Chapter 5, Problem 5.1P

Consider a thin electrical heater attached to a plate and backed by insulation. Initially, the heater and plate are at the temperature of the ambient air, $T_{\infty }.$ Suddenly. the power to the heater is activated, yielding a constant heat flux $q_o^n\left({\text{W/m}}^{\text{2}}\right)$ at the inner surface of the plate.

1. Sketch and label, on $T-x$ coordinates, the temperature distributions: initial, steady-state, and at two intermediate times.
2. Sketch the heat flux at the outer surface $q_x^n\left(L,t\right)$as a function of time.

To determine

Sketch the temperature distributions $T\left(x,t\right)$ at initial, steady and two intermediate times.

Answer to Problem 5.1P

Temperature distributions $T\left(x,t\right)$ at initial, steady and two intermediate times is shown below:

  

Explanation of Solution

Given:

  

1. There is One-dimensional conduction.
2. A property of distribution is considered as constant.
3. Heat loss through insulation of heater is negligible.

Calculation:

On the backside of plate, an electrical heater is attached while front surface is exposed to convection process $\left(T_{\infty },h\right)$

There is a uniform temperature for an ambient air. At this temperature the plate is placed and by providing a constant $q_0^{\ast }$ on switching on the heater power.

In the following diagram, all temperature distributions for four-time conditions including steady-state distribution and the initial distribution, $T\left(x,t\right)$

and are shown

As the flux is a constant the temperature is also a constant at $q_0^{\ast }$ at $x=0,-dT/d{x}_{x=0}$ for $t>0$ .

The steady state temperature distribution will be liner such that:

  $T_0=T\left(0,x\right)$

  $q_0^{\ast }=k\frac{T_0-T\left(L,\infty \right)}{L}=h\left[T\left(L,\infty \right)-T_{\infty }\right]$

Conclusion:

Therefore, temperature distributions $T\left(x,t\right)$

at initial, steady and two intermediate times is shown below:

  

To determine

Sketch the heat flux at the outer surface, $q^{\ast }x\left(L,t\right)$ as a function of time.

Answer to Problem 5.1P

Heat flux at the outer surface, $q"x\left(L,t\right)$

as a function of time is shown as below:

  

Explanation of Solution

Given:

  

1. There is One-dimensional conduction.
2. A property of distribution is considered as constant.
3. Heat loss through insulation of heater is negligible.

Calculation:

On the backside of plate, an electrical heater is attached while front surface is exposed to convection process $\left(T_{\infty },h\right)$

There is a uniform temperature for an ambient air. At this temperature the plate is placed and by providing a constant $q_0^{\ast }$ on switching on the heater power.

At front surface $x=L$ , the heat flux at the is given by,

  $q^{\ast }x\left(L,t\right)=-k{\left(dT/dx\right)}_{x=L}$

Now, by using the temperature distribution, plot of heat flux-time is shown as below:

  

At early times, the heat flux and temperature values are not change from their initial values at $x=L$

That means, at early times there is zero slope for $q^{\ast }x\left(L,t\right)$ .

At last, the value of $q^{\ast }x\left(L,t\right)$ goes to the steady-state value $q^{\ast }0$

Conclusion:

Therefore, heat flux at the outer surface, $q"x\left(L,t\right)$

as a function of time is shown as below: