Explanation Assume that the potential and kinetic energy changes are negligible. Assume the pump to be in control volume for its analysis. Write the expression for rate of work transfer per unit mass flow rate for the pump. ( ( W ˙ c v ) int rev m ˙ ) = − υ ( p 2 − p 1 ) ( W ˙ c v ) int rev = − υ m ˙ ( p 2 − p 1 ) (I) Here, the mass flow rate is m ˙ , the rate of reversible work transfer is ( W ˙ c v ) int rev , the specific volume is υ , the initial pressure is p 1 and the final pressure is p 2 . Negative sign indicates power is consumed by the pump. Write the efficiency of the pump. η p = ( W ˙ c v ) int rev W ˙ W ˙ = ( W ˙ c v ) int rev η p (II) Here, the rate of work transfer through the pump is W ˙ . Conclusion: Refer to Table A-2E “Properties of Saturated Water (Liquid–Vapor): Temperature Table” to obtain specific volume at saturated liquid as 1.0018 × 10 − 3 m 3 / kg with respect to 20 ° C temperature. Substitute 1.0018 × 10 − 3 m 3 / kg for υ , 5 MPa for p 2 and 100 kPa for p 1 , 53 kg / min for m ˙ in Equation (I)

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 6.13, Problem 178P

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Chapter 6 Solutions

Fundamentals of Engineering Thermodynamics

Ch. 6.13 - Prob. 1E Ch. 6.13 - Prob. 2E Ch. 6.13 - Prob. 3E Ch. 6.13 - Prob. 4E Ch. 6.13 - Prob. 5E Ch. 6.13 - 6. Is entropy produced within a system undergoing...Ch. 6.13 - 7. When a mixture of olive oil and vinegar...Ch. 6.13 - Prob. 8E Ch. 6.13 - Prob. 9E Ch. 6.13 - 10. Is Eq. 6.51a restricted to adiabatic processes...

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