l The rod which is heated electrically has an electrical hea

l. The rod which is heated electrically (has an electrical heater inside) by discharging electrical energy at the rate of g- 26w connects two plates maintained at 123 C (at x 0) and 73oC (at X-L (Fig. 21) The material of the rod has thermal of 41 K). vs x); find Derive formula for the calculation of the temperature distribution along the rod (T the location (Xmax) of the temperature maximum find the value of the temperature at this location; prove that it is maximum (not minimum) temperature; and find the heat flux at x 0 The length of the rod (L) is 45cm, and diameter of the rod is 32mm. Figure 1

For one dimensional, steady state heat transfer with heat generation generalized equation =

T ^** = - q/K

Where T^** = second order derivative of T (temperature with respect to X = d²T/dX²

Given Length of rod L = 45cm = 0.45m

Diameter of rod = 32mm = 0.032m

Volume of rod V = pi*L*D²/4 = 3.61 *10^-4m³ ; where pi =22/7

q= heat generated in the rod/volume = 26W / volume of rod = 26 /V = 71840W/m³

K = thermal conductivity of rod = 41W/mK

Given Length of rod L = 45cm = 0.45m

Diameter of rod = 32mm = 0.032m

after substituting values of q,k

T^** = - q/K = - 1752.2 K/m²

To get temperature distribution across the rod integrate two times the above equation with X

T^* = (-1752.2X) + C₁ = (- 2qX/K) +C₁----------------- Equation 1

Integrate once again

T = (-1752.2 *X²/2) + XC₁ + C₂

T = (-876.1 *X²) + XC₁ + C₂

To get the values for constants C₁, C₂

Apply boundary conditions at X= 0, T= 123⁰C = 396K

Applying boundary conditions we get

1^st boundary condition, 396 = (-876.1 *0) +0 + C₂

So C₂ = 396

Apply second boundary condition

X= L, T= 73⁰C = 346K

346 = (-876.1L²) + L*C₁ + 396

C₁ = (-50+ 876.1L²)/L

Temperature distribution equation =

T = (-876.1X²) + X (-50+ 876.1L²)/L + 396

Substitute values for L

T = (-876.1*X²) + X (-50+ 876.1*0.45²)/0.45 + 396

Temperature distribution T = - 876.1X² +283.13X + 396 ----------------- Equation 2

To get the location for maximum temperature, differentiate T with respect to X

dT/dX = (-876.1*2*X) +283.3 ----------------- Equation 3

At maximum temperature dT/dX = 0 as the slope changes at that point so

(-1752.2*X) – 283.3 = 0

X= 0.161m

At X= 0.161m Temperature will be maximum

T_max = - 876.1(0.161)² +283.13*(0.161) + 396 = 418 K

Tmax = 418 K is greater than the temperatures 396 K and 346 K (at X=0 and X=L) so, we can say that it is maximum temperature and not minimum temperature.

Let Heat flux at X= 0 be Q W/m²

Q = K dT/dX ----------------According to Fourier’s law of heat conduction

= 41 *( (-876.1*2*X) +283.3) from equation 3

So Q at X=0

Q = 41*((-876.1*2*0) +283)

= 41*283 =11603 W/m²

Q= 11603 W/m²

Heat flux at X= 0 = 11603 W/m²