Explanation Write the expression for the velocity at exit of the duct. V 2 = 2 ( h 1 − h 2 ) + V 1 2 (I) Here, the specific enthalpy at inlet of the duct is h 1 , the specific enthalpy at exit of the duct is h 2 . Write the expression for sound velocity at inlet. c 1 = k 1 R T 1 (II) Here, the specific heat ratio is k 1 , the gas constant for air is R and the temperature at state 1 is T 1 . Write the expression for mach number at inlet state. M 1 = V 1 c 1 (III) Here, the velocity of fluid is V 1 , the sonic velocity is c 1 . Write the expression for stagnation enthalpy. h o 1 = h 1 + V 1 2 2 (IV) Here, the enthalpy at state 1 is h 1 . Write the expression for stagnation pressure. p o 1 = p 1 ( p r 01 p r 1 ) (V) Here, the reduced pressure at state 01 is p r 01 and the reduced pressure at state 1 is p r 1 . Write the expression for sound velocity at exit state. c 2 = k 2 R T 2 (VI) Here, the specific heat ratio is k 2 , the gas constant for air is R and the temperature at state 2 is T 2 . Write the expression for mach number at exit state. M 2 = V 2 c 2 (VII) Here, the velocity of fluid is V 2 , the sonic velocity is c 2 . Write the expression for stagnation pressure. p o 2 = p 2 ( p r 02 p r 2 ) (VIII) Here, the reduced pressure at state 02 is p r 02 and the reduced pressure at state 2 is p r 2 . Conclusion: Refer to table A-22 “ideal gas properties of air” to obtain the specific enthalpy at temperature T 1 = 500 K as 503.02 kJ / kg . Refer to table A-22 “ideal gas properties of air” to obtain the specific enthalpy at temperature T 2 = 440 K as 441.61 kJ / kg . Here, the universal gas constant is 0.287 kJ / kg ⋅ K . Substitute 441.61 kJ / kg for h 2 , 503.02 kJ / kg for h 1 and 50 m / s for V 1 in Equation (I). V 2 = 2 ( 503.02 kJ / kg − 441.61 kJ / kg ) + ( 50 m / s ) 2 = 2 ( 61.41 ) ( kJ / kg ) ( 1 kg ⋅ m / s 2 1 N ) ( 10 3 N ⋅ m 1kJ ) + 2500 ( m / s ) 2 = 125320 ( m / s ) 2 = 354 m / s Refer table A-20 “Ideal gas specific heats of some common gases” to obtain the specific heat ratio at temperature 500 K as k = 1.387 . Substitute 1.387 for k , 287 N ⋅ m / kgK for R and 500 K for T 1 in Equation (II). c 1 = 1.387 ( 287 N ⋅ m / kgK ) ( 500 K ) = ( 199034.5 N ⋅ m / kg ) ( 1 kg ⋅ m / s 2 1 N ) = 446.132 m / s Substitute 446.132 m / s for c 1 and 50 m / s for V 1 in Equation (III)

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ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Chapter 9.14, Problem 118P

To determine

The Mach number at inlet and outlet.

The stagnation temperature at inlet and outlet.

The stagnation pressure at inlet and outlet.

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Chapter 9.14, Problem 117P

Chapter 9.14, Problem 120P

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