One wall of a house consists of 0019 m thick plywood backed...

One wall of a house consists of 0.019 m thick plywood backed by 0.076 m thick insulation. The temperature at the inside surface is 25 °C, while the temperature at the outside surface is 4 °C. The thermal conductivities of the insulation and plywood are 0.03 J/(s·m·°C) and 0.08 J/(s·m·°C) respectively. The area of the wall is 35 m^2. What is the temperature at the insulation-plywood interface? (4) At what rate is heat conducted through the wall? (4) By what factor would you expect the cost of heating this home to increase if the insulation were removed? Justify your response. (4)

Transcript

00:01 Hi everyone! so the question is saying that the temperature inside a bowl, a one wall open house consists of 0 .401 and meter thick and the temperature inside the bowl that is equal to 25 degree and downside is 4 degree and the thermal conductivity of insulation that is equal to 0 .0 .0.

00:21 Jule and the plywood is 0 .05 jule.

00:25 Area of the bowl is 35 meter square and the thickness of the plywood and there is a which also given then what we have to find out first of all we have to find out what is the temperature at the insulation flyboard interface okay so the rate of flow of heat as we know the rate of heat flow from insulation hd1i that is theta 1 okay and this is equal to k1, theta 2 minus theta, multiply by 825.

01:09 Okay, so this is the rate of insolidation.

01:13 So, as you know here, the rate of heat flow through the flyboard.

01:21 The rate of heat flow, the rate of heat flow, so flyboard will be equal to as theta to row to and will be equal to k2 multiply by theta minus theta 1 verified 3 2 into 8.

01:49 Okay, so as we know here, all at the steady rate, at the steady rate, what will happen here? that the value of theta 1 node will be equal to theta 2 dot.

02:08 Okay, so that means the value of k1, theta 1, theta 2 minus theta, multiplied by a, divided d, d.

02:23 Okay, and this is equal to k2, theta minus theta 1, multiplied by a, 22.

02:35 Okay, this will be d1 and this will be d2.

02:38 So by softening we got the value of theta from here, i got the value of theta from here, and the value theta will be equal to k1, t2, theta 2 plus k2 t1, theta 1, d2, d2, d2, d2 plus k2 t1.

03:02 Okay, this will be the formula of 4 theta here.

03:06 So theta will be equal to k1, k1, k1.

03:09 0 .03 multiplied by d to 8 0 .019 multiplied by 25 plus 0 .08 multiplied by 0 .076 multiplied by 4.

03:26 And you added this time by a1 which equal to 0 .03 multiplied by d is equal to 0 .019 plus 0 .08 and multiply by zero point zero seven six so by solving this you wrote that this is equal to 5 .8 degree calorie okay so this is the method of here so the temperature of the temperature of insulation live here will be here phase is 5 .8 degrees celsius so by this we go that the value of a theta -1 -teta -1 -teta -2 minus theta into a 30 -1.

04:27 So this will be equal to 0 .03 into theta 2 is 25 minus 5 .8, 225 multiplied by lh equal to 35.

04:41 5251 -degree is equal to 0 .0076.

04:45 So by so being to all of this value, we go to that the value of theta -1 -4 is equal to 265 .264.

04:54 Okay, so this is the rate at which heat is conducted through the form...

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