Explanation The below figure shows the circular well insulated horizontal duct. Figure-(1) Write the expression for specific volume of air at inlet. v 1 = R T 1 p 1 (I) Here, the gas constant is R , the temperature at inlet is T 1 and the pressure at inlet is p 1 . Write the area of the duct. A = π 4 d 2 (II) Here, the diameter of the duct is d . Write the expression for the velocity of air at inlet. V 1 = m ˙ v 1 A (III) Here, the mass flow rate is m ˙ , the specific volume at point 1 is v 1 , the area of duct is A . Write the expression for exit velocity of air. V 2 = 2 ( h 1 − h 2 ) + V 1 2 (IV) Write the expression for specific volume of air at exit. v 2 = A V 2 m ˙ (V) Write the expression for pressure at exit. p 2 = R T 2 v 2 (VI) Write the expression for the force of the duct on the air. F x = m ˙ ( V 2 − V 1 ) + ( p 2 − p 1 ) A (VII) Here, the force of the duct on the air is F x , the velocity at inlet of the duct is V 1 , the velocity at exit of the duct is V 2 , the area of the duct at inlet is A 1 , the area of the duct at outlet is A 2 , the pressure at inlet is p 1 and the pressure at outlet is p 2 . Conclusion: Substitute 0.287 kJ / kgK for R , 300 K for T 1 and 6.8 bar for p 1 in Equation (I). v 1 = ( 0.287 kJ / kgK ) ( 300 K ) 6.8 bar = 86.1 kJ / kg 6.8 bar × ( 1 bar 10 5 N / m 2 ) × ( 10 3 N ⋅ m 1 kJ ) = 0.1266 m 3 / kg Substitute 0.1 m for d in Equation (II). A = π 4 ( 0.1 m ) 2 = 7.8539 × 10 − 3 m 2 Substitute 270 kg / min for m ˙ , 0.1266 m 3 / kg for v 1 and 7.8539 × 10 − 3 m 2 for A in Equation (III). V 1 = ( 270 kg / min ) ( 0.1266 m 3 / kg ) 7.8539 × 10 − 3 m 2 = ( 270 kg / min ) ( 0.1266 m 3 / kg ) ( 1 min 60 s ) 7.8539 × 10 − 3 m 2 = 72.57 m / s Refer table A-22 “Ideal gas properties of air” to obtain the specific enthalpy at 300 K as 300.19 kJ / kg . Refer table A-22 “Ideal gas properties of air” to obtain the specific enthalpy at 250 K as 250.05 kJ / kg . Substitute 300.19 kJ / kg for h 1 , 250.05 kJ / kg for h 2 and 72.57 m / s for V 1 in Equation (IV). V 2 = 2 ( 300.19 kJ / kg − 250.05 kJ / kg ) + ( 72.57 m / s ) 2 = ( 100

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 9.14, Problem 110P

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The net horizontal force exerted by the duct wall on the air.

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Chapter 9.14, Problem 109P

Chapter 9.14, Problem 111P

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