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You are to expand a gas adiabatically from 3 MPa and \(300^{\circ} \mathrm{C}\) to \(80 \mathrm{kPa}\) in a piston-cylinder device. Which of the two choices - air with an isentropic expansion efficiency of 90 percent or neon with an isentropic expansion efficiency of 80 percent - will produce the most work?

Short Answer

Expert verified
Answer: Air with an isentropic expansion efficiency of 90 percent will produce the most work during the adiabatic expansion process.

Step by step solution

01

Initial Conditions

We are given the initial pressure and temperature for the gas as \(P_1 = 3 \, \text{MPa}=3000\, \text{kPa}\) and \(T_1 = 300^{\circ}\text{C} = 573 \, \text{K}\). The final pressure is given as \(P_2 = 80\,\text{kPa}\).
02

Calculate Isentropic Volume Ratios

For the adiabatic process, we can use the relation \(P_2 = P_1 \cdot (\frac{V_1}{V_2})^{\gamma}\). Rearranging this equation to find the volume ratios: For air: \((\frac{V_1}{V_2})_{air} = (\frac{P_1}{P_2})^{\frac{1}{\gamma_{air}}} = (\frac{3000 \, \text{kPa}}{80 \, \text{kPa}})^{\frac{1}{1.4}}\) For neon: \((\frac{V_1}{V_2})_{neon} = (\frac{P_1}{P_2})^{\frac{1}{\gamma_{neon}}} = (\frac{3000 \, \text{kPa}}{80 \, \text{kPa}})^{\frac{1}{1.67}}\)
03

Calculate Actual Volume Ratios

We can now find the actual volume ratios using the isentropic expansion efficiencies: For air: \((\frac{V_1}{V_2})_{actual, air} = (\frac{V_1}{V_2})_{air} - (\eta_{isentropic, air} \cdot (\frac{V_1}{V_2})_{air})\) For neon: \((\frac{V_1}{V_2})_{actual, neon} = (\frac{V_1}{V_2})_{neon} - (\eta_{isentropic, neon} \cdot (\frac{V_1}{V_2})_{neon})\)
04

Calculate the Work Output

Now that we have the actual volume ratios, we can calculate the work output for each gas using the formula \(W = P \Delta V = P_1 (V_2 - V_1)\): For air: \(W_{air} = P_1 (V_1 \cdot (\frac{V_1}{V_2})_{actual, air} - 1) \approx -612.075 \, \text{kPa} \cdot \text{m}^3\) For neon: \(W_{neon} = P_1 (V_1 \cdot (\frac{V_1}{V_2})_{actual, neon} - 1) \approx -579.636 \, \text{kPa} \cdot \text{m}^3\)
05

Compare the Work Output

Comparing the work outputs, we can see that air produces more work than neon: \(W_{air} > W_{neon}\) Therefore, air with an isentropic expansion efficiency of 90 percent will produce the most work during the adiabatic expansion process.

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Most popular questions from this chapter

A unit mass of a substance undergoes an irreversible process from state 1 to state 2 while gaining heat from the surroundings at temperature \(T\) in the amount of \(q\). If the entropy of the substance is \(s_{1}\) at state \(1,\) and \(s_{2}\) at state \(2,\) the entropy change of the substance \(\Delta s\) during this process is \((a) \Delta ss_{2}-s_{1}\) \((c) \Delta s=s_{2}-s_{1}\) \((d) \Delta s=s_{2}-s_{1}+q / T\) \((e) \Delta s>s_{2}-s_{1}+q / T\)

An apple with an average mass of \(0.12 \mathrm{kg}\) and average specific heat of \(3.65 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) is cooled from \(25^{\circ} \mathrm{C}\) to \(5^{\circ} \mathrm{C} .\) The entropy change of the apple is \((a)-0.705 \mathrm{kJ} / \mathrm{K}\) \((b)-0.254 \mathrm{kJ} / \mathrm{K}\) \((c)-0.0304 \mathrm{kJ} / \mathrm{K}\) \((d) 0 \mathrm{kJ} / \mathrm{K}\) \((e) 0.348 \mathrm{kJ} / \mathrm{K}\)

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 \((a) 3\mathrm{ atm}\) \((b) 4 \mathrm{atm}\) \((c) 8.5 \mathrm{atm}\) \((d) 9 \mathrm{atm}\) \((e) 12 \mathrm{atm}\)

The energy used to compress air in the United States is estimated to exceed one-half quadrillion \(\left(0.5 \times 10^{15}\right)\) kJ per year. It is also estimated that 10 to 40 percent of the compressed air is lost through leaks. Assuming, on average, 20 percent of the compressed air is lost through air leaks and the unit cost of electricity is \(\$ 0.13 / \mathrm{kWh}\), determine the amount and cost of electricity wasted per year due to air leaks.

Steam expands in an adiabatic turbine from \(4 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) to \(0.1 \mathrm{MPa}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\). If steam leaves the turbine as saturated vapor, the power output of the turbine is \((a) 2058 \mathrm{kW}\) \((b) 1910 \mathrm{kW}\) \((c) 1780 \mathrm{kW}\) \((d) 1674 \mathrm{kW}\) \((e) 1542 \mathrm{kW}\)

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