Question

Consider a fluid with mean inlet temperature Ti flowing through a tube of diameter D and length L, at a mass flow rate m'. The tube is subjected to a surface heat flux that can be expressed as q(x)= a + bsin(xpi/L), where a and b are constants. Determine an expression for the mean temperature of the fluid as a function of the x-coordinate.

Answer 1

Answer:

$T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))$

Explanation:

Our data given are:

$T_i =$ Mean temperature (inlet)

D = Diameter

L = Length

$\dot{m}=$Mass flow rate

Equation to surface flow as,

$q(x) = a+bsin(x\pi/L)$

We need to consider the perimeter of tube (p) to apply the steady flow energy balance to a tube , that is

$q(x)pdx=\dot{m}c_p dT_m$

Where $p=\pi D$

Re-arrange for $dT_m,$

$dT_m = \frac{q(x)pdx}{\dot{m}c_p}$

Integrating from 0 to x (the distance intelt of pipe) we have,

$\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x q(x)dx$

Replacing the value of q(x)

$\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x (a+bsin(x\pi/L))dx$

$T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}\int_0^x(a+bsin(x\pi/L))dx$

$T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}(ax-\frac{bL}{\pi}cos(\frac{x\pi}{L}))^x_0$

$T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))$