# Cooling of sphere with pseudo steady state condition

1. Nov 22, 2014

### Maylis

1. The problem statement, all variables and given/known data
A hot solid sphere of initial radius $a$ with a uniform initial temperature $T_{0}$ is allowed to
cool under stagnant air at ambient temperature, $T_{\infty}$ . Assume the temperature within
the sphere is uniform throughout the cooling process. Show that under pseudo-steady
state conditions, the temperature of the solid sphere decreases with time according to
$$T - T_{\infty} = (T_{0} - T_{\infty}) \hspace{0.05 in} exp \Big( - \frac {3kt}{\rho C_{p} a^2} \Big)$$
where $k$ is the thermal conductivity of the surrounding air and $\rho$ and $C_{p}$ are the
density and specific heat of the solid sphere, respectively.
2. Relevant equations

3. The attempt at a solution
I do a general energy balance on the sphere,
$$\frac {dE}{dt} = \dot Q_{in} - \dot Q_{out} + \dot Q_{gen} + \dot W_{s}$$
I assume that there is no heat generation, no shaft work, and that no heat enters the sphere
$$\rho V C_{p} \frac {dT}{dt} = - \dot Q_{out}$$
From Newton's Law of cooling,
$$\dot Q_{out} = h_{\infty}A (T - T_{\infty})$$
$$\rho V C_{p} \frac {dT}{dt} = -h_{\infty}A (T - T_{\infty})$$
$$\frac {dT}{dt} + \frac {h_{\infty}A}{\rho V C_{p}} (T - T_{\infty}) = 0$$
Now that the assumption is pseudo steady state, I say $\frac {dT}{dt} = 0$, so I end up with
$$\frac {h_{\infty}A}{\rho V C_{p}} (T - T_{\infty}) = 0$$
And from here, I have no idea how I will be able to get the expression that I am supposed to derive now that I have no derivative to integrate and use boundary conditions to derive the expression.

2. Nov 22, 2014

### rude man

Pseudo-steady-state does not mean that dT/dt= 0. If it did the cottonpicker would never cool down!

3. Nov 22, 2014

### Maylis

Alright, so then to continue with my derivation,
$$\frac {dT}{dt} + \frac {h_{\infty}A}{\rho V C_{p}} (T - T_{\infty}) = 0$$
I define $\theta = T - T_{\infty}$, and the boundary condition at $t = 0$ is $\theta_{0} = T_{0} - T_{\infty}$. Then I take the derivative of theta with respect to time, $\frac {d \theta}{dt} = \frac {dT}{dt}$
$$\frac {d \theta}{dt} + \frac {h_{\infty}A}{\rho V C_{p}} \theta = 0$$
$$\int \frac {d \theta}{\theta} = - \frac {h_{\infty}A}{\rho V C_{p}} \int dt$$
$$ln \hspace{0.05 in} \theta = - \frac {h_{\infty}A}{\rho V C_{p}}t + C$$
Using my boundary condition, I find $C = ln(T_{0} - T_{\infty})$. I then subtract to the LHS of the equation and get
$$ln \hspace{0.05 in} \Big(\frac {T - T_{\infty}}{T_{0} - T_{\infty}} \Big) = - \frac {h_{\infty}A}{\rho V C_{p}} t$$
I know the characteristic length for a sphere, $L_{c} = V/A = a/3$. So I substitute and get
$$ln \hspace{0.05 in} \Big(\frac {T - T_{\infty}}{T_{0} - T_{\infty}} \Big) = - \frac {3h_{\infty}}{\rho C_{p}a} t$$
Now, somehow I need to get rid of the heat transfer coefficient, $h_{\infty}$ using the pseudo steady state condition. So I use Fourier's Law at the interface of the sphere and air,
$$\dot Q_{out} = -kA \frac {dT}{dr} \bigg |_{r=a}$$
$$h_{\infty}A(T - T_{\infty}) = -kA \frac {dT}{dr} \bigg |_{r=a}$$
However, I don't know what the derivative at r = a will be

Last edited: Nov 22, 2014
4. Nov 22, 2014

### Staff: Mentor

They're trying to get you to use the steady state heat conduction solution for a sphere immersed in an infinite medium to get the heat transfer coefficient to the air.

Suppose you have a sphere at a constant surface temperature T* immersed in an infinite medium of thermal conductivity k, and suppose that the temperature far from the sphere is T. What is the steady state heat conduction temperature profile surrounding the sphere?

Chet

5. Nov 22, 2014

### Maylis

Okay, so I will do another energy balance at the surface
$$\frac {dE}{dt} = \dot Q_{r} - \dot Q_{r + \Delta r} + \dot e_{gen}A \Delta r + \dot W_{s}$$
Assume no work is done,
$$\rho \hat C_{p} A \Delta r \frac {dT}{dr} = \dot Q_{r} - \dot Q_{r + \Delta r} + \dot e_{gen}A \Delta r$$
$$\rho \hat C_{p} \frac {dT}{dt} = - \frac {1}{A} \frac {d}{dr} \Big( -kA \frac {dT}{dr} \Big) + \dot e_{gen}$$
Assume no heat generation and steady state,
$$\frac {d}{dr} \Big(4 \pi k r^2 \frac {dT}{dr} \Big) = 0$$
$$4 \pi k r^2 \frac {dT}{dr} = C_{1}$$
I know at $r = 0$, then $\frac {dT}{dr} = 0$, therefore $C_{1} = 0$
$$k(4 \pi r^2) \frac {dT}{dr} = 0$$
So I guess the temperature profile is T(r) = constant.

6. Nov 23, 2014

### Staff: Mentor

No. You should be integrating over the region from "a" to infinity (the "stagnant" air). The key word is stagnant.

Chet

7. Nov 23, 2014

### Maylis

Okay, so I get down to
$$4 \pi k r^2 \frac {dT}{dr} = C_{1}$$
So then at $r = \infty$, $\frac {dT}{dr} = 0$ and $T = T_{\infty}$, therefore $C_{1} = 0$. So then this reduces to
$$4 \pi k \int dT = C_{1} \int \frac {dr}{r^2}$$
$$4 \pi k T = C_{1} \Big (- \frac {1}{r} \Big) + C_{2}$$
I use my boundary conditions, at $r = \infty$, $T = T_{\infty}$
$$4 \pi k T_{\infty} = C_{1} \Big (- \frac {1}{\infty} \Big) + C_{2}$$
Therefore $C_{2} = 4 \pi k T_{\infty}$. So now I have
$$4 \pi k (T - T_{\infty}) = C_{1} \Big (- \frac {1}{r} \Big)$$
$$T = \frac {C_{1}}{4 \pi k} \Big(\frac {-1}{r} \Big) + T_{\infty}$$
$$\frac {dT}{dr} = \frac {C_{1}}{4 \pi k} \Big(\frac {1}{r^2} \Big)$$
For my condition $r = \infty$, $\frac {dT}{dr} = 0$,
$$0 = \frac {C_{1}}{4 \pi k} \Big(\frac {1}{\infty^2} \Big)$$
This just leads me nowhere, it gives $C_{1} = 0$ and therefore $T = T_{\infty}$, as if T is not a function of radius at all.

Last edited: Nov 23, 2014
8. Nov 23, 2014

### Staff: Mentor

The other boundary condition is T = T* at r = a.

I get:

$$T=T_∞+(T^*-T_∞)\frac{a}{r}$$

where T* is the instantaneous temperature at the surface of the sphere. The T here is the temperature of the air, not the temperature of the sphere. What you called T in your earlier equations is what I call T* here. From this equation, you should be able to derive the heat transfer coefficient h. What do you get?

Chet

9. Nov 23, 2014

### Maylis

I got $h_{\infty} = \frac {k}{a}$, so now I get the expression. I was able to redo my derivation and get the same expression you derived. Thanks a lot!