Question 4 consider one dimensional conduction in a 3...

QUESTION 4 Consider one-dimensional conduction in a 3 layered plane composite wall. The outer surface of layer A is exposed to a air at T-1= 25°C and inner surface of layer C is exposed to a fluid at T-2 = 250°C. The middle layer B experiences uniform heat generation, while there is no generation in layers A and C. The temperatures at the interfaces A-B and B-C are T1= 46 °C and T2=134 °C, respectively. Calculate the heat generation of layer B, B in W/m3. (kA= 20 W/K, kC= 50 W/K, LA= 30 mm, LB= 42 mm, LC= 30 mm, h1= 15 W/m2-K, h2= 760 W/m2-K ) Please report your answer with one decimal figure.

Transcript

00:01 So, here in this question we are considering the composite function which we are given here.

00:05 So, let's say we are considering this are the walls a, this is the wall b, this is the c and this is the wall d from here.

00:12 So, this is what we are considering from here.

00:15 This is given that is 1 centimeter, this from here is 8 centimeter and this from here is 10 centimeter.

00:21 So, this is the distance which we are given here.

00:23 We are having the value of dt infinity that is at 40 degree celsius, h is equals to 25 weber per meter square kelvin, where the value of t1 at the surface is equals to 25 degree celsius.

00:35 And here we are having the value of t2 that is equals to 10 degree celsius, where we are having the value of ka that is equals to 0 .5 weber per meter kelvin.

00:45 In the same way, the value of kb is equals to 0 .03 weber per meter kelvin.

00:50 The value of kc from here is equals to 50 weber per meter degree celsius and that of ab is equal to 1 meter square and that of ac is equal to 0 .3 meter square where the temperature t2 is of the surface given in the first part we have to find out the value of the heat transfer rate.

01:10 So heat transfer rate is given as a naught that is equal to a of b plus a of c that from here is equal to 1 plus 0 .3 that from here is equal to 1 .3 that from here is equal to so q from here is equal to h of a which is multiplied by the delta of t.

01:28 So plugging into the value that is 25 multiplied by the 1 .3 that is further multiplied by the t of environment minus t of the wall a.

01:36 So this from here is equal to 25 multiplied by the 1 .3 that is further multiplied by the 40 minus 25.

01:43 Simplifying the term we get the value of q that is equal to 487 .5 watt.

01:47 Hence the answer to the part a of the question.

01:49 Now in the part b of the question we are considering about the thermal resistance of the wall a.

01:54 So this is given as rth that is equal to l which is divided by k of a.

01:58 Plugging into the value that is 0 .01 divided by 0 .5 which is further multiplied by the 1 .3.

02:04 So simplifying the term from here so we get the value of rth 48 that is equal to 0 .0153 kelvin per watt...

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