A rod of diameter D = 25 m and thermal conductivity of 60 W/m·K protrudes from a furnace with a wall temperature of 200ºC. The rod is welded to the furnace wall and is used as a hangar for instrumentation cables. To avoid damaging the cables, the surface temperature of last 100 m of the rod must be kept below 100ºC. The ambient air temperature is 25ºC and the convection coefficient is 15 W/m2K. A. Write the finite-difference equation for an internal node with x = 2 mm. B. Write the finite difference equation for the node at the fin tip.

(a)T₀= T∞ + Rfin/ Rins +Rfin (Tw -T∞) (b) T = log. 16º C. therefore the length rod cannot meet specified limit, so we can use stainless steel to alter thermal conductivity.

Explanation:

Solution

Recall that:

A diameter of rod D = 25m with a Thermal conductivity of  60 W/m·K

Protrudes from a furnace with a  temperature wall of  200ºC

The rod is welded to the furnace wall and is used as a hangar for instrumentation cable.

The surface of he temperature of last 100m is kept below 100ºC

So,

The ambient temperature is = 25ºC

Convection coefficient = 15 W/m2K.A.

Diameter = 25mm

K = 60 W/m·K

Tw = 200 ºC.

L ins = 200mm

Now,

Tmax = 100ºC

T∞ = 25C

h = 15w/m²k

Note: Kindly find an attached copy of part of the solution to this given example

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

answer:

yes

explanation: