Explanation Consider the system is in steady state. Hence, the inlet and exit mass flow rates are equal. The mass flow rate of air ( m ˙ a ) is as follows. m ˙ 1 = m ˙ 2 = m ˙ a The mass flow rate of water ( m ˙ w ) is as follows. m ˙ 3 = m ˙ 4 = m ˙ w The flow work of air is expressed as follows. w flow,a = P 2 v 2 − P 1 v 1 (I) Here, the pressure is P , the specific volume is v and the subscripts 1 and 2 indicates the initial and final states of air. Write the ideal gas relation. P v = R T Here, the gas constant is R and the temperature is T . Since, the air is an ideal gas. Rewrite the Equation (I) with reference to the ideal gas relation as follows. w flow,a = R T 2 − R T 1 = R ( T 2 − T 1 ) (II) The flow work of water is expressed as follows. w flow,w = P 4 v 4 − P 3 v 3 (III) Write the formula total flow power required. W ˙ flow = m ˙ a w flow,a + m ˙ w w flow,w (IV) Write the formula for mass flow rate of air. m ˙ a = V ˙ P 1 R T 1 (V) Here, the volumetric flow rate of air is V ˙ , the pressure is P , the gas constant of air is R and the temperature is T ; the suffix 1 indicates the inlet condition of air. For water: At inlet: The water is at the state of saturated water. Refer Table A-4E, “Saturated water–Temperature table”. Obtain the inlet specific volume ( v 3 ) corresponding to the temperature of 50 °F . v f @ 50 ° F = v 3 = 0.01602 ft 3 /lbm At exit: Still, the water is at the state of saturated water. Refer Table A-4E, “Saturated water–Temperature table”. Obtain the outlet specific volume ( v 4 ) corresponding to the temperature of 90 °F . v f @ 90 ° F = v 4 = 0.01610 ft 3 /lbm Refer Table A-1E, “Molar mass, gas constant, and critical-point properties”. The gas constant ( R ) of air, in Btu/lbm ⋅ R is 0.06855 Btu/lbm ⋅ R . The gas constant ( R ) of air, in psia ⋅ ft 3 /lbm ⋅ R is 0.3704 psia ⋅ ft 3 /lbm ⋅ R . Conclusion: Substitute 0.06855 Btu/lbm ⋅ R for R , 100 ° F for T 2 , and 200 °F for T 1 in Equation (II). w flow,a = 0.06855 Btu/lbm ⋅ R ( 100 °F − 200 ° F ) = 0.06855 Btu/lbm ⋅ R [ ( 100 + 460 ) R − ( 200 + 460 ) R ] = 0.06855 Btu/lbm ⋅ R ( − 200 R ) = − 6

9th Edition

ISBN: 9781259822674

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 5.5, Problem 149RP

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The total flow power, in hp, required and the flow work, in Btu/lbm for both air and water streams.

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Chapter 5.5, Problem 148RP

Chapter 5.5, Problem 150RP

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The total flow power, in hp , required and the flow work, in Btu/lbm for both air and water streams.

Solution Summary: The author explains the flow power and flow work required for both air and water streams. The mass flow rate of air is as follows.

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The total flow power, in hp, required and the flow work, in Btu/lbm for both air and water streams.

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