Explanation 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 ) ] (I) 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. Write the formula for change in enthalpy in terms of liquid condition. h 1 − h 2 = [ h f ( T 1 ) + v f ( T 1 ) [ p 1 − p sat ( T 1 ) ] ] − [ h f ( T 2 ) + v f ( T 2 ) [ p 2 − p sat ( T 2 ) ] ] (II) Here, the enthalpy of liquid corresponding to the temperature of T 1 is h f ( T 1 ) , the specific volume corresponding to the temperature of T 1 is v f ( T 1 ) , the pressure is p and the saturation pressure is p sat ; the suffixes 1 and 2 indicates the inlet and exit states of the liquid. It is given that there is no significant changes in the temperature and kinetic energy of the water. Hence, neglect the kinetic energy effect i.e. ( V 1 2 − V 2 2 ) 2 = 0 . When, there is no significant changes in the temperature ( T 1 = T 2 ) of the water between the inlet and outlet, the change in enthalpy becomes as follows. h 1 − h 2 = v f ( T ) [ p 1 − p 2 ] (III) The heat transfer between the pump and surroundings id negligible i.e. Q ˙ cv = 0 . Then the Equation (I) becomes as follows to obtain the work transfer rate ( W ˙ cv ) . 0 = − W ˙ cv + m ˙ [ ( h 1 − h 2 ) + g ( z 1 − z 2 ) ] W ˙ cv = m ˙ [ ( h 1 − h 2 ) + g ( z 1 − z 2 ) ] (IV) Here, the exit pipe of the pump is at the elevation of 15 m from the inlet pipe (datum)

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Author: MORAN

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Chapter 4.12, Problem 66P

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