00:01
Hello students in this question given us that consider a plane composite wall that is composed of two materials of thermal conductivity is k .a is equal to 0 .1 watt per meter and kb is equal to 0 .04 watt per meter.
00:15
So actually in this diagram given as two materials and the diagram of this problem looks like this, this type of diagram here.
00:25
And so this type of diagram here this is 40 degree actually so this is 40 degree so the direct diagram of this following looks like this here two metals first is a and second is b the thermal conductivity of a is k a is equal to 0 .1 watt per meter and k b is 0 .04 watt per meter okay and here also given us the distances like thickness not distance thickness thickness so the thickness of a plate a is l a equal to because given as 10 millimeter but we already know that 1 millimeter is equal to 10 to 10 to power minus 3 meter so we can write this as a 10 mitre 10 to power minus 3 and l b that means thickness of let we is also given us it is 70 million 20 multiplied by 10 to the power minus 3.
01:33
Okay.
01:35
And after that the question says that the contact resistance at the interface between the two metals is known as it's known to be 0 for 30 matrix.
01:47
So the contact resistance also given here and after that, meter square will be like kilo and after that given us the metal a ad joins a fluid at 200 degrees inches.
01:58
So here the temperature is 200 degrees celsius and after that given us that the material b adjoins a fluid at 4 degrees celsius.
02:08
Okay and here the heat h also is given for a the value of h is 10 watt per meter square according to position and for b the value of h is 20 watt per meter square.
02:22
Okay so here we like in this question given us two parts in first part to solve part a of this question and in part of this question the question says that find the rate of heat transfer through a wall that is 2 meter high by 2 .5 meter wide so here we need to find the heat transfer so in order to find the value of heat transfer first we need to calculate the thermal resistant due to convection at fluid adjoining a and fluid adjoining b so first we find thermal resistance are due to convention convention at fluid adjoining a is equal to it's equal to 1 divided by h a multiply by cross sectional area and the value of h a is also given in cuisn it is 10 watt for matrix square multiply by k so put the all value so we get right now we put all village 1 divided by this is 10 and multiply by a and a is the cross -section area and we already know age equal to length multiply width so the like given us that the width is 2 .5 meter and the length is 2 meter okay so the value of length is 2 meter multiplied the value of width is 2 .5 meter according to this so from here the value of thermal resistance due to convection at fluid adjoining a is 0 .02 kilo per watt okay now we need to find thermal resistance for at fluid adjoining b so thermal resistance due to convection at fluid adjoining b is equal to 1 divided by hb and the value of xb according to question is 40 multiply by a and the value of a is 2 multiply 2 .5 so when we solve this then we get 0.
04:29
Sorry this is not 40 this is according to this is 20 this is 20 because in question also given that b ad joining a fluid at 40 degrees celsius for which h is equal to 20 okay so the value of hv here is 20 so 20 so 20 milcholive to milk to level 2 .5 so this when we solve this then this becomes 0 .01 kilo for now we calculate the thermal conductivity in the walls so thermal conductivity thermal conductivity in wall a is equal to thermal conductivity is equal to l a divided by if we find the semiconductor in wall a then it's formalized thermal conduct within a equal to l a divided by k a multiply by a okay so now put the all value so what is the value of l and the value of l incision given us that l a is equal to 10 m here l a is the thickness of flat a okay so this is 10 millimeter that means 10 multiplied 10 to the power minus 3 meter divided by ka and the value of k is 0 .1 so this is value is 0 .1 multiplied by a and what is the value of a sorry the value of a is 0 to multiply by 2 .5 so it's 2 multiplied by 2 .5 so it's 2 multiplied 2.
05:57
So when we solve this then we get that thermal conductivity for a is 0 .02, 0 .02 kilo for watt.
06:17
Now we find that thermal conductivity for platte v and its alb divided by kb multiplied by a.
06:28
So the value of lb inclusion is 20 multiplied by 10 to the power minus 3 divided by and kv and the value of kb is 0 .04 multiply by a and the value of h2 multiply by 2 .5 so when we solve this then we get our conductivity for v is equal to 0 .1 kilowatt okay and now we calculate the rate of heat transfer because in this question our main motive is to find the rate of heat transfer so rate of heat transfer q is equal to rate of heat transfer q is equal to t a that means temperature of flat a minus temperature of plate b divided by thermal conductivity due to convection for a plus conductivity of flight a plus r in plus thermal conduct sorry thermal resistance due to convection at adjoining b due to convection at adjoining b at adjoining fluid b plus thermal conductivity for b and r in is internal resistance and this is also given in kitchen inkisson given us that like a in case and given us that the constant resistance at the interface between the two metals is known to be 0 .30 which is called kilo for watt.
08:23
So the value of r in is also the amount.
08:27
So now put all the values...