversible engines have equal efficiencies, so

$$E_{th1} = E_{th2}$$

and hence

$$\frac{T_{L1}}{T_{H1}} = \frac{T_{L2}}{T_{H2}}, \quad \text{but} \quad T_{H2} = T_{L1} \text{ and so}$$ $$\frac {T _ {L 1}}{T _ {H 1}} = \frac {T _ {L 2}}{T _ {H 2}}, \text{ or } T _ {L 1} ^ {2} = T _ {H 1} T _ {L 2} \text{ and so}$$ $$T _ {L 1} = \sqrt {T _ {H 1} T _ {L 2}}$$

Equation 2, which only applies when the thermal efficiencies of the two heat engines are equal, can be used to find the temperature of the common, intermediate, thermal reservoir.

Applying equation 1 to engine 1

$$\frac {Q _ {\mathrm {L} _ {1}}}{Q _ {\mathrm {H} _ {1}}} = \frac {\bar {T} _ {\mathrm {L} _ {1}}}{\bar {T} _ {\mathrm {H} _ {1}}}, \text{ or }$$ $$Q _ {H _ {1}} = Q _ {L _ {1}} \frac {\bar {T} _ {H _ {1}}}{\bar {T} _ {L _ {1}}}$$

Equation 3 can be used to find the heat input when the heat rejection and the temperatures of the thermal reservoirs are known.

There are various ways in which you could go about calculating the net work outputs $\overline{W}_1$ and $\overline{W}_2$ using the expressions already mentioned above. Once you have found the temperature of the intermediate thermal reservoir, and as the temperatures of the other two reservoirs are given, you can calculate their thermal efficiency, which is the same for both heat engines. Once you have found the heat transfer to the first heat engine and as the heat transfer to the second heat engine is given, you can multiply each heat input amount by the thermal efficiency to find the work output of the engine.

Alternatively you could find the work quantities as the difference between the heat input and the heat rejection of each engine. For the second engine the heat rejection could be calculated from the heat input using equation 1, applied to that heat engine.

The temperatures in the expressions are all absolute temperatures, so don't forget to convert the Celsius values to Kelvin (0°C corresponds to 273.15 K).

$$\bar {T} _ {\mathrm {H} _ {1}} = 5 2 7 + 2 7 3. 1 5 [ \mathrm {K} ] = 8 0 0. 1 5 [ \mathrm {K} ]$$ $$\bar {T} _ {\mathrm {L} _ {2}} = 1 7 + 2 7 3. 1 5 [ \mathrm {K} ] = 2 9 0. 1 5 [ \mathrm {K} ]$$

From equation 2

$$\bar {T} _ {\mathrm {L} _ {1}} = \sqrt {\bar {T} _ {\mathrm {H} _ {1}} \bar {T} _ {\mathrm {L} _ {2}}} = \sqrt {8 0 0 . 1 5 \cdot 2 9 0 . 1 5} = 4 8 1. 8 3 4 [ \mathrm {K} ].$$ $$Q _ {L _ {1}} = 4 0 0 [ \mathrm {k J} ]$$

From equation 3

$$Q _ {H _ {1}} = Q _ {L _ {1}} \frac {\bar {T} _ {H _ {1}}}{\bar {T} _ {L _ {1}}} = 4 0 0 \times \frac {8 0 0 . 1 5}{4 8 1 . 8 3 4} [ \mathrm {k J} ] = 6 6 4. 2 5 4 [ \mathrm {k J} ].$$ $$E _ {\mathrm {t h} _ {1}} = 1 - \frac {\bar {T} _ {L _ {1}}}{\bar {T} _ {H _ {1}}} = 1 - \frac {4 8 1 . 8 3 4}{8 0 0 . 1 5} = 0. 3 9 7 8 2 1$$ $$W _ {1} = Q _ {H _ {1}} E _ {t h _ {1}} = 6 6 4. 2 5 4 \times 0. 3 9 7 8 2 1 [ \mathrm {k J} ] = 2 6 4. 2 5 4 [ \mathrm {k J} ]$$ $$Q _ {H _ {2}} = Q _ {L _ {1}} = 4 0 0 [ \mathrm {k J} ]$$ $$E _ {\mathrm {t h} _ {2}} = E _ {\mathrm {t h} _ {1}} = 0. 3 9 7 8 2 1$$ $$W _ {2} = Q _ {H _ {2}} E _ {t h _ {2}} = 4 0 0 \times 0. 3 9 7 8 2 1 [ \mathrm {k J} ] = 1 5 9. 1 2 8 [ \mathrm {k J} ]$$

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