5 a silver ingot 25 mm x 40 mm x 150 mm with a surface...

5. A silver ingot (25 mm x 40 mm x 150 mm) with a surface temperature of 40°C is cooling in air at 20°C. If the ingot is sitting on an insulating surface that completely stops heat flow through one of the 40 mm x 150 mm faces and the heat transfer coefficient between the ingot and the air is 18 W/m²K, find the rate at which the ingot is cooling. Assume the temperature within the ingot is uniform.

Transcript

00:01 Hi, in the question it is given that we have thermal, we have thermal conductivity, we have thermal conductivity that is k is equal to 180 watt per meter kelvin and we have density, density of plate, density of plate that is rho is equal to 2800 kg upon meter cube and we have heat capacity that is cp is equal to 880 joule upon kg kelvin and we have thickness, thickness of plate is equal to 2 centimeter and we have initial temperature, we have initial temperature that is ti is equal to 155 degree celsius and we have velocity of air is equal to, velocity of air that is v is equal to 20 meter upon second and we have temperature, temperature of air that is t1 is equal to 10 degree celsius and we have temperature, temperature of plate t2 is equal to 45 degree celsius.

02:03 So, properties, properties of air that is we have p is equal to 1 .52 multiplied by 10 to the power minus 5 meter square upon second.

02:21 So, since we know that re is equal to vl upon p, so this will be further equal to 20 multiplied by 1 multiplied by 10 to the power 5 upon 1 .52, so this is equal to 13 .57 multiplied by 10 to the power 5.

02:42 So, since ra is greater than 5 multiplied by 10 to the power 5, so nu will be equal to 0 .037 rel to the power 0 .8 minus 871 multiplied by pr to the power 1 by 3 and this is the mixed, this is the mixed condition.

03:18 So, we will have pr is equal to 0 .7336 and we will have nu will be equal to 0 .037 multiplied by 13 .57 multiplied by 10 to the power 5 to the whole power 0 .8 minus 871 multiplied by 0 .7336 to the whole power 1 by 3.

03:54 So, nu will be equal to 1902 .58 and this further equal to 2hl upon k.

04:05 Since we have k is equal to 0 .02439 watt per meter kelvin and we will have h is equal to 46 .40 watt per meter square kelvin.

04:23 Since bi is equal to hlc upon k, so bi is equal to 46 .40 multiplied by 0 .02 upon 180 multiplied by 2, so this is equal to 0 .002578.

04:46 So, this is less than, this is less than 0 .01, then lambda is valid since t of t minus tp upon ti minus tp is equal to e to the power minus hast upon svc...

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