Explanation The following figure shows the two sections of the pump where we apply steady flow energy equation. Figure-(1) Write the expression for steady flow energy equation for energy balance. − W ˙ C V + m ˙ ( h 1 − h 2 ) = 0 (I) Here, the enthalpy at entry is h 1 , enthalpy at exit is h 2 , mass flow rate is m ˙ and work done by 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 ) (II) Here, temperature at entry is T 1 , temperature at exit is T 2 ,specific heat at constant volume is c ,pressure at entry is p 1 , pressure at exit is p 2 and specific volume of the fluid is v . Substitute T 2 for T 1 in Equation (II). ( h 1 − h 2 ) = − v ( p 2 − p 1 ) Substitute v ( p 2 − p 1 ) for ( h 1 − h 2 ) in Equation (I). − W ˙ C V − m ˙ v ( p 2 − p 1 ) = 0 − W ˙ C V + m ˙ v ( p 1 − p 2 ) = 0 (III) Write the expression for continuity equation m ˙ = A V v m ˙ v = A V Here, cross sectional area of the pump hose is A and velocity of discharge is V

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

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

Publisher: WILEY

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Chapter 4.12, Problem 69P

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The change in pressure of the blood as it flows through the pump.

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