Explanation Here, the refrigerant operated at through the compressor. Hence, the inlet and exit mass flow rates are equal. m ˙ 1 = m ˙ 2 (I) Write the formula for inlet mass flow rate. m ˙ 1 = A 1 V 1 v 1 (II) Here, the volumetric flow rate is A 1 V 1 and the specific volume is v ; the suffix 1 indicates the inlet state. Write the energy rate balance equation. 0 = Q ˙ cv − W ˙ cv + m ˙ [ ( h 1 − h 2 ) + ( V 1 2 − V 2 2 ) 2 + g ( z 1 − z 2 ) ] (III) Here, the rate of heat transfer is Q ˙ cv , rate of work transfer is W ˙ cv , rate of mass flow is m ˙ , the enthalpy is h , velocity is V , gravitational acceleration is g , the elevation is z and the suffixes cv, 1 and 2 indicates control volume, initial and final state of the system respectively. Neglecting the potential and kinetic energies, i.e. g ( z 1 − z 2 ) = 0 and ( V 1 2 − V 2 2 ) 2 = 0 . Then the Equation (III) becomes as follows to obtain the heat transfer rate. Q ˙ cv = W ˙ cv + m ˙ ( h 2 − h 1 ) (IV) Rewrite the Equation (IV) in terms of heat transfer per kg of mass flow of refrigerant. Q ˙ cv m ˙ = W ˙ cv m ˙ + ( h 2 − h 1 ) Q ˙ cv = m ˙ [ W ˙ cv m ˙ + ( h 2 − h 1 ) ] (V) Here, the work input to the compressor is given as 60 kJ per kg of mass flow of refrigerant. W ˙ cv m ˙ = 60 kJ/kg At inlet: The refrigerant is at the state of saturated vapor

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 4.12, Problem 59P

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The heat transfer rate, in kW.

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Chapter 4.12, Problem 58P

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Fundamentals of Engineering Thermodynamics

Ch. 4.12 - Prob. 1E Ch. 4.12 - 2. When a drip coffeemaker on-off switch is turned...Ch. 4.12 - Prob. 3E Ch. 4.12 - Prob. 4E Ch. 4.12 - Prob. 5E Ch. 4.12 - Prob. 6E Ch. 4.12 - Prob. 7E Ch. 4.12 - Prob. 8E Ch. 4.12 - Prob. 9E Ch. 4.12 - 10. How does the operator of a pumper-tanker fire...

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