Question

a. Draw the carnot cycle on P-V diagram. Show that the amount of heat absorbed ( rejected ) in a reversible cycle is proportional to the temperature of source (sink).

b. The efficiency of a carnot engine is 30% . Its efficiency is to be raised to 60% by how much the temperature of the source be increased if the sink is at 27°?

Answer 1

Answer on Question #66749-Physics-Molecular Physics-Thermodynamics

a. Draw the carnot cycle on P-V diagram. Show that the amount of heat absorbed ( rejected ) in a reversible cycle is proportional to the temperature of source (sink).

Solution

Carnot cycle is composed of four processes:

1-2. Isothermal heat addition (T=const)

2-3. Isentropic expansion (S=const)

3-4. Isothermal heat rejection (T=const)

4-1. Isentropic compression (S=const)

P-V and T-S diagrams of Carnot Cycle

The amount of heat absorbed from the T-S diagram is

$$Q _ {H} = T _ {s o u r c e} (S _ {2} - S _ {1}) \sim T _ {s o u r c e}.$$

The amount of heat rejected from the T-S diagram is

$$Q _ {L} = T _ {s i n k} (S _ {2} - S _ {1}) \sim T _ {s i n k}.$$

b. The efficiency of a Carnot engine is $30\%$ . Its efficiency is to be raised to $60\%$ by how much the temperature of the source be increased if the sink is at $27{}^{\circ}$ ?

Solution

$$\eta = 1 - \frac {T _ {s i n k}}{T _ {s o u r c e}} = 0. 3 \rightarrow T _ {s o u r c e} = \frac {T _ {s i n k}}{0 . 7}.$$ $$\eta^ {\prime} = 1 - \frac {T _ {s i n k}}{T _ {s o u r c e} ^ {\prime}} = 0. 6$$ $$T _ {s o u r c e} ^ {\prime} = \frac {T _ {s i n k}}{0 . 4}$$

The change in temperature of the source is

$$T _ {s o u r c e} ^ {\prime} - T _ {s o u r c e} = \frac {T _ {s i n k}}{0 . 4} - \frac {T _ {s i n k}}{0 . 7} = T _ {s i n k} \left(\frac {1}{0 . 4} - \frac {1}{0 . 7}\right) = (2 7 3 + 2 7) \left(\frac {1}{0 . 4} - \frac {1}{0 . 7}\right) = 3 2 1 K.$$

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