Answer:
Total rate of heat transfer for the tube. Q = 13.1 kW
Explanation:
Given data:
Diameter of the tube, D = 0.025 m
Lengthofthetube, L=IOm
Mass flow rate, m = 0.5 kg/s
Glycerin inlet temperature, 1 = 25°C
Tube surface temperature, 7 = 140°C
The expression for the outlet temperature Td is
(see image 1 for equation).
where I, - Heat transfer coefficient
A, - Heat transfer surface area
m.- Mass flow rate
Cp- Specific heat at constant pressure
Properties of glycerine:
Density, = 1258 kg/rn
Specific heat at constant pressure, c, —2447 i/kg. K
Thermal conductivity, k = 0.2860 W/m. K
Dynamic viscosity. p = 0.6582 kg/rn. s
Prandtl number. Pr = 5631
Now Reynolds number can be calculated using the equation,
(See image 2).
Since the value is less than 2300, the flow is laminar.
Now, the hydrodynamic entry length is given by,
Lk = 0.05 Re D
= 0.05 x 38.69 X 0.025
= 0.04836 m
Thermal entry length is given by.
Lt= Lh x Pr
0,04X36x5631
= 272.32 m
As the thermal entry length is greater than the given tube length. the flow is not thermally
developed but developing. Therefore, the average Nusselt number for the thermal entrance
region can be determined using the following equation.
See image 3.
The convective heat transfer coefficient.
h= k•Nu
/D
.
= 152.29 W/m2.K
Heat transfer surface area.
A = πDL
= 0.7854 m2
The mean temperature at outlet.
see image 4 for explanation.
Hence the mean temperature at outlet of the tube. I = 35.7C1
Hence the mean temperature at outlet of the tube. I = 35.7C1
The total rate of heat transfer for the tube.
Q = mcp (To - T1)
=O.5x2447x(35.71—25)
= O.5x2447x10.7
= 13091.45 W
= 13.1 kW
Therefore, the total rate of heat transfer for the tube. IQ = 13.1 kW
