Question

Air is to be compressed steadily and isentropically from 1 atm to 16 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be \((a) 3\mathrm{ atm}\) \((b) 4 \mathrm{atm}\) \((c) 8.5 \mathrm{atm}\) \((d) 9 \mathrm{atm}\) \((e) 12 \mathrm{atm}\)

Short answer

a) 3 atm b) 4 atm c) 8.5 atm d) 9 atm e) 12 atm Answer: c) 8.5 atm

Step-by-step solution

Recognize the knowns and unknowns

We know the initial pressure (1 atm) and final pressure (16 atm). We need to find the optimal intermediate pressure from the given options (3 atm, 4 atm, 8.5 atm, 9 atm, and 12 atm) that minimizes the total work input in the two-stage compressor.

Find the work input for each stage in terms of intermediate pressure

The minimum work input for an isentropic process is given by the following relationship: \(W_{min} = \int_{P_1} ^{P_2} v dP\) Where, \(W_{min}\) is the minimum work input, \(P_1\) and \(P_2\) are initial and final pressures, respectively, and \(v\) is the specific volume. For isentropic process, specific volume remains constant. We also know that the work input for both the stages is the same. Let us denote the intermediate pressure as \(P_{int}\). The total work input will be, \(W_{total} = W_{1 \to int} + W_{int \to 16}\) Since, specific volume is constant for isentropic process, \( \int_{P_1} ^{P_{int}} v dP + \int_{P_{int}} ^{P_2} v dP = v(P_{int} - P_1) + v(P_2 - P_{int})\)

Minimize the total work input

In order to minimize the total work input, we should minimize the expression for \(W_{total}\). \(W_{total} = v(P_{int} - P_1) + v(P_2 - P_{int})\) Taking the derivative of \(W_{total}\) with respect to \(P_{int}\) and equate it to zero: \(\frac{dW_{total}}{dP_{int}} = v - v = 0\) This expression shows that the work input is minimized when the work input in both stages is equal. Now, in order to find the optimal intermediate pressure, we can equate the work input of the first stage (\(W_{1 \to int}\)) to work input of the second stage (\(W_{int \to 16}\)): \(v(P_{int} - P_1) = v(P_2 - P_{int})\) Solving for \(P_{int}\): \(P_{int} = \frac{P_1 + P_2}{2} = \frac{1 \mathrm{atm} + 16 \mathrm{atm}}{2} = 8.5 \mathrm{atm}\)

Conclusion

The optimal intermediate pressure that minimizes the total work input in a two-stage isentropic compressor is 8.5 atm. Hence, the correct answer is (c) 8.5 atm.