The temperature distribution across a wall 0.5 m thick at a certain instant of time is T(x) = a + bx + cx^2, where T is in degrees Celsius and x is in meters, a = 250°C, b = -250°C... The temperature distribution across a wall 0.5 m thick at a certain instant of time is T(x) = a + bx + cx^2, where T is in degrees Celsius and x is in meters, a = 250°C, b = -250°C/m, and c = 30°C/m². The wall has a thermal conductivity of 1 W/mK. (a) Determine the rate of heat transfer into and out of the wall and the rate of change of energy stored by the wall on a unit surface area basis. (b) What is the convection coefficient if the cold surface is exposed to a fluid at 150°C? Read more

Steps to Solve

1. Temperature Distribution Analysis

   The temperature distribution across the wall is given by the equation:

   $$ T(x) = a + bx + cx^2 $$

   where:

   • $a = 250^\circ C$
   • $b = -250^\circ C/m$
   • $c = 30^\circ C/m^2$

   We will evaluate this function at both sides of the wall, $x = 0$ and $x = 0.5$.

2. Calculating Temperatures at the Boundaries

   For $x = 0$ (hot side): $$ T(0) = 250 + (-250)(0) + 30(0)^2 = 250^\circ C $$

   For $x = 0.5$ (cold side): $$ T(0.5) = 250 + (-250)(0.5) + 30(0.5)^2 $$ $$ = 250 - 125 + 7.5 = 132.5^\circ C $$

3. Heat Transfer Calculation

   The rate of heat transfer $q$ through the wall can be calculated using Fourier's law of heat conduction:

   $$ q = -k \frac{dT}{dx} $$

   First, we need to find $\frac{dT}{dx}$:

   $$ \frac{dT}{dx} = b + 2cx = -250 + 60x $$

   At $x = 0.5$:

   $$ \frac{dT}{dx} \bigg|_{x=0.5} = -250 + 60(0.5) = -250 + 30 = -220 ,^{\circ}C/m $$

   Now substitute this into Fourier's law:

   $$ q = -1 \cdot (-220) = 220 , W/m^2 $$

4. Rate of Energy Stored

   The rate of change of energy stored in the wall per unit surface area, given the specific heat capacity $C_p$, can be calculated using:

   $$ \frac{dQ}{dt} = \rho C_p \frac{dT}{dt} $$

   Assuming a reasonable specific heat capacity (e.g., concrete) is around $C_p = 840 , J/(kg \cdot K)$ and density $\rho = 2400 , kg/m^3$. Substituting this into the equation, we evaluate:

   $$ \frac{dQ}{dt} = 2400 \cdot 840 \cdot \frac{(T(0.5) - T(0))}{\Delta t} $$

   Without a specific $\Delta t$, we can just note that the energy storage needs a time context.

5. Calculating Convective Heat Transfer Coefficient

   The convection coefficient $h$ can be determined using Newton's Law of Cooling:

   $$ q = h (T_{\infty} - T_{\text{surface}}) $$

   Rearranging for $h$ gives:

   $$ h = \frac{q}{T_{\infty} - T_{\text{surface}}} $$

   Where:

   • $T_{\infty} = 150^\circ C$
   • $T_{\text{surface}} = T(0.5) = 132.5^\circ C$

   Substitute the known values:

   $$ h = \frac{220}{150 - 132.5} = \frac{220}{17.5} \approx 12.57 , W/(m^2 \cdot K) $$