Thermal conductivity through a cylinder?

[nelson_gslc]

Homework Statement

A copper tube, of internal radius 6.959mm, overall radius 7.975mm, thickness 1.016mm, contains water at 68 deg Celsius, and is present in a room with an ambient temp of 26.65 deg Celsius. If the tube is 1000mm long, what will be the rate of heat transferred from the copper tube?
k copper = 401 W/mK, A of tube = 0.050m, change in T=42.35 deg C, thickness =0.001016m.

Homework Equations

Q/t=kAΔT/d... or Heat transfer across length of cylindrical tubing equation (not specifically known)

The Attempt at a Solution

I have entered into the Q/t=kAΔT/d equation and received an answer of about 850 000 W, or 850kW, which seems entirely unrealistic. what am I doing wrong?

 

[Astronuc]

One needs to use the appropriate formula for heat conduction in the radial direction for a cylinder. One is using the formula for heat conduction normal to a plane in Cartesian coordinates.

See - http://www.engineersedge.com/heat_transfer/conduction_cylidrical_coor.htm

 

[m2003]

There are a few things to consider when calculating the rate of heat transfer through a cylinder. First, you have correctly identified the equation to use, Q/t = kAΔT/d, where Q is the rate of heat transfer, t is time, k is the thermal conductivity, A is the surface area, ΔT is the temperature difference, and d is the thickness of the material.

However, in order to get a more realistic answer, it is important to convert all units to the same system. In this case, it appears that you have mixed units of millimeters and meters. The radius of the tube is given in millimeters, while the thickness is given in meters. It is important to convert all units to either millimeters or meters before plugging them into the equation.

Additionally, the given values for the tube's dimensions and temperature difference seem quite small for a copper tube containing water at 68 degrees Celsius. It is possible that there is a mistake in the given values, which is resulting in an unrealistic answer.

In order to get a more accurate answer, it may be helpful to double check the given values and make sure they are all in the same unit system. You may also want to consider using a heat transfer calculator or consulting a heat transfer expert for more precise calculations.