Explanation The following figure shows the two sections of the pump. Figure-(1) Write the expression for steady flow energy equation for energy balance for pump. − W ˙ C V + m ˙ ( h 1 − h 2 ) = 0 (I) Here, mass flow rate of water is m ˙ , enthalpy at inlet is h 1 , enthalpy at exit is h 2 , and work done on the control volume is W ˙ C V . Write the expression for change in specific enthalpy of an incompressible substance. ( h 2 − h 1 ) = ∫ T 1 T 2 c d T + v ( p 2 − p 1 ) − ( h 1 − h 2 ) = c ( T 2 − T 1 ) + v ( p 2 − p 1 ) ( h 1 − h 2 ) = − c ( T 2 − T 1 ) − v ( p 2 − p 1 ) ( h 1 − h 2 ) = − c ( T 2 − T 1 ) + v ( p 1 − p 2 ) (II) Here, temperature at inlet is T 1 , temperature at exit is T 2 , pressure at inlet is p 1 , pressure at exit is p 2 , specific heat at constant volume is c and specific volume of control volume is v . Substitute Equation (II) in Equation (I). − W ˙ C V + m ˙ [ − c ( T 2 − T 1 ) + v ( p 1 − p 2 ) ] = 0 − W ˙ C V − m ˙ c ( T 2 − T 1 ) + m ˙ v ( p 1 − p 2 ) = 0 ( T 2 − T 1 ) = − W ˙ C V + m ˙ v ( p 1 − p 2 ) m ˙ c ( T 2 − T 1 ) = − W ˙ C V m ˙ c + v ( p 1 − p 2 ) c × 1 / ρ v ( T 2 − T 1 ) = [ − W ˙ C V m ˙ + ( p 1 − p 2 ) ρ ] c (III) Here, the mass rate flow is m ˙ , the density of water is ρ . Write the expression for continuity equation. m ˙ = ρ Q (IV) Here, discharge rate of fluid flow is Q . Conclusion: Substitute 60.58 lb / ft 3 for ρ and 0.42 gal / min for Q in Equation (IV). m ˙ = ( 60.58 lb / ft 3 ) ( 0.42 gal / min ) = ( 60.58 lb / ft 3 ) ( 0

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Chapter 4.12, Problem 70P

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The change in temperature of the water as it flows through the pump.

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Chapter 4.12, Problem 69P

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