Cooling of sphere with pseudo steady state condition

[gfd43tg]

Homework Statement

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.

Homework Equations

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.

 

[rude man]

\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

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

Solve your diff. eq.

 

[gfd43tg]

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

 

[Chestermiller]

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

 

[gfd43tg]

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.

 

[Chestermiller]

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.

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

Chet

 

[gfd43tg]

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.

 

[Chestermiller]

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

 

[gfd43tg]

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!