1- Temperature distribution in a part of a heat exchanger in the steady state co
ID: 1767035 • Letter: 1
Question
1- Temperature distribution in a part of a heat exchanger in the steady state condition should be determined. The part is shown in Figure-1 has a thermal conductivity of k 25W/m.K and dimensions and boundary conditions are specified in Figure-1 and Figure-2 respectively. Develop a computer code to find the temperature distribution in the part (use 4yImm). Under operating conditions for which he-1000 W/m2K, T1700 K, h,-200 W/m2K, and T400 K. Also, at what location is the temperature a maximum? 5mm 2mm 3mm 2mm Figure-1. Dimensions Tnoho Figure-2. Boundary ConditionsExplanation / Answer
Ans: This problem is based on flat heat exhanger. We have to find dt/dx as well as position where the temperature is maximum.
Given T x=0 = 1700 K , Tx=L = 400K , K=25W/mK hi= 200 W/m2K.
There are two ways in order to find temperature distribution .
Method 1: Equation ofone dimensional steady state tranfer with heart generation .
d2t/dx2+q/K=0.....
We have to find q.
q=ma(T1-T2). For a unit mass and Area a=20mm2 so
q=20*1*(1700-400)= 26000W/mm3
Put the value in euqation we get
d2T/dx2+26000/25=0
d2T/dx2+1040=0
On integrating we get
dT/dx+1040x=C1
again integrating we get
T+1040 x2/2=C1x+C2
here C1 and C2 are two constants of integration that can be dtermined from boundary condition.
At x=0 , Tx=0=1700 K so, C2= 1700
At x=L Tx=l= 400 K so, x= 5mm
400+1040(5)2/2=5C1+1700
C1= 2340.
Substituing the values of C1 and C2 we get,
T+520 x2=2340 x +1700
T= 2340x +1700-520 x2
The condition for maximum tempetaure.
dT/dx =0
2340-1040 x = 0
X= 1300 or 1.3 mm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.