Thermodynamics HW Solutions 4

Thermodynamics HW Solutions 4 - Chapter 1 Basics of Heat...

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Unformatted text preview: Chapter 1 Basics of Heat Transfer 1-16 A 15 cm × 20 cm circuit board houses 120 closely spaced 0.12 W logic chips. The amount of heat dissipated in 10 h and the heat flux on the surface of the circuit board are to be determined. Assumptions 1 Heat transfer from the back surface of the board is negligible. 2 Heat transfer from the front surface is uniform. Analysis (a) The amount of heat this circuit board dissipates during a 10-h period is & Q = (120)(0.12 W) = 14.4 W Chips, 0.12 W & Q = QΔt = (0.0144 kW)(10 h) = 0.144 kWh & Q (b) The heat flux on the surface of the circuit board is As = (0.15 m )(0.2 m ) = 0.03 m 2 15 cm & Q 14.4 W & = = 480 W/m 2 qs = 2 As 0.03 m 20 cm 1-17 An aluminum ball is to be heated from 80°C to 200°C. The amount of heat that needs to be transferred to the aluminum ball is to be determined. Assumptions The properties of the aluminum ball are constant. Properties The average density and specific heat of aluminum are given to be ρ = 2,700 kg/m3 and C p = 0.90 kJ/kg.°C. Metal ball Analysis The amount of energy added to the ball is simply the change in its internal energy, and is determined from Etransfer = ΔU = mC (T2 − T1) where E m = ρV = π 6 ρD3 = π 6 (2700 kg / m3 )(015 m)3 = 4.77 kg . Substituting, Etransfer = (4.77 kg)(0.90 kJ / kg. ° C)(200 - 80)° C = 515 kJ Therefore, 515 kJ of energy (heat or work such as electrical energy) needs to be transferred to the aluminum ball to heat it to 200°C. 1-18 The body temperature of a man rises from 37°C to 39°C during strenuous exercise. The resulting increase in the thermal energy content of the body is to be determined. Assumptions The body temperature changes uniformly. Properties The average specific heat of the human body is given to be 3.6 kJ/kg.°C. Analysis The change in the sensible internal energy content of the body as a result of the body temperature rising 2°C during strenuous exercise is ΔU = mCΔT = (70 kg)(3.6 kJ/kg.°C)(2°C) = 504 kJ 1-4 ...
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