(a) Explanation Write the expression for the specific enthalpy at state 1. h 1 = h a + V a 2 2 (I) Here, the specific enthalpy at state 1 is h 1 , the specific volume at state a is V a and the specific enthalpy at state a is h a . Write the expression for the pressure at state 1. p 1 = ( p r 1 p r a ) p a (II) Here, the pressure at state 1 is p 1 , the pressure at state a is p a , the reduced pressure at state 1 is p r 1 and the reduced pressure at state a is p r a . Write the expression for the saturated pressure at state 2. p r 2 s = ( p 2 p 1 ) p r 1 (III) Here, the pressure at state 2 is p 2 and the saturated pressure at state a is p r 2 s . Write the expression for the specific enthalpy in terms of compressor efficiency at state 2. h 2 = h 1 + ( h 2 s − h 1 η c ) (IV) Here, the compressor efficiency is η c , the specific enthalpy at state 2 is h 2 and the saturated specific enthalpy at state 2 is h 2 s . Write the expression for the saturated reduced pressure at state 4. p r 4 s = ( p 4 p 3 ) p r 3 (V) Here, the pressure at state 4 is p 4 , the reduced pressure at state 4 is p r 4 s and the reduced pressure at state 3 is p r 3 . Write the expression for the specific enthalpy at state 4. h 4 = h 3 − η t ( h 3 − h 4 s ) (VI) Here, the specific enthalpy at state 4 is h 4 and the specific enthalpy at state is h 3 . Write the expression for the pressure at state 4. p 4 = ( p r 4 s p r 3 ) p 3 (VII) Here, the pressure at state 4 is p 4 , the pressure at state 3 is p 3 , the reduced pressure at state 3 is p r 3 and the saturated reduced pressure at state 4 is p r 4 s . Write the expression for the reduced pressure at state 5. p r 5 = ( p 5 p 4 ) p r 4 (VIII) Here, the pressure at state 5 is p 5 , the reduced pressure at state 5 is p r 5 and the reduced pressure at state 4 is p r 4 . Write the expression for the mass flow rate of air. m ˙ = ( A V ) a p a R T a (IX) Here, the area is A , the volume of the air is V , the rate of the mass flow is m ˙ , the ambient temperature is T a and the universal gas constant is R . Write the expression for the power delivered to propeller. W ˙ p = m ˙ [ ( h 3 − h 4 ) − ( h 2 − h 1 ) ] (X) Here, the power delivered to propeller is W ˙ p . Conclusion: Refer to table A-22E “ideal gas properties of air” to obtain the specific enthalpy and reduced pressure at temperature T a = 460 ° R as 109.90 Btu / lb and 0.7913 . Substitute 109.90 Btu / lb for h a , 520 ft / s for V a in Equation (I). h 1 = ( 109.90 Btu / lb ) + ( 520 ft / s ) 2 2 = ( 109.90 Btu / lb ) + 135200 ( ft / s ) 2 ( 1 lbf 32.2 lb ⋅ ft / s 2 ) ( 1 Btu 778 ft ⋅ lbf ) = 115.296 Btu / lb ≈ 115.30 Btu / lb Refer to table A-22E “ideal gas properties of air” to obtain the reduced pressure at specific enthalpy h 1 = 115.30 Btu / lb as 0.93604 . Substitute 0.93604 for p r 1 , 0.7913 for p r a and 12 lbf / in 2 for p a in Equation (II). p 1 = ( 0.93604 0.7913 ) ( 12 lbf / in 2 ) = 1.1829 ( 12 lbf / in 2 ) ≈ 14.19 lbf / in 2 Substitute 9 for p 2 p 1 and 0.93604 for p r 1 in Equation (III). p r 2 s = ( 0.93604 ) ( 9 ) = 8.4243 Refer to table A-22E “ideal gas properties of air” to obtain the saturated specific enthalpy at reduced pressure p r 2 s = 8.4243 as 216.36 Btu / lb . Substitute 216.36 Btu / lb for h 2 s , 115.30 Btu / lb for h 1 and 0.87 for η c in Equation (IV). h 2 = 115.30 Btu / lb + ( 216.36 Btu / lb − 115.30 Btu / lb 0.87 ) = 115.30 Btu / lb + 116 (b) To determine The velocity at nozzle exit.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 9.14, Problem 93P

(a)

To determine

The power delivered to the propeller.

(b)

To determine

The velocity at nozzle exit.

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Chapter 9.14, Problem 92P

Chapter 9.14, Problem 94P

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Fundamentals of Engineering Thermodynamics

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