A spherical vessel containing hot fluid at 160c in a...

A spherical vessel containing hot fluid at 160°C (in a chemical process) is of 0.4 m OD and is made of Titanium of 25 mm thickness. The thermal conductivity is 20 W/mK. The vessel is insulated with two layers of 5 cm thick insulations of thermal conductivities 0.06 and 0.12 W/mK. There is a contact resistance of 6 × 10–4 and 5 × 10–4 m2 °C/W between the metal and first insulation and between the insulating layers. The outside is exposed to surrounding at 30°C with a convection coefficient of 15 W/m2-K. Determine the rate of heat loss, the interface temperatures, and the overall heat transfer coefficient based on the metal surface area.

Transcript

00:01 Here in this question we are given the inside fluid temperature which is equal to 160 degree celsius.

00:11 The outer diameter of vessel is given which is equal to 0 .4 meter.

00:21 The thickness of metal is given which is equal to 25mm the thermal conductivity is given which is equal to 25mm the thermal conductivity is given which is equal to 20 watts per meter kelvin the thickness of insulation is given which is equal to 5 centimeter contact resistances are given there are two contact resistance first 1 r1 is equal to 6 into 10 power minus 4 meter square per sorry meter square degree celsius per watt and r2 2 is equal to 5 into 10 power minus 4 meter square degree celsius per watt.

01:09 Also the outside temperature is given which is equal to 30 degrees celsius and convection coefficient is given which is equal to 15 watts per meter square kelvin.

01:25 Here first we need to calculate total thermal thermal.

01:30 Resistance.

01:31 So here the total thermal resistance is given by capital r is equal to r1 plus r2 plus r3 plus r c1 which is the first contact resistance plus rc2 here this is actually r c1 and this is actually r c2 since they are contact resistances plus r convection.

02:00 So this is equal to instead of r1, the equation is r2 minus r1 divided by 4 pi k into r1 r2 plus instead of r2 we can write a r3 minus r2 divided by 4 pi k into r3 into r2 plus instead of r3 we can write a r4 minus r3 divided by 4 pi k into r3.

02:32 Plus contact resistance 1 plus contact resistance 2 plus instead of r convection we can write 1 divided by h into so this is equal to r2 is given in the question which is 0 .2 minus r1 is 0 .175 divided by 4 into 5 into value of k is 20 into 0 .2 into 0 .175.

03:04 Plus r3 is given as 0 .25 minus 0 .2 divided by 4 pi into 0 .06 into 0 .25 into 0 .2 plus 0 .3 minus 0 .25 divided by 4 pi into 0 .12 into 0 .12 into 0 .3 minus 0 .25 divided by 4 pi into 0 .12 into 0 .10.

03:33 0 .3 into 0 .0 .5.

03:36 0 .25 plus contact resistance 1 which is 6 into 10 power minus 4 plus 5 into 10 power minus 4 plus 1 divided by 15 into 4 pi into 0 .3 square.

04:00 4 pi r square is the area.

04:02 This is the area.

04:05 Upon solving we will get a total thermal resistance.

04:09 To be 1 .83128.

04:13 So this is the value of total thermal resistance.

04:17 Now we need to calculate total heat transfer which is represented as q is equal to delta t divided by r.

04:32 Here the temperatures given are 160 degrees celsius minus 30 degrees celsius divided by total thermal resistance which is 1 .83128.

04:44 Upon solving we will get total heat transfer to be 70 .98 watts...