Heat Transfer, Convective and Conductive Rates

[chriskay301]

Homework Statement

A 1.0 cm diameter steel rod with k = 20W/(m-K) is 20 cm long. It has one end maintained at 35°C and the other at 100°C. It is exposed to convection heat transfer with h=65W/(m^2-K) and an ambient air stream at 20°C. a) Sketch the distribution of temp within the rod. b) Determine the rate of convective heat transfer from the rod. c) What are the conductive heat transfer rates at each end of the rod?

Homework Equations

Rate of Convection = (h)(Area)(T-T∞)
Rate of Conduction = -k(Area)(dT/dx)

The Attempt at a Solution

I'm not really sure how to attempt this problem. I know that using an energy balance E(in)=E(out) I could solve for a diff. eq. for the temperature anywhere in the rod. But I'm not sure how that would be useful. For conduction, if its at the end of the rod how is there a change in x direction? And for convection, I'm not sure what temperature I would use for T to find the rate since the entire rod is going to be at different temperatures.

 

[Chestermiller]

Hi chriskay301. Welcome to physics forums!

You are going to be assuming that the temperature within the rod is a function only of x, where x is the distance from the 35C end. The idea is to first find this temperature profile. Do a heat balance on the section of rod between x and x + dx. In terms of the cross sectional area A and the thermal conductivity k, what is the rate of heat flow by conduction into this control volume at x? What is the rate of heat flow by conduction out of this control volume at x + dx? What is the rate of heat flow by convection out of the control volume (in terms of the perimeter P= πD)? Combine these into your differential heat balance. You should have a second order ODE that you can solve subject to the temperature boundary conditions at the two ends. Solve this set of equations for the temperature as a function of x. Now you have what you need to answer the rest of the questions.

Chet

 

[chriskay301]

Okay, so I got a temperature distribution equation in respect to my left and right boundary conditions. I'm still not entirely sure what to do from here.

I know I can use this equation to get the temperature of any x, but how does that help me for convection or conduction? Do I just use the center of the rod for the convection temperature? Do I use a point such as x=.1 and use that to determine the conduction at the left end?

Still a bit confused!

 

[Chestermiller]

Okay, so I got a temperature distribution equation in respect to my left and right boundary conditions. I'm still not entirely sure what to do from here.

I know I can use this equation to get the temperature of any x, but how does that help me for convection or conduction? Do I just use the center of the rod for the convection temperature? Do I use a point such as x=.1 and use that to determine the conduction at the left end?

Still a bit confused!

The rate of convective heat transfer from the rod is equal to the net rate of conductive heat transfer into the rod at its two ends. These are determined by the temperature gradient at the ends times the thermal conductivity times the cross sectional area. You can also get the rate of convective heat transfer another way, by integrating the local rate of convective heat transfer with respect to x over the surface of the rod. Both these methods will give you the same answer.
For part (c), you already calculated the heat transfer rates at the ends in part (b).

 

[rezeew]

I would first review the information given in the problem and identify the key variables and equations needed to solve it. From the given information, we can see that the problem involves heat transfer through a steel rod, with one end being maintained at a constant temperature of 35°C and the other end at 100°C. The rod is also exposed to convection heat transfer with a known convective heat transfer coefficient (h) and an ambient air temperature (T∞) of 20°C. We are also given the length and diameter of the rod, as well as the thermal conductivity (k) of the steel.

To solve this problem, we will need to consider both conductive and convective heat transfer. Conductive heat transfer occurs within the rod itself, while convective heat transfer occurs between the rod and the surrounding air. We can use the equations for these two types of heat transfer to determine the temperature distribution within the rod and the rate of heat transfer.

a) To sketch the distribution of temperature within the rod, we can use the equation for conduction heat transfer:

Rate of Conduction = -k(Area)(dT/dx)

Since the rod has a constant diameter, we can simplify this equation to:

Rate of Conduction = -k(Area)(ΔT/Δx)

We can see that the rate of conduction is directly proportional to the temperature difference (ΔT) and inversely proportional to the distance (Δx) along the rod. This means that the temperature will decrease as we move along the rod from the hot end to the cold end. A sketch of the temperature distribution within the rod would show a gradual decrease in temperature from 100°C to 35°C.

b) To determine the rate of convective heat transfer from the rod, we can use the equation for convection heat transfer:

Rate of Convection = (h)(Area)(T-T∞)

Since we are given the values for h, Area, and T∞, we can simply plug them into the equation to find the rate of convective heat transfer. It is important to note that the temperature (T) in this equation refers to the surface temperature of the rod, which will vary along its length. Therefore, we will need to use an average temperature for T in order to find the overall rate of convective heat transfer from the rod.

c) The conductive heat transfer rates at each end of the rod