Explanation Write the formula for mass flow rate ( m ˙ ) . m ˙ = AV v (I) Rearrange the Equation (I) to obtain velocity ( V ) . V= m ˙ v A (II) Here, the cross sectional area is A , the velocity is V and the specific volume is v . Write the formula for cross sectional area ( A ) of the cylindrical pipe. A = π d 2 4 . (III) Here, the diameter is d . The given temperature ( − 15 ° F ) and quality ( 0.05 ) of refrigerant-134a says it is in liquid-vapor mixture. Write the formula for specific volume of the refrigerant. v = v f + x ( v g − v f ) (IV) Here, the specific volume at fluid condition is v f , the specific volume at gaseous condition is v g and the quality of the refrigerant is x . Refer Table A-10E, “Properties of saturated refrigerant-134a (liquid vapor): Temperature table”. The specific volume at fluid condition ( v f ) corresponding to the temperature of − 15 ° F is 0.01163 ft 3 /lb and the specific volume at gaseous condition ( v g ) corresponding to the temperature of − 15 ° F is 3.0286 ft 3 /lb . Conclusion: Substitute 0.75 in . for d in Equation (III). A = π ( 0.75 in . × 1 ft 12 in . ) 2 4 = 3

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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