Explanation Here, the pump operates at steady state. Hence, the inlet and exit mass flow rates are equal. m ˙ 1 = m ˙ 2 Write the formula for inlet mass flow rate. m ˙ 1 = A 1 V 1 v 1 (I) Here, the volumetric flow rate is expressed as AV , the specific volume is v and 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 ) ] (II) 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 ) ] ] (III) 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 Equation (III) becomes as follows. h 1 − h 2 = v f ( T ) [ p 1 − p 2 ] (IV) The heat transfer between the pump and surroundings is negligible i.e. Q ˙ cv = 0 . Refer Fig. P4.67. The pump is immersed into the reservoir up to the depth of 10 m . Hence, the pressure ( p 1 ) of the water is expressed as follows. p 1 = p atm + ρ g L (V) Here, the atmospheric pressure is p atm , the density is ρ , the acceleration due to gravity is g and the immersed depth of the pump into the reservoir is L . The exit pressure of the water is, p 2 = 1 atm = p atm The density is also expressed as, ρ = 1 v Here, the specific volume is v . Substitute p 2 for p a t m and 1 v for ρ in Equation (V). p 1 = p 2 + 1 v g L p 1 − p 2 = 1 v g L v ( p 1 − p 2 ) = g L Here, the inlet and exit temperature of the water is same. Hence, the specific volume corresponding to the temperature at 15 ° C is v f ( T ) = v . Substitute v for v f ( T ) in Equation (IV)

Fundamentals of Engineering Thermodynamics

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ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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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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