Answer:
[tex]T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))[/tex]
Explanation:
Our data given are:
[tex]T_i =[/tex] Mean temperature (inlet)
D = Diameter
L = Length
[tex]\dot{m}=[/tex]Mass flow rate
Equation to surface flow as,
[tex]q(x) = a+bsin(x\pi/L)[/tex]
We need to consider the perimeter of tube (p) to apply the steady flow energy balance to a tube , that is
[tex]q(x)pdx=\dot{m}c_p dT_m[/tex]
Where [tex]p=\pi D[/tex]
Re-arrange for [tex]dT_m,[/tex]
[tex]dT_m = \frac{q(x)pdx}{\dot{m}c_p}[/tex]
Integrating from 0 to x (the distance intelt of pipe) we have,
[tex]\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x q(x)dx[/tex]
Replacing the value of q(x)
[tex]\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x (a+bsin(x\pi/L))dx[/tex]
[tex]T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}\int_0^x(a+bsin(x\pi/L))dx[/tex]
[tex]T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}(ax-\frac{bL}{\pi}cos(\frac{x\pi}{L}))^x_0[/tex]
[tex]T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))[/tex]
