Q. 4.1

Question

Recall Problem 1.34, which concerned an ideal diatomic gas taken around a rectangular cycle on a $P V$ diagram. Suppose now that this system is used as a heat engine, to convert the heat added into mechanical work.

(a) Evaluate the efficiency of this engine for the case $V_{2} = 3 V_{1}, P_{2} = 2 P_{1}$ .

(b) Calculate the efficiency of an "ideal" engine operating between the same temperature extremes.

Step-by-Step Solution

Verified

Answer

(a) The efficiency of the given cycle is $12 %$ .

(b) The efficiency of an "ideal" engine operating between the same temperature extremes is $83 %$

1Part (a) Step 1 : Given Information and formula used

$\begin{aligned} V_{2} = 3 V_{1} \\ P_{2} = 2 P_{1} \end{aligned}$

Formula used:

Efficiency of the engine can be written as:

$e = \frac{W}{Q_{h}}$

Where,

$W$ is the work done.

$Q_{h}$ is the total heat absorbed.

2Part (a) Step 2 : Calculation

Work done can be calculated as area under the curve.

$W = (V_{2} - V_{1}) (P_{2} - P_{1})$

Plugging in the given values in the equation,

$\begin{aligned} W = (3 V_{1} - V_{1}) (2 P_{1} - P_{1}) \\ W = 2 V_{1} P_{1} \end{aligned}$

From first law of thermodynamics, heat absorbed during the process A can be calculated as:

$\begin{aligned} Q_{A} = n C_{V} Δ T \\ Q_{A} = n \frac{5}{2} R (T_{2} - T_{1}) \\ Q_{A} = \frac{5}{2} R n (\frac{P_{2} V_{1}}{n R} - \frac{P_{1} V_{1}}{n R}) \\ Q_{A} = \frac{5}{2} R n (\frac{2 P_{1} V_{1}}{n R} - \frac{P_{1} V_{1}}{n R}) \\ Q_{A} = \frac{5}{2} P_{1} V_{1} \end{aligned}$

3Part (a) Step 3 : Total heat absorbed

Heat absorbed during the process $B$ can be calculated as:

$\begin{aligned} Q_{B} = n C_{P} Δ T \\ Q_{B} = n \frac{7}{2} R (T_{3} - T_{2}) \\ Q_{B} = \frac{7}{2} R n (\frac{P_{2} V_{2}}{n R} - \frac{P_{2} V_{1}}{n R}) \\ Q_{B} = \frac{7}{2} R n (\frac{2 P_{1} \times 3 V_{1}}{n R} - \frac{2 P_{1} V_{1}}{n R}) \\ Q_{B} = \frac{7}{2} \times 2 P_{1} (3 V_{1} - V_{1}) \\ Q_{B} = \frac{7}{2} \times 4 P_{1} V_{1} \\ Q_{B} = \frac{28}{2} P_{1} V_{1} \end{aligned}$

Total heat absorbed during the complete cycle is

$\begin{aligned} Q_{B} = n C_{P} Δ T \\ Q_{B} = n \frac{7}{2} R (T_{3} - T_{2}) \\ Q_{h} = Q_{A} + Q_{B} \\ Q_{h} = \frac{5}{2} P_{1} V_{1} + \frac{28}{2} P_{1} V_{1} \\ Q_{h} = \frac{33}{2} P_{1} V_{1} \end{aligned}$

4Part (a) Step 4 : Efficiency and conclusion

Efficiency of heat engine can be calculated as:

$\begin{aligned} e = \frac{W}{Q_{h}} \\ e = \frac{2 P_{1} V_{1}}{\frac{33}{2} P_{1} V_{1}} \\ e = 0.12 \\ e = 12 % \end{aligned}$

Thus, the efficiency of the given cycle is $12 %$ .

5Part (b) Step 1 : Formula used

Let us use the formula

e = 1 - \frac{T_{c}}{T_{h}}

$T_{c}$ temperature of cold reservoir

$T_{h}$ temperature of hot reservoir

6Part (b) Step 2 : Calculation

Highest value of temperature can be calculated as:

Temperature doubles as the pressure double and get tripled when the volume triples.

$\begin{aligned} T_{1} \propto P_{1} V_{1} \\ T_{2} \propto P_{2} V_{2} \\ T_{2} \propto 2 P_{1} \times 3 V_{1} \\ T_{2} \propto 6 P_{1} V_{1} \end{aligned}$