Thermodynamics HW Solutions 73

Thermodynamics HW Solutions 73 - 6 977 3 3 tank = = ρ = V...

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Chapter 1 Basics of Heat Transfer 1-136 Somebody takes a shower using a mixture of hot and cold water. The mass flow rate of hot water and the average temperature of mixed water are to be determined. Assumptions The hot water temperature changes from 80 ° C at the beginning of shower to 60 ° C at the end of shower. We use an average value of 70 ° C for the temperature of hot water exiting the tank. Properties The properties of liquid water are C p = 4.18 kJ/kg. ° C and ρ = 977.6 kg/m 3 (Table A-2). Analysis We take the water tank as the system. The energy balance for this system can be expressed as [] ) ( ) ( 1 2 tank out in hot in e, sys out in T T C m t T T C m W E E E = Δ + Δ = where T out is the average temperature of hot water leaving the tank: (80+70)/2=70 ° C and kg 656 . 58 ) m 06 . 0 )( kg/m
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Unformatted text preview: 6 . 977 ( 3 3 tank = = ρ = V m Substituting, [ ] kg/s 0.0565 = ° ° × ° ° + hot hot C 80)-C)(60 kJ/kg. kg)(4.18 (58.656 = s) 60 (8 C 70)-C)(20 kJ/kg. (4.18 kJ/s 6 . 1 m m & & To determine the average temperature of the mixture, an energy balance on the mixing section can be expressed as C 44.2 ° = ° + = ° ° + ° ° + = + = mixture mixture mixture cold hot cold cold hot hot out in C) kJ/kg. kg/s)(4.18 0.06 (0.0565 C) C)(20 kJ/kg. kg/s)(4.18 (0.06 C) C)(70 kJ/kg. kg/s)(4.18 (0.0565 ) ( T T CT m m CT m CT m E E & & & & & & T mixture =? T in = 20 ° C m cold = m hot T cold = 20 ° C m cold =0.06 kg/s T 1 = 80 ° C T 2 = 60 ° C T out = 70 ° C m hot = ? W e =1.6 kW 1-73...
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