PROBLEM 6.35

PROBLEM 6.41

PROBLEM 6.43 Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an ideal gas and kinetic and potential energy effects are negligible. Determine the amount of entropy produced, in kJ/K per kg of air, for the compression. What is the minimum theoretical work input, in kJ per kg of air, for an adiabatic compression from the given initial state to a final pressure of 10 bar? KNOWN: Air is compressed adiabatically in a piston-cylinder. The initial and final states are specified. FIND: Determine the amount of entropy produced and the minimum theoretical work input for adiabatic compression from the initial state to the given final pressure, each per unit mass of air. SCHEMATIC AND GIVEN DATA:

ANALYSIS: To find the entropy produced, begin with the entropy balance: ΔS =

+ σ

Thus σ/m = s2 – s1 = so(T2) - so(T1) – R ln(p2/p1) With data from Table A-22

σ/m = (2.40902 – 1.70203)kJ/kg∙K –

ln = 0.04618 kJ/kg∙K

The work is determined using the energy balance, as follows: W/m = (u1 – u2) As u varies with T for an ideal gas, the work input decreases as T2 decreases. From the entropy balance s2 – s1 = σ/m ≥ 0 → s2 ≥ s1

Air p1 = 1 bar T1 = 300 K p2 = 10 bar T2 = 600 K

T

1 bar

s

10 bar

300 K

600 K

(1)

(2) (2s)

T2s . .

. ENGINEERING MODEL: (1) The air is a closed system. (2) Q = 0 and kinetic and potential energy effects are negligible. (3) The air is modeled as an ideal gas.

states not accessible

adiabatically

PROBLEM 6.43 (CONTINUED) As shown on the accompanying T-s diagram, only states at 10 bar to the right of state 2s (isentropic compression – with σ/m = 0) are accessible in an adiabatic compression, and the corresponding temperature T2s is the lowest possible temperature. Hence, compression to state 2s gives the minimum theoretical work input. For s2 – s1: so(T2s) = so(T1) + R ln(p2/p1) = 1.70203 + (8.314/28.97) ln (10/1) = 2.36284 kJ/kg∙K Interpolating in Table A-22: T2s ≈ 564.1 K and u2 ≈ 407.55 kJ/kg. With u1 = 214.07 kJ/kg

#1 (W/m)min input = (u2s – u1) = 407.55 – 214.07 = 193.48 kJ/kg

1. Note: The work for actual process from state 1 to state 2 is (W/m) input = (u2 – u1) = 434.78 – 214.07 = 220.71 kJ/kg which is greater than the theoretical minimum, as expected.

PROBLEM 6.62

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11.21

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11.22

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11.24

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11.37

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11.37

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11.43