DQ/dt=kA(dT/dx) = Heat Transfer. What is Q(t) Solution ?

[morrobay]

Homework Statement

With the time rate of change for heat transfer:
dQ/dt = kA(dT/dx)
k = thermal conductivity constant
A = boundary surface area
dT/dx = temp gradient

Homework Equations

Can the solution for Q(t) be obtained from modifying Newtons Law of Cooling since dQ/dt is proportional to T₀ - T_ambient and with A , x and k as constants ?

The Attempt at a Solution

[/B]
Q(t) = T_a + (T_a-T₀)e^-ktAx

 

[Simon Bridge]

What sort of heat transfer does Newton's law of cooling work for?
What sort of heat transfer is the equation above for?

 

[morrobay]

Newtons Law of Cooling can apply for the cooling of a hot liquid in cooler surroundings.
The equation above for dQ/dt applies to conductive/contact heat transfer.(edit)
So why not similar solutions, T(t) for Newton solution and Q(t) for the above equation

 

[Simon Bridge]

Newtonian cooling only applies where the temperature closest to the object being cooled stays the same.
Is that the case for the other relation?
Is the solution you proposed from considering the Newton case a solution to the DE for the contact-conduction case?

 

[morrobay]

For Newtons DE, dT/dt = -k(T-T_a)
with T_ambient remaining the same. And dQ/dt = -kA(dT/dx) there is heat transfer with temp change. for two objects. But in both cases does this apply ? : y(t) = y₀ e^-kt and therefore my proposed solution.with additional constants A,and x.
If not then what function is Q(t) ?

 

[Simon Bridge]

That's what you have to figure out.
Generally "it depends".

 

[morrobay]

Ok, this is a more complete description ; Two 1kg cubes of copper one at 10⁰ C (930 calories) the other at 40⁰C (3720 calories) are in contact with a .5 meter conduction bar with area .0025 m² area. Area of one face of copper cube, other faces are insulated. This is not steady state sinceT₂ temp decreasing and T₁ temp increasing. So dQ/dt = kA dT/dx may not apply. I assume the DE is related to dQ/dt = -k(Q₂-Q₁) And if the solution is related to y(t) = y₀ e^-kt then therefore proposed solution : Q(t) = 930 +2790e^-ktAx
Note: I could not find anything in Halliday Resnick text for this.

 

[morrobay]

I plugged in all the values for k(t₃)Ax and above is obviously not the solution for either Q(t) or T(t) for the copper cube that is 40⁰ C at t₀

 

[Chestermiller]

Ok, this is a more complete description ; Two 1kg cubes of copper one at 10⁰ C (930 calories) the other at 40⁰C (3720 calories) are in contact with a .5 meter conduction bar with area .0025 m² area. Area of one face of copper cube, other faces are insulated. This is not steady state sinceT₂ temp decreasing and T₁ temp increasing. So dQ/dt = kA dT/dx may not apply. I assume the DE is related to dQ/dt = -k(Q₂-Q₁) And if the solution is related to y(t) = y₀ e^-kt then therefore proposed solution : Q(t) = 930 +2790e^-ktAx
Note: I could not find anything in Halliday Resnick text for this.

Is the thermal inertia of the conduction bar significant (i.e.,does it have significant heat capacity) or can its heat capacity be neglected?

In this problem, you assume that the resistance to heat transfer within the copper cubes is negligible, so that the two copper cubes are each at a uniform temperature at all times. Is this assumption a problem for you?

Chet

 

[morrobay]

Is the thermal inertia of the conduction bar significant (i.e.,does it have significant heat capacity) or can its heat capacity be neglected?

In this problem, you assume that the resistance to heat transfer within the copper cubes is negligible, so that the two copper cubes are each at a uniform temperature at all times. Is this assumption a problem for you?

Chet

Yes , heat capacity of conduction bar is negligible. By uniform if you mean even heat distribution in cubes then yes
The cubes at t₀ are 10⁰ and 40⁰ with T₁ absorbing heat and T₂ transferring heat. I would appreciate the DE and Q(t) or T(t) solution. Thanks

 

[Chestermiller]

Yes , heat capacity of conduction bar is negligible.

Then the temperature profile along the bar responds instantaneously to the changes in the temperature of the cubes at each of its ends, and the temperature profile along the bar is always linear. So, if T_H is the temperature of the hotter cube at any time, and T_C is the temperature of the colder cube at any time, then the rate of heat flow from the hotter cube to the colder cube at any time is equal to $kA\frac{(T_H-T_C)}{L}$. Since this is the rate of heat loss of the hot cube, how is the rate of change of the hot cube temperature related to this rate of heat loss?

Chet

 

[morrobay]

Then the temperature profile along the bar responds instantaneously to the changes in the temperature of the cubes at each of its ends, and the temperature profile along the bar is always linear. So, if T_H is the temperature of the hotter cube at any time, and T_C is the temperature of the colder cube at any time, then the rate of heat flow from the hotter cube to the colder cube at any time is equal to $kA\frac{(T_H-T_C)}{L}$. Since this is the rate of heat loss of the hot cube, how is the rate of change of the hot cube temperature related to this rate of heat loss?

Chet

So dQ/dt = kA/dT/dx applies to both steady( T₁ and T₂ fixed) and non steady state time rate of change heat transfer which of course decreases with heat loss of hotter cube. In steady state with thermal conductivity coefficient Q(t) for hotter cube can be obtained. But in non steady state situation dQ/dt rate is not linear. So how do you get Q(t) for hotter cube ? Thanks

 

[Chestermiller]

So dQ/dt = kA/dT/dx applies to both steady and non steady state time rate of change heat transfer which of course decreases with heat loss of hotter cube. In steady state with thermal conductivity coefficient Q(t) can be obtained. But in non steady state situation dQ/dt rate is not linear. So how do you get Q(t) ? Thanks

Maybe this will help:
$$mC\frac{dT_H}{dt}=-kA\frac{(T_H-T_C)}{L}$$
$$mC\frac{dT_C}{dt}=+kA\frac{(T_H-T_C)}{L}$$
where m is the mass of each cube and C is the heat capacity of copper. I hope this makes sense to you, because it is correct.

Chet

 

[morrobay]

Maybe this will help:
$$mC\frac{dT_H}{dt}=-kA\frac{(T_H-T_C)}{L}$$
$$mC\frac{dT_C}{dt}=+kA\frac{(T_H-T_C)}{L}$$
where m is the mass of each cube and C is the heat capacity of copper. I hope this makes sense to you, because it is correct.

Chet

The question is not only about rate of heat transfer, dQ/dt, which is constant in steady state where T₁ and T₂ are fixed.
And in a non steady state where T₂ decreasing and T₁ increasing and therefore dQ/dt is not constant. The question is for the function Q(t) , the quantity of calories in the hotter cube at any time (t) for non steady state. Since heat transfer rate dQ/dt in non steady state is not constant where both cubes have changing temperatures , I don't understand how equation above can obtain Q(t) ? For example in a non steady state at t₀ T₂ = 40⁰ C. Then at suppose t₈ where T₂ is less than 40 ⁰ how many calories are left in hotter cube or what is its temperature ? Even for the case in Newtons Cooling where only T₂ is changing (cooling) T(t) is a DE solution. In the case above where T₂ > T₁ and both are changing it appears a DE and
solution for Q(t) is involved.

 

[Chestermiller]

The question is not only about rate of heat transfer, dQ/dt, which is constant in steady state where T₁ and T₂ are fixed.
And in a non steady state where T₂ decreasing and T₁ increasing and therefore dQ/dt is not constant. The question is for the function Q(t) , the quantity of calories in the hotter cube at any time (t) for non steady state. Since heat transfer rate dQ/dt in non steady state is not constant where both cubes have changing temperatures , I don't understand how equation above can obtain Q(t) ? For example in a non steady state at t₀ T₂ = 40⁰ C. Then at suppose t₈ where T₂ is less than 40 ⁰ how many calories are left in hotter cube or what is its temperature ? Even for the case in Newtons Cooling where only T₂ is changing (cooling) T(t) is a DE solution. In the case above where T₂ > T₁ and both are changing it appears a DE and
solution for Q(t) is involved.

That's right. And all this become clearer to you once you solve those two differential equations I wrote down, subject to the initial conditions on TH and TC. Do you know how to solve those two simultaneous first order ordinary differential equations for TH and TC as functions of time? If so, let's see what you obtain.

Chet

 

[Chestermiller]

If I subtract the cold equation from the hot equation, I get:
$$mC\frac{d(T_H-T_C)}{dt}=-2kA\frac{(T_H-T_C)}{L}$$
where m is the mass of each cube and C is the heat capacity of copper. The solution to this differential equation is:

$$(T_H-T_C)=(T_{H0}-T_{C0})e^{-\frac{2kAt}{mCL}}$$

So, Q(t) (as you call it) is given by:

$$\frac{dQ}{dt}=\frac{kA}{L}(T_{H0}-T_{C0})e^{-\frac{2kAt}{mCL}}$$

$$Q(t)=\frac{mC}{2}(T_{H0}-T_{C0})\left(1-e^{-\frac{2kAt}{mCL}}\right)$$

Any questions?

Chet[/QUOTE]