(a) To determine The final temperature in the tank. Explanation Write the expression for energy balance for air entering in the tank. m 2 u 2 − m 1 u 1 = h 1 ( m 2 − m 1 ) (I) Here, initial mass air is m 1 , the initial internal energy of air is u 1 , final internal energy of air is u 2 , final mass of air is m 2 and initial enthalpy of air is h 1 . There is no mass of air present in the rigid tank because initially tank is evacuated. Substitute 0 g for m 1 in Equation (I). m 2 u 2 − ( 0 g ) u 1 = h 1 ( m 2 − 0 g ) m 2 u 2 = m 2 h 1 u 2 = h 1 (II) Conclusion: Convert the surrounding temperature T from Centigrade to Kelvin. T = 25 ° C = ( 25 ° C + 273 .15 ) K = 298.15 K Refer to Table A-22 “Ideal Gas Properties of Air”, at surrounding temperature 298.15 K , to obtain enthalpy ( h 1 x ) as 295.17 kJ/kg corresponding to temperature ( T 1 ) as 295 K , enthalpy ( h 2 x ) as 300.19 kJ/kg corresponding to temperature ( T 2 ) as 300 K . Write the expression for interpolation to obtain initial enthalpy. h 1 = ( T − T 1 T 2 − T 1 ) ( h 2 x − h 1 x ) + h 1 x (III) Substitute 298.15 K for T , 295 K for T 1 , 300 K for T 2 , 300.19 kJ/kg for h 2 x and 295.17 kJ/kg for h 1 x in Equation (III). h 1 = ( 298.15 K − 295 K 300 K − 295 K ) ( 300 (b) To determine The amount of air leaks into the tank. (c) To determine The amount of entropy produced.

8th Edition

ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 6.13, Problem 113P

(a)

To determine

The final temperature in the tank.

(b)

To determine

The amount of air leaks into the tank.

(c)

To determine

The amount of entropy produced.

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Chapter 6.13, Problem 112P

Chapter 6.13, Problem 114P

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