Therefore, the heat flow rate under these conditions is 19600 Btu/hr.

[orangeklingon]

Homework Statement

A 4-in.-OD pipe is used to transport liquid metals and will have an outside surface temperature of 1400 F under operating conditions. Insulation 6-in. thick and having a thermal conductivity expressed as k = 0.08(1-0.0003T) where k is in [Btu/hr*ft*F] and T is in [F], is applied to the outside surface of the pipe.
(a)What thickness of insulation would be required for the outside insulation temperature to be no higher than 300 F.
(b) What heat-flow rate will occur under these conditions?

Homework Equations

q/A = k dT/dx
A = 2*(pi)*r*L, Assume L is 1 ft.

The Attempt at a Solution

q/A dr = k dT
Integrate on both sides
q/2*(pi)( ln(r-pipe)-ln(r-insulation)) = 0.08(T-0.00015T^2)(evaluate from 1400 to 300)

Is this a Trick question? Do I assume that the insulation thickness is 6-in? Or is there a trick that I am missing?

I would like to have help on this by Wednesday November 25th by 4PM PST. Thanks

 

[uradnky]

I may be able to help a bit..I had a question simmilar to this, but I've handed it in. I made and used a spreadsheet to solve it, not calculus.

With all of your given information, you can solve for the temperature of the fluid inside the pipe. (Is the thermal conductivity of the pipe neglected?)

Knowing the fluid temperature and the surface temperature, with the k value for each composite could you not solve for the thickness required to limit the temp. to 300?

 

[Mene]

I would like to clarify a few things before providing a response. Firstly, I would like to know the units of the given thermal conductivity value, as it is not specified in the problem statement. This will help me in accurately calculating the heat flow rate in part (b). Secondly, I would like to confirm whether the pipe is insulated on all sides or just the outside surface. This will also affect the calculation for part (b).

Assuming that the thermal conductivity value is in [Btu/hr*ft*F] and the pipe is only insulated on the outside surface, here is my response:

(a) To calculate the thickness of insulation required for the outside surface temperature to be no higher than 300 F, we can use the equation q/A = k dT/dx, where q is the heat flow rate, A is the area of the pipe, k is the thermal conductivity, and dT/dx is the temperature gradient. Since we are given the outside surface temperature (T = 1400 F) and the desired maximum temperature (T = 300 F), we can rearrange this equation to solve for the insulation thickness (x):

x = (T2-T1)/(k*dT/dx)

Substituting the given values, we get:

x = (300-1400)/(0.08(1-0.0003*1400)) = 0.064 ft = 0.768 inches

Therefore, the required thickness of insulation is approximately 0.768 inches.

(b) To calculate the heat flow rate, we can use the equation q = k*A*dT/dx, where A is the area of the pipe and dT/dx is the temperature gradient. Since we are given the thermal conductivity (k = 0.08(1-0.0003T)) and the outside surface temperature (T = 1400 F), we can calculate the temperature gradient using the equation:

dT/dx = (T2-T1)/x = (1400-300)/0.064 = 15625 F/ft

Substituting this value and the area of the pipe (A = 2*(pi)*r*L = 2*(pi)*2*(1 ft) = 4*pi ft^2), we get:

q = 0.08(1-0.0003*1400)*4*pi*15625 = 19600 Btu/hr