(a) Explanation Given: The inlet pressure of air ( P 1 ) is 280 kPa . The inlet temperature of air ( T 1 ) is 77 ° C . The inlet velocity of air ( V 1 ) is 50 m/s . The exit pressure of air ( P 2 ) is 85 kPa . The surrounding temperature of air ( T s u r r ) is 20 ° C . The exit velocity of air ( V 2 ) is 320 m/s . The heat lost to the surroundings ( q o u t ) is 3.2 kJ/kg . Calculation: Refer Table A-17, “Ideal gas properties of air”, obtain the following properties at the temperature of 350 K. s 1 = 1.85708 kJ / kg ⋅ K h 1 = 350.49 kJ / kg System is under steady state condition, thus rate of change of internal energy is zero. Δ E ˙ s y s t e m = 0 Write the expression for the energy balance equation of the system. E ˙ i n − E ˙ o u t = Δ E ˙ s y s t e m m ˙ ( h 1 + V 1 2 2 ) − ( m ˙ ( h 2 + V 2 2 2 ) + Q o u t ) = 0 h 2 = h 1 − q o u t − V 2 2 − V 1 2 2 (I) Here, rate of net energy transfer inside the system is E ˙ i n , rate of net energy transfer from outside to system is E ˙ o u t , and rate of change of internal energy in the system is Δ E ˙ s y s t e m . Substitute h 1 = 350.49 kJ / kg , q o u t = 3.2 kJ / kg , V 2 = 320 m / s , and V 1 = 50 m / s in Equation (I). h 2 = 350.49 kJ / kg − 3 (b) To determine The total entropy change of the system

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ISBN: 9780078027680

Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala

Publisher: McGraw-Hill Education

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Chapter 8, Problem 81P

(a)

To determine

The exit temperature of the air.

(b)

To determine

The total entropy change of the system

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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