FTFS Chap05 P045 - Chapter 5 The First Law of...

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Chapter 5 The First Law of Thermodynamics Closed System Energy Analysis: Solids and Liquids 5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the brass balls are given to be ρ = 8522 kg/m 3 and C p = 0.385 kJ/kg. ° C. Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as EE E QU m u u Qm C T T in out out ball out −= = =− Net energy transfer by heat, work, and mass system Change in internal, kinetic, potential, etc. energies 12 43 4 4 Δ Δ () 21 Brass balls, 120 ° C Water bath, 5 ° C The total amount of heat transfer from a ball is mV D C T T out == = = = °− ° = ρρ ππ 3 6 8522 005 0558 0 558 0 385 120 74 9 88 (. . ( .) ( . ( ). kg /m m) 6 kg kg kJ /kg. C) C kJ /ball 3 3 Then the rate of heat transfer from the balls to the water becomes & & (( . ) Qn Q total × = ball ball balls/min) kJ /ball 100 9 88 988 kJ / min Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50 ° C since energy input must be equal to energy output for a system whose energy level remains constant. That is, E in out = = when system Δ 0 . 5-42
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Chapter 5 The First Law of Thermodynamics 5-46 A number of aluminum balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of aluminum at the average temperature of (120+74)/2 = 97 ° C = 370 K are ρ = 2700 kg/m 3 and C p = 0.937 kJ/kg. ° C (Table A-3). Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as EE E QU m u u Qm C T T in out out ball out −= = =− Net energy transfer by heat, work, and mass system Change in internal, kinetic, potential, etc. energies 12 43 4 4 Δ Δ () 21 Aluminum balls, 120 ° C Water bath, 50 ° C The total amount of heat transfer from a ball is kJ/ball 62 . 7 C ) 74 120 ( C) kJ/kg. 937 . 0 )( kg 1767 . 0 ( ) ( kg 1767 . 0 6 m) 05 . 0 ( ) kg/m 2700 ( 6 2 1 3 3 3 = ° ° = = = = = = T T mC Q D V m out π ρρ Then the rate of heat transfer from the balls to the water becomes kJ/min 762 = × = = ) kJ/ball 62 . 7 ( balls/min) 100 ( ball ball Q n Q total & & Therefore, heat must be removed from the water at a rate of 762 kJ/min in order to keep its temperature constant at 50 ° C since energy input must be equal to energy output for a system whose energy level remains constant. That is, E in out = = when system Δ 0 .
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FTFS Chap05 P045 - Chapter 5 The First Law of...

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