To determine The heat transfer in the steady state. Explanation Write the expression for the specific water vapor. v 1 = V m v 1 (I) Here, volume of the tank is V and mass of the water vapor is m v 1 in the state-1. Write the expression for the quality of the saturated water vapor in the state-2. x 2 = v 1 − v f v g − v f (II) Here, specific volume of the saturate water is v f and specific volume of the saturate water vapor is v g Write the expression for the mass of the water vapor in the state-2. m v 2 = x 2 m v 1 (III) Write the expression for the heat transfer. Q = [ m C O 2 ( u ¯ C O 2 ( T 2 ) − u ¯ C O 2 ( T 1 ) ) M C O 2 + ( m v 2 u g ( T 2 ) − m v 1 u g ( T 1 ) + m v 2 u f ) ] (IV) Here, molar specific internal energy of the C O 2 at state-1 is u ¯ C O 2 ( T 1 ) , molar specific internal energy of the C O 2 at state-2 is u ¯ C O 2 ( T 2 ) , specific internal energy of the saturated water vapor at state-2 is u g ( T 2 ) , specific internal energy of the saturated water vapor at state-1 is u g ( T 1 ) , and specific internal energy of the saturated liquid water is u f . Conclusion: Substitute 1 m 3 for V and 0.05 kg for m v in the Equation (I). v = 1 m 3 0.05 kg = 20 m 3 / kg Refer to TableA-2 “Properties of Saturated Water (Liquid–Vapor)” to obtain the v g as 57.791 m 3 / kg at temperature 20 ° C and v f as 0.0010018 m 3 / kg at temperature 20 ° C . Substitute 20 m 3 / kg for v 1 , 0.0010018 m 3 / kg for v f , 57.791 m 3 / kg for v g in the Equation (II). x 2 = 20 m 3 / kg − 0.0010018 m 3 / kg 57.791 m 3 / kg − 0.0010018 m 3 / kg = 19.9989 m 3 / kg 57.7899 m 3 / kg = 0.3461 Substitute 0.3461 for x 2 and 0.055 kg for m v 1 in Equation (III) m v 2 = 0.3461 ( 0.055 kg ) = 0.01731 kg Refer to Table A-2 “Properties of Saturated Water (Liquid–Vapor)” to obtain u g ( T 1 ) as 2475.9 kJ / kg at temperature 20 ° C , u g ( T 2 ) as 2402.9 kJ / kg at temperature 75 ° C , u f as 83.95 kJ / kg at temperature 20 ° C Refer to Table A-23 “Ideal Gas Properties of Selected Gases” to obtain u ¯ C O 2 ( T 1 ) at temperature 20 ° C using interpolation method of two variables. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (V) Show the temperature at 290 K and 298 K as in Table (1)

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

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

Publisher: WILEY

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Chapter 12.9, Problem 64P

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aversity of Baoyion aculty of Engineering-AIMusyab Automobile Eng. Dep. Year: 2022-2023, st Course, 1st Attempt Stage: 3rd Subject: Heat Transfer I Date: 2023\01\23- Monday Time: 3 Hours Q4: A thick slab of copper initially at a uniform temperature of 20°C is suddenly exposed to radiation at one surface such that the net heat flux is maintained at a constant value of 3×105 W/m². Using the explicit finite-difference techniques with a space increment of Ax = = 75 mm, determine the temperature at the irradiated surface and at an interior point that is 150 mm from the surface after 2 min have elapsed. Q5: (12.5 M) A) A steel bar 2.5 cm square and 7.5 cm long is initially at a temperature of 250°C. It is immersed in a tank of oil maintained at 30°C. The heat-transfer coefficient is 570 W/m². C. Calculate the temperature in the center of the bar after 3 min. B) Air at 90°C and atmospheric pressure flows over a horizontal flat plate at 60 m/s. The plate is 60 cm square and is maintained at a…

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