In: Physics
A thin-walled capillary tube (diameter=5mm, length=1m) carries steam at 100degreesC The capillary is covered by a later of insulation (thermal conductivity=0.10W/mK) and thickness delta. Assume that the heat transfer coefficient in the outside ir is 25W/m2K and its temp is 25degreesC. Compute and plot the steady state heat loss from the tube as a function of the thickness of insulation used for the range 0<delta<2cm
Solution :
Since the capillary tube is given as thin-walled, it is assumed its wall thickness is negligible. The insulation is covered over the capillary tube. Hence the inner diameter of the insulation is 5mm
Given:
d1 =
5mm
[Inner diameter of the insulation]
r1 = 2.5mm = 2.5 x 10-3 m
[Inner radius of the insulation]
= thickness of the insulation
L =
1m
[Length of the capillary tube]
K = 0.1 W/m
K
[Thermal conductivity of the insulation material]
h = 25 W/m2K [Heat transfer coefficient at
the outside surface]
T2 =
250C
[Temperature at the outside]
T1 =
1000C
[Temperature at the inside ]
We know that,
Heat transfer Q through a cylindrical wall is given by,
Where is is the effective thermal resistance between the two temperatures T1 and T2 and is given by they sum of the thermal resistance between inner radius r1 and outer radius r2 given by R1 and the thermal resistance between the outer surface and the surrounding Air given by R2.
Hence, , and,
Hence,
Substituting the values, We get,
From the given data,
is the thickness of the insulation. Hence,
Substituting the value of r2 in the above equation of Q,
The above equation gives the relation between the heat transfer and the thickness of insulation . All other values are known from the given data. Substituting the values of all other variables,
The above equation has two variables - Q and . This equation can be plotted in and Excel sheet for different values of to get corresponding values of Q.
The following plot is represented below in Excel :
It can be noted from the plot that the heat transferred increases in the beginning, reaches a maximum value and then reduces as the insulation thickness increases. This maximum value is called critical thickness.