sec1-1 - ChE 120B Heat Transfer Connection to...

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ChE 120B 1 - 1 Heat Transfer — Connection to Thermodynamics Thermos – How much heat is removed from 1 liter of water as it cools from 90 ° C to 80 ° C? This is a question that we can answer via thermodynamics. Energy conservation states that energy in – energy out = net change in energy of system if UU U −= Δ HH H Δ and from definitions of heat capacity vv fp p dU mC dT dU mC dT == v ff ii TT p UmC d T HmC d T Δ= ∫∫ which for constant v , p CC gives () ( ) v f ip f i Um CTT Hm Values of v , p depend on the material. For an ideal gas v p CCR =+ Most liquids and solids, under normal conditions, are incompressible, or have a constant density as a function of pressure. This means that constant volume and constant pressure changes are the same and v p = and v p UC d T HC d T = Often the heat capacity of a material changes with temperature; hence, we often need an average p C for calculations. From the 1 st Law, UQW Δ , and mainly we will be dealing with cases where 0 W = . However, back to the thermos. We know that eventually the hot drink in the thermos will cool off. Thermodynamics will not distinguish between a good thermos that takes a long time to cool off and a poor thermos. What is important is the heat transfer rate – not just Q , but Q ± to be labeled q or Q later today – somewhat confusing), the rate of heat/time being exchanged. 0 t QQ d t Δ = ± Once we know Qt ± we can calculate Q if Q ± is constant QQt = Δ ±
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ChE 120B 1 - 2 The rate of heat transfer per unit area normal to the direction of the heat transfer is heat flux 2 (W/m ) Q q A = ± ± in which A is the heat transfer area. In English, watts 2 is Btu/hr ft . q ± The heat flux can vary with time, temperature, and position. Example 1: A 10 cm diameter copper ball is to be heated from 100 ° C to 150 ° C in 30 minutes. C u ρ = 8950 kg/m 3 and p C = 0.395 kJ/kg ° C. What is: a) the total amount of heat transfer to the ball, b) the average heat transfer rate, and c) the average heat flux at the ball surface? a) The total energy requirement is from thermodynamics ( ) pf i UH Q m C T T Δ≅ Δ== () 3 4 VV 3 p f i CT T C rT T ρρ π = −= ( ) 3 33 4 v= 8950 .1m 4.7kg 3 66 mr D ππ ⎛⎞ == = = ⎜⎟ ⎝⎠ k J g k 4.7 kg 0.395 C 150 100 C 92.6 kJ Q ° = b) While the average rate of heat transfer is ave kk 92.6 J J s 1800 s .0514 51.4 Q QW t = = Δ ± c) The heat flux is defined as the heat transfer/time area 2 ave ave ave 2 2 51.4 1640 / m .1 QQ W qW AD = = ±± These are average values of Q ± and q are likely not to be equal to the actual instantaneous value Q ± or q during the process. Example 2: Heat loss from ducts in basement A 5 m long section of an air heating system of a house passes through an unheated basement. The duct is 20 x 25 cm. Hot air enters the duct at 200 kPa and 60 ° C at an average velocity of 5 m/s. The temperature of the air drops to 54 ° C through the loss of heat to the basement. What is the rate of heat loss?
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sec1-1 - ChE 120B Heat Transfer Connection to...

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