Three slabs are joined together as shown in the figure. There is no thermal contact resistance at the interfaces. The center slab experiences a non-uniform internal heat generation with an average value equal to 10000 Wm^–3, while the left and right slabs have no internal heat generation. All slabs have thickness equal to 1 m and thermal conductivity of each slab is equal to 5 Wm^–1K^–1. The two extreme faces are exposed to fluid with heat transfer coefficient

100Wm^–2K^–1 and bulk temperature 30°C as shown. The heat transfer in the slabs is assumed to be one dimensional and steady, and all properties are constant. If the left extreme face temperature T₁ is measured to be 100°C, the right extreme face temperature T₂ is __________°C.

                                          

The correct answer is 60.

Step-by-Step Explanation:

To determine the temperature of the right extreme face, $T_2$, we can apply the principle of conservation of energy to the entire three-slab system under steady-state conditions.

1. Define the Control Volume and System Parameters:
Based on the provided problem description and image, the system consists of three adjacent slabs, each of thickness 1 m:
- Left slab: Thickness $L_1=1\text{m}$, thermal conductivity $k=5\text{W/(m·K)}$, and no internal heat generation.
- Center slab: Thickness $L_2=1\text{m}$, thermal conductivity $k=5\text{W/(m·K)}$, with a non-uniform internal heat generation having an average value of ${\dot{q}}_{}\text{avg}=10000{\text{W/m}}^3$.
- Right slab: Thickness $L_3=1\text{m}$, thermal conductivity $k=5\text{W/(m·K)}$, and no internal heat generation.
- The left extreme face is at temperature $T_1=100\text{°C}$.
- The right extreme face is at temperature $T_2$.
- Both outer surfaces are exposed to a fluid at bulk temperature $T_{\infty }=30\text{°C}$ with a convective heat transfer coefficient $h=100{\text{W/(m}}^2\text{·K)}$.

2. Calculate the Total Heat Generation Rate:
The heat is generated only in the middle slab. The total rate of heat generated per unit cross-sectional area ($q^{"}{}_{\text{gen}}$) is given by:
$q^{"}{}_{\text{gen}}={\dot{q}}_{}\text{avg}\times L_2$
Substituting the given values:
$q^{"}{}_{\text{gen}}=10000{\text{W/m}}^3\times 1\text{m}=10000{\text{W/m}}^2$

3. Convective Heat Flux from the Left Face:
The heat flux leaving the left extreme face of the system to the surrounding fluid by convection is:
$q^{"}{}_1=h(T_1-T_{\infty })$
Substituting the given values:
$q^{"}{}_1=100\times (100-30)=100\times 70=7000{\text{W/m}}^2$

4. Energy Balance on the Entire System:
Under steady-state conditions, the rate of heat generated within the system must equal the net rate of heat leaving the boundary surfaces of the system. Therefore:
$q^{"}{}_{\text{gen}}=q^{"}{}_1+q^{"}{}_2$
where $q^{"}{}_2$ is the convective heat flux leaving the right extreme face:
$q^{"}{}_2=h(T_2-T_{\infty })$
Substituting the known values into the energy balance equation:
$10000=7000+q^{"}{}_2$
$q^{"}{}_2=10000-7000=3000{\text{W/m}}^2$

5. Solve for $T_2$:
Now, using the expression for $q^{"}{}_2$:
$3000=100\times (T_2-30)$
Divide both sides by 100:
$30=T_2-30$
$T_2=30+30=60\text{°C}$