A steam pipe inner diameter 150 mm and outer diameter 160 mm...

A steam pipe (inner diameter = 150 mm and outer diameter = 160 mm) having thermal conductivity 58 W/m. °C is covered with two layers of insulation, of thickness 30 mm and 50 mm respectively and thermal conductivities 0.18 W/m. °C and 0.09 W/m. °C respectively. The temperature of the inner surface of the steam pipe is 320°C and that of the outer surface of the insulation layers is 40°C. Determine the quantity of heat lost per meter length of the steam pipe and the layer contact temperature.

Transcript

00:01 Here we are given t1, that is temperature 1 to be 320 degrees celsius, temperature 2 is equal to 40 degrees celsius, and here length is equal to 1 meter, that is per meter length.

00:18 And here, k of pipe, k means the thermal conductivity of pipe is given, which is 58 watt per meter degree celsius.

00:28 Thermal conductivity of the insulation 1 is given which is equal to 0 .18 watt per meter degree celsius.

00:39 Thermal conductivity of insulation 2 is given which is equal to 0 .09 watt per meter degree celsius.

00:50 Now internal radius r1 is equal to the internal diameter is given.

00:58 So internal diameter by 2 which is 150mm by 2 which becomes 75mm is the internal radius.

01:08 Now the outer radius or the external radius r2 will be equal to the outer diameter which is d2 by 2.

01:18 The outer diameter is given which is 160 m divided by 2 which becomes 80mm.

01:23 Now the thickness of insulation 1 that is t1 thickness of insulation 1 is t1 is equal to 30 millimeter so the outer radius of insulation 1 is represented as r3 will be equal to r2 which is the outer radius plus the thickness that is thickness of insulation 1 which is equal to outer radius is 80 mm plus the thickness of insulation 1 is 30 mm so 80 plus 30 becomes 110 mm will be the outer radius of insulation 1 now the thickness of insulation 2 is given which is represented as t 2 is equal to 50 mum if t 2 is equal to 50 mum then we can calculate calculate the outer radius of insulation 2.

02:30 Outer radius of insulation 2, r4 will be equal to r3, that is outer radius of insulation 1 plus the thickness of insulation 2.

02:41 Here r3 is 110 mm plus thickness 2 is 50 which becomes 160 m m is the outer radius of insulation 2.

02:53 Also here in this case we must know that here the three resistances are in series between the inner surface of pipe and the outer surface of insulation too or we can say that here t1 temperature first flows through the resistance of pipe which is in series with the resistance of of insulation 1 which is again in series with the resistance of insulation 2 and it flows out as temperature 2...

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