Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 5 (Q.No. 18)
18.
A composite flat wall of a furnace is made of two materials 'A' and 'B'. The thermal conductivity of 'A' is twice of that of material 'B', while the thickness of layer of 'A' is half that of B. If the temperature at the two sides of the wall are 400 and 1200°K, then the temperature drop (in °K) across the layer of material 'A' is
Discussion:
2 comments Page 1 of 1.
Khalid.abed said:
8 years ago
The resistances are in series and sum to R = R1 + R2. If TL is the temperature at the left, and TR is the temperature at the right, the heat transfer rate is given by;
q = (TL-TR)/R1+R2.
q = (1200 - 400)/(XA/KA+XB /KB),
q= 800/(0.5 XB /2KB + XB /KB),
q= 800/ (1.25 XB /KB)---> (1)
q = T at A/( XA/KA)---> (2)
Sub eq. (1) in eq. (2) then;
800/ (1.25 XB /KB) =T at A/( XA/KA),
T at A = 160.
q = (TL-TR)/R1+R2.
q = (1200 - 400)/(XA/KA+XB /KB),
q= 800/(0.5 XB /2KB + XB /KB),
q= 800/ (1.25 XB /KB)---> (1)
q = T at A/( XA/KA)---> (2)
Sub eq. (1) in eq. (2) then;
800/ (1.25 XB /KB) =T at A/( XA/KA),
T at A = 160.
(4)
M.Faisal said:
8 years ago
The concept of a thermal resistance circuit allows ready analysis of problems such as a composite slab (composite planar heat transfer surface). In the composite slab shown in Figure 2.5, the heat flux is constant with x.
The resistances are in series and sum to R = R1 + R2. If TL is the temperature at the left, and TR is the temperature at the right, the heat transfer rate is given by;
q = (TL-TR)/R1+R2
where R = L/KA SO PUTTING VALUES,
(1200-400)/(.5/1 * 1) + (1/0.5) = 160.
The resistances are in series and sum to R = R1 + R2. If TL is the temperature at the left, and TR is the temperature at the right, the heat transfer rate is given by;
q = (TL-TR)/R1+R2
where R = L/KA SO PUTTING VALUES,
(1200-400)/(.5/1 * 1) + (1/0.5) = 160.
(1)
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