The final piping section of the feedwater system in a power plant runs from a cl

Question

The final piping section of the feedwater system in a power plant runs from a closed feedwater heater to the boiler. The pressurized liquid water in the pipe is at 240 degress celsius. The lenght of the pipe is 10 m and it has an inner and our radius of ri=0.5m and ro=0.53m respectivily. The pipe is made out of plain carbon steel with a thermal conductivity of k=48 W/m k, the ambient temp is 25 degrees celsius, and the inside and outside heat transfer coefficients are h1= 100 W/m^2 K and h2= 3 W/m^2K respectively.

Determine the following:

A) express the differential equation and the boundary condidtions for steady 1-D heat conduction through the pipe

B) show the general solution and solve for the unknown constants to obtain T(r), the temperature distribution within the pipe wall as a function of radical location in the pipe wall. Make sure to indictae the units for the T(r) equation, i.e. does it give temperature in degrees celsius or kelvin?

C) what is the temperature of the outside of the pipe in degrees celsius?

D) what is the total rate of heat transfer through the pipe in kilowatts?

Explanation / Answer

Let pipe inner raidus be ri and outer radius of pipe be ro

and thermal conductivity of pipe K

and heat transfer coefficient for inside of pipe be h1 and outside heat transfer coefficient be h2

The tempertature of inside pipe be Ti and To be out side temperature

Then the differential equation of heat transfer be Q = Ti - To / R

where R = thermal resisitance = 1 / 2 Pi x L ( 1/ K1 ln ( r0 / ri ) + 1 /hi ri + 1 h0 r0 )

T(r) = Q x R

T(r ) in K need to transfer C to K

because heat trasfer coefficient in K kelvin degrees

Temperature of outside of pipe is To = 25 C

Themal resistance T(r ) R = 1 /2 x 3.14 x 10( 1 / 48 ln ( 0.53 / 0.5 ) + 1 / 100 x 0.5 + 1 / 3 x 0.53 )

= 0.01035

Totalt heat transfer rate Q = Ti - T0 / R

= 240 - 25 / 0.01035

= 20773 W

= 20.773 KW

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