# Heat transfer in series and parallel

• knc
In summary, a homework equation states that a transfer of energy between two objects occurs at the same rate in all directions. Assuming a steady state system, two variables must be set equal to find the temperature. The rate of heat flow through a rod is found by multiplying the power of the heat source by the distance between the heat source and the object.
knc

## Homework Equations

$$P = k A \frac{dT}{dx}$$

## The Attempt at a Solution

a)
Assuming steady state transfer, energy transfers through rods at the same rate everywhere.
Letting T be the temperature at the point of welding.

$$P_1 = k_1 A \frac{T_h-T}{L} \\ P_2 = k_2 A \frac{T - T_c}{L}$$
Setting rates equal to represent steady-state system:
$$k_1 A \frac{T_h-T}{L} = k_2 A \frac{T - T_c}{L}$$
Solving for T:
$$T = \frac{k_1 T_h + k_2 T_c}{k_1 + k_2}$$
Plugging this into either equation for individual rod:
$$P = \frac{T_h - T_c}{L} \frac{k_1k_2}{k_1 + k_2} A$$

a)
I don't know where to begin with part B. I would imagine I would try to find a ratio between the conductors in series and parallel.

Setting up the ratio of the powers for the two cases would be a reasonable approach. See if you can show that part (b) does not have a definite answer. So, the question is not well posed. [EDIT: Consider the ratio of the powers for the case where k1 and k2 are equal and the case where k1 is much smaller than k2 (i.e., k1 is now a good insulator.)]

Last edited:
You will need to give the answer in terms of k1 and k2 rather than a numerical answer.

In b, what is the rate of heat flow through rod A? What is the rate of heat flow through rod B. What is the total rate of heat flow through the two rods in parallel.

I can easily find the heat flow through each individual rod, I am just not sure if I can add them together to represent the overall rate of heat...

knc said:
I can easily find the heat flow through each individual rod, I am just not sure if I can add them together to represent the overall rate of heat...
Yes, for the parallel rods, the total rate of heat flow is the sum of the individual rates. What is your reservation about this?

Chestermiller
Finding the total heat flow in configuration b)
$$P = P_1 + P_2$$
$$P_1 = k_1 A \frac{T_h - T_c}{L} \\ P_2 = k_2 A \frac{T_h - T_c}{L}$$
$$P = (k_1 + k_2) A \frac{T_h - T_c}{L}$$

We are trying to find time taken to conduct 10J, difference in temperature is 100C.

$$P = \frac{\Delta Q}{\Delta t} \\ \frac{10}{t} = (k_1+k_2)A \frac{100}{L}$$
Rearranging:
$$t = \frac{L}{10 A (k_1 + k_2)}$$

I don't see how the first bit of information regarding the heat flow rate of the first configuration is relevant.

knc said:
$$t = \frac{L}{10 A (k_1 + k_2)}$$

I don't see how the first bit of information regarding the heat flow rate of the first configuration is relevant.
Your expression for t is OK. You could use the first bit to eliminate L/A so that the answer for t is expressed in terms of just k1 and k2.
I found the question to be a little odd. When I first read it, I thought that the answer to part (b) should be a numerical answer. But, the best you can do is express the answer in terms of k1 and k2. Actually, you can express it in terms of just the ratio k1/k2. But you can't get a numerical answer.

Here is what I got:
$$P_a = \frac{10}{120} = \frac{100}{L}\frac{k_1 k_2}{k_1 + k_2} A$$
$$\frac{L}{A} = 1200 \frac{k_1 k_2}{k_1 + k_2}$$
Plugging into t:
$$t = \frac{120 (k_1 k_2)}{(k_1 + k_2)^2}$$

That looks correct.

knc
One thing we can conclude is that, irrespective of k1 and k2, the time for (b) is less than 30 seconds.

knc
Thanks for the help.

## 1. What is the difference between heat transfer in series and parallel?

In series heat transfer, the heat flows through a series of materials or components one after the other, whereas in parallel heat transfer, the heat flows through multiple paths simultaneously.

## 2. Which method is more efficient for heat transfer: series or parallel?

It depends on the specific system and conditions. In some cases, series heat transfer may be more efficient as the heat has to travel through fewer materials, but in other cases, parallel heat transfer may be more efficient as it can distribute the heat more evenly.

## 3. How does the thermal conductivity of materials affect heat transfer in series and parallel?

The thermal conductivity of a material determines how easily heat can flow through it. In series heat transfer, the total resistance to heat flow is the sum of the resistances of each material, so a higher thermal conductivity will result in more efficient heat transfer. In parallel heat transfer, the total resistance to heat flow is lower, so the effect of thermal conductivity on efficiency is not as significant.

## 4. Is there a limit to the number of materials that can be used in series or parallel for heat transfer?

Yes, there is a limit to the number of materials that can be used in series or parallel for heat transfer. In series heat transfer, each material adds its own resistance to the overall heat flow, so adding too many materials can significantly decrease efficiency. In parallel heat transfer, the heat flow will eventually become evenly distributed among the materials, so adding more materials will not significantly increase efficiency.

## 5. What are some real-world applications of heat transfer in series and parallel?

In series heat transfer, this concept is used in heat exchangers, where fluids flow through a series of tubes to transfer heat. In parallel heat transfer, this concept is used in cooling systems for electronic devices, where multiple fans or heat sinks work simultaneously to dissipate heat. It is also used in solar panel arrays, where multiple panels are connected in parallel to maximize energy production.

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