Question

A 1-kW commercial electrical heating rod, 8 mm in diameter and 0.3 m long, is to be used in a highly corrosive gaseous environment. Therefore, it is to be provided with a cylindrical sheath of fireclay (k = 0.9 W/m.K). The gas flows by at 120⁰C, and h is 230 W/m2 .K outside the sheath. The surface of the heating rod cannot exceed 800⁰C. a. Derive the differential equation (DE) to determine the temperature profile in the sheath. b. Solve the DE of part (a) using proper boundary conditions and obtain the temperature profile.

Answer 1

Answer:

A) The differential equation that describes the temperature profile in the sheath is:

$\displaystyle \frac{dQ}{dt}=\dot Q_0=-\lambda_{sh} (2\pi rL) \frac{dT}{dr}$

B) The temperature profile follows the equation:

$T(r)=T_{rod}-\displaystyle \frac{Q_0}{2\pi L \lambda_{sh}}ln(\frac{r}{r_0})$

with Trod of 800ºc and 0.004m<r<0.00755m

Explanation:

The differential equation is obtained from the Fourier law, where:

$\displaystyle \frac{dQ}{dt}\vec{u}=-\lambda_{sh} S_t \vec{\nabla} T$ (1)

In this case, we can consider this problem as a unidimensional problem where heat transfer is done in a radial direction. Because the radius is much smaller than the length of the rod, we can consider the system as an infinite tube with an internal radius of 4mm (r₀) and an unknown external radius Re. Therefore the direction of the heat transfer is radial and the gradient can be transformed in a simple derivate. Also, we have the information that the heat transfer is done in a steady state. The equation (1) can be transformed in:

$\displaystyle \frac{dQ}{dt}\vec{u}=\dot Q\vec{r}=-\lambda_{sh} (2\pi rL) \frac{dT}{dr}\vec{r}$

$\dot Q_0=-\lambda_{sh} (2\pi rL) \frac{dT}{dr}$ (2)

To solve this equation we need to know the boundaries within this is valid. We know that the surface of the heating rod cannot exceed 800⁰C. Therefore the maximum temperature T(r₀) must be 800ºC.

Solving the equation (2) for r₀<r<rmax (we still don't know rmax):

$\dot Q_0dr=-\lambda_{sh} (2\pi rL) \frac{dT}{dr}dr\\\displaystyle \frac{\dot Q_0}{\lambda_{sh} (2\pi rL)}dr=-dT\\\int_{r_0}^{r} \frac{\dot Q_0}{\lambda_{sh} (2\pi rL)}\,dr=-\int_{T_0}^{T(r)}\,dT$

$T_{rod}-T(r)=\displaystyle \frac{Q_0}{2\pi L \lambda_{sh}}ln(\frac{r}{r_0})$ (3)

To determinate the external radius of the sheath of fireclay we use the Newton Convection law:

$\dot Q_0=hs_{ext}\Delta T=h(2\pi L r_{ext}) (T_{ext}-T_{fluid})$

$\displaystyle \frac{\dot Q_0}{h(2\pi L r_{ext})}=T_{ext}-T_{fluid}$(4)

Using equation (3) with this external unknown radius:

$T_{rod}-T_{ext}=\displaystyle \frac{Q_0}{2\pi L \lambda_{sh}}ln(\frac{r_{ext}}{r_0})$(5)

With (4)+(5):

$T_{rod}-T_{fluid}=(800-120)\ºc=\displaystyle Q_0(\frac{1}{2\pi L \lambda_{sh}}ln(\frac{r_{ext}}{r_0})+\frac{1}{2\pi Lhr_{ext} })$ (6)

Solving this equation gives us the Maximus radius. In this case, is 0.00755m.