Explanation The diffuser with entry point 1 and exit point 2 is shown below. Figure-(1) Write the expression for the density at inlet of the diffuser. ρ 1 = P 1 R T 1 (I) Here, the density of the nitrogen at inlet is ρ 1 , the inlet pressure is P 1 , the gas constant for the nitrogen is R and the inlet temperature is T 1 . Write the expression for the density at outlet of the diffuser. ρ 2 = P 2 R T 2 (II) Here, the density of the nitrogen at outlet is ρ 2 , the outlet pressure is P 2 , the gas constant for the nitrogen is R and the outlet temperature is T 2 . Write the expression for the specific heat of the nitrogen. c p = k R k − 1 (III) Here, polytropic index is k and the gas constant is R . Write the expression for the continuity equation for the nitrogen through diffuser. ρ 1 A 1 V 1 = ρ 2 A 2 V 2 (IV) Here, the density of the nitrogen at inlet and outlet is ρ 1 and ρ 2 respectively, the area at the inlet and outlet of the diffuser is A 1 and A 2 respectively, velocity of the nitrogen at inlet and outlet is V 1 and V 2 respectively. Write the expression for the steady flow energy equation for the flow of the nitrogen in the diffuser with no heat and work transfer and potential energy effects are negligible. c p ( T 1 − T 2 ) + ( V 1 2 − V 2 2 2000 ) = 0 (V) Here, the specific heat of the nitrogen is c p , the temperature of the nitrogen at inlet and outlet of the diffuser is T 1 and T 2 , and the velocity of the nitrogen at inlet and outlet of the diffuser is V 1 and V 2 respectively. Write the expression for the mass flow rate. m ˙ = ρ 1 A 1 V 1 (VI) Here, the mass flow rate of the nitrogen is m ˙ Write the expression for entropy rate balance for nitrogen. σ ˙ = m ˙ [ c p ln T 2 T 1 − R ln P 2 P 1 ] (II) Here, the entropy generation is σ ˙ , the mass flow rate of the nitrogen in the diffuser is m ˙ , , the specific heat of the nitrogen is c p , the gas constant is R , the inlet and outlet pressure of the nitrogen is P 1 and P 2 respectively and the inlet and outlet temperature of the nitrogen is T 1 and T 2 respectively. Conclusion: Substitute 1.4 for k , 0.296 kJ / kg K for R in Equation (III). c p = 1.4 × ( 0.296 kJ / kg ⋅ K ) 1.4 − 1 = ( 0.4144 0.4 ) kJ / kg ⋅ K = 1.036 kJ / kg ⋅ K Substitute 1.036 kJ / kg K for c p 300 K for T 1 , 282 m / s for V 1 and 130 m / s for V 2 . ( 1.036 kJ / kg ⋅ K ) ( 300 K − T 2 ) + ( ( 282 m / s ) 2 − ( 130 m / s ) 2 2000 ) = 0 T 2 = ( 310.8 + 31.312 1.036 ) K T 2 = 330.22 K Thus, the exit temperature is 330 .22 K . Substitute 0.656 bar for P 1 , 0.296 kJ / kg K for R and 300 K for T 1 , 282 m / s for V 1 and 130 m / s for V 2 , 4

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 6.13, Problem 104P

To determine

The exit temperature in K.

The exit area in m2.

The rate of entropy production in kJ/kg K.

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Chapter 6.13, Problem 103P

Chapter 6.13, Problem 105P

Chapter 6 Solutions

Fundamentals of Engineering Thermodynamics

Ch. 6.13 - Prob. 1E Ch. 6.13 - Prob. 2E Ch. 6.13 - Prob. 3E Ch. 6.13 - Prob. 4E Ch. 6.13 - Prob. 5E Ch. 6.13 - 6. Is entropy produced within a system undergoing...Ch. 6.13 - 7. When a mixture of olive oil and vinegar...Ch. 6.13 - Prob. 8E Ch. 6.13 - Prob. 9E Ch. 6.13 - 10. Is Eq. 6.51a restricted to adiabatic processes...

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