Consider a 5 m high, 8 m long (into the page) and 0.22 m thick wall whose repres

ID: 2991888 • Letter: C

Question

Consider a 5 m high, 8 m long (into the page) and 0.22 m thick wall whose representative cross-section (12 cm high and repeated vertically like this over the whole height of the wall) is as given in the diagram below. The thermal conductivities of the different materials used in the wall construction (in W/mK) are kA = kF = 2, kC = 8, kD = 20, kD = 15 and kE = 35. The left and right surfaces of the wall are exposed to fluids with conditions 300 degree C, h = 100 degree W/m2K and 100 degree C, h = 500 W/m2K, respectively. Assuming the heat transfer to be one-dimensional determine the rate of heat transfer through the entire wall, q, and the temperature between sections CBC and DE. [Answers: 147.7 kW, 233 degree C]

Explanation / Answer

R = L/KA

Ra = .01 / 2 x (.12 x 8) = .00521

Rb = .05 / 8 x (.04 x 8) = .01953

Rc = .05 / 20 x (.04 x 8) = .0078125

Rd = .1 / 15 x (.06 x 8) = .013889

Re = .1 / 35 x (.06 x 8) = .005952

Rf = .06 / 2 x (.12 x 8) = .03125

C - B - C are in parallel

1 / R3 = 1/Rc + 1/Rb + 1/Rc = 307.2

R3 = .003255

D - E are in parallel

1 / R4 = 1 / Rd + 1/ Re = 240

R4 = .004166

Rconv1 = 1 / hA = 1 / 100 x (.12 x 8) = .0104

Rconv2 = 1 / 500 (.12 x 8) = .00208

Rtotal = Rconv1 + Ra +R3 +R4 +Rf +Rconv2 = .05636

Q = (T1 - T2) / Rtotal = (300 - 100) / .05636 = 3548.55 W

This is transferred through 1 block of 12 cm height

So total Qt = 5 / .12 x 3548.55 = 147 . 8 kW

[Here we are not using parallel rule because we have already calculated the heat through one section, it will directly add over entire height]

b.Let T3 be the temp of the reqd section

Resistance upto reqd section,R' = Rconv1 + Ra +R3 = .018865

This is the Restance through one section of 12cm wall.

R'total (over the height 5 m )

Using parallel rule

1 / R'total = 1 / R(one section) + 1 / R(one section) +...n times,

[where n = 5m / .12 m = 41.66 ]

1 / R'total = n / R(one section)

R'total = R(one section ) / 41.66 = 0.0004526

(T1 - T3) = Qt x R'total = 147.7 x 10^3 x .0004526 = 66.94

T3 = 233.06 C

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