00:01
For the given problem, the diameter of the tube is 4 cm.
00:05
4 cm is the diameter of the tube that is also given as 0 .04 in meters.
00:11
As well as the length of the tube is given as 14 in meters.
00:16
In the problem it has been told that the temperature of water, that the inlet temperature of water, that is inlet temperature of water.
00:27
This is considered as t1 is given as 20 degrees.
00:30
Celsius the mass flow rate is given as 0 .8 kilogram per second as well as the steam temperature that is t of steam is given as 165 degrees celsius so we can say that as steam temperature is given as 165 degrees celsius it will be condensing over the tubes therefore if we assume that the surface temperature that a surface temperature is also equal to 165 degrees celsius therefore we can say that t s is equal to 165 degrees celsius so we can say that the tube surface is 165 and the entrance of the temperature of water that is entrance or the temperature of water is given as 20 degrees celsius now we assume that the mean temperature if we are assuming that the mean temperature is 90 degrees celsius so at 90 degrees celsius, the properties that we have is that density is 965 .3 kilogram per meter cube.
01:44
The specific heat comes out to be 4206 jule per k g kelvin.
01:51
The thermal conductivity comes out to be 0 .675 watt per meter kelvin.
01:58
The dynamic viscosity comes out to be 0 .315 multiplied to 10 to the per minus 3 .5 kilo.
02:05
Gram meter second and the prandal number comes out to be 1 .96 this is the property at 90 degrees celsius thus the chirobatic viscosity the value of kinematic viscosity will be given as new will be equals to mu upon row so for mu we have 0 .315 multiplied to 10 to the power minus 3 divided upon row at lancity is 965 .3 this comes out to be 3 .263 multiplied to 10 to the power minus 7 meter square per second also the value of average velocity the average velocity will be equal to mass flow rate divided upon the row ac so from here the value of mass flow rate will be equal to row ac v average where ac is the area of cross -section that will be pi d square upon 4 so we can say that velocity average will be equal to as this is by d square upon 4 so this will come out to be four times of mass floor rate divided upon row pi d square so if we substitute we will have four multiplied to 0 .8 divided upon 965 .3 multiplied to pi multiplied to 0 .0 4 whole square so from here the value of average velocity will come out to be 0 .6595 meter per second hence in order to conclude about the reynolds number we know that it is vd upon new so this will be equal to 0 .6525 multiplied to 0 .04 divided upon 3 .263 multiplied to 10 to minus 7 and has the value of renounce number comes out to be 80845 .84 this is larger than 10 ,000 it is greater than 10 ,000 so we can check the entry length that will be lh will be approximately equal to lt this will be approximately equal to 10 times the diameter that will be equal to 10 multiplied to 10 .0 .04.
04:10
This is also equal to 0 .4 meter...