Respuesta :
Explanation:
The given data is as follows.
Flow of mass rate of hydrocarbon, m = 0.126 kg/s
Heat of condensation, ([tex]h_{f}[/tex]) = 335 kJ/kg
For water [tex]t_{inlet}= 10^{o}C[/tex], and [tex]t_{out} = 32^{o}C[/tex]
Formula for transfer of heat is as follows.
Q = [tex]m \times h_{f}[/tex]
= [tex]0.126 kg/s \times 335 kJ/kg[/tex]
= 41.875 kJ/s
Also, it is known that Q = [tex]h \times A \times (\Delta T)_{lmtd}[/tex]
As it is given that for condensing T = [tex]88^{o}C[/tex].
[tex](\Delta T)_{lmtd}[/tex] = [tex]\frac{\Delta T_{1} - \Delta T_{2}}{ln \frac{\Delta T_{1}}{\Delta T_{2}}}[/tex]
= [tex]\frac{(78 - 56)}{ln \frac{78}{56}}[/tex]
= [tex]\frac{(78 - 56)}{0.33}[/tex]
= [tex]66.67 ^{o}C[/tex]
As it is given that inside and outside diameters of the tube are as follows.
[tex]D_{in}[/tex] = 0.0127, and [tex]D_{out}[/tex] = 0.0152 m
Formula for A = [tex]\pi (D_{out} - D_{in}) \times L[/tex] ......... (1)
Hence, putting value of equation (1) in the following formula.
Q = [tex]h \times A \times (\Delta T)_{lmtd}[/tex]
41.875 kJ/s = [tex]1420 W/m^{2} \times 3.14 \times (0.0152 - 0.0127)m \times L \times 66.67 ^{o}C[/tex]
L = \frac{41.875 \times 1000 J/s}{743.17 W^{o}C}
= 56.34 h
Thus, we can conclude that length of copper tubing will be required to accomplish the desired heat transfer 56.34 h.
