PS7_sol_S08 - Problem Set 7(2.006 Spring 08 Problem 1 a The...

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Problem Set 7 (2.006, Spring 08) Problem 1 a) The E-NTU single stream heat exchanger expression is used to calculate the temperature of the blood leaving the foot (T f,out ). () 63 3 min * min max min 25 10 1000 / 4000 / 1.667 / 60 1.5 / /0 , 0 . 9 0 1.667 / hb p foot m C C Vc kg m J kgK W K s UA WK CC C N T U CW K ρ ⎛⎞ × == = = ⎜⎟ ⎝⎠ = = = & Use 1e x p NTU ε =− , configuration #1 from Table 12.1 of the course note. Then h C = ,, min fi n fo u t TT C , , , 42 1 exp 0.90 42 ( 7) 12.9 o f out o o n i c e o f out CT C C TC −− = The heat lost to the ice per foot is 1.667 / 42 12.9 48.5 oo hf i n f o u t QCT T C C W = −= & b) In this case we have two heat exchangers. The first heat exchanger is the single stream heat exchanger that models the heat transfer from the duck’s foot to the ice. The second heat exchanger is a counter flow heat exchanger that models heat transfer between the arteries. To find a value for T c,out we first guess a value for the blood temperature leaving the duck’s foot (T f,out in the figure). We then use that temperature and the E-NTU expression for a counter flow concentric tube heat exchanger to calculate the blood temperature entering the duck’s foot, T f,in . Next, the calculation done in part (a) is repeated to calculate T f,out . This temperature is then compared to the guessed temperature. The desired result is obtained when the guessed temperature equals the calculated temperature exiting the duck’s foot. Once T f,out is converged, then calculate T c,out . T f,in =T h,out T f,out =T c,in T c,out T h,in= 42 o C T f, in T f, out
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The two arteries have matching heat capacities therefore the counter flow heat exchanger expression (Using L’hospital’s rule on the correlation #3 in Table 12.1 of the course note. Or the formula in the right column for the case C* =1) reduces to: () min ,, , , 0.073 / ,0 . 0 4 3 8 11 . 6 6 7 / 0.0438 0.0420 10 . 0 4 3 8 leg h in f in c out f out h in f out h in f out UA NTU W K NTU NTU C W K TT T T ε == = = + −− ⇒= = = = + The iterative process can be done in excel. For example, if you start with , 12.9 o f out TC = , you get , 40.78 o fi n = and , 12.43 o fo u t = . By repeating you will get , 12.42 o f out = (and , 40.76 o n = ). Then , , 0.0420 13.66 o c out f out h in f out T T C =+ = c) The heat loss from the duck’s foot to the ice under the condition of a finite heat transfer between the two arteries in the duck’s leg is 1.667 / 40.76 12.42 47.2 oo hf i n f o u t QCT T WK C C W =− = = & d) A geometric model for the two arteries in the duck’s leg is illustrated in the figure 2b. In calculating the value for (UA) leg we ignore any thermal resistance in the tissue that makes up the wall of the artery. This is not a good assumption since we expect that tissue to have a thermal conductivity close to that of water. However, we make this assumption here since the thickness of the walls is not given to us. The calculated (UA) leg will be an overestimation the actual value.
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PS7_sol_S08 - Problem Set 7(2.006 Spring 08 Problem 1 a The...

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