Many hot-water heating systems have a reservoir tank connected directly to the pipeline, so as to allow for expansion when the water becomes hot. The heating system of a house has 74 m of copper pipe whose inside radius is 9.10 10-3 m. When the water and pipe are heated from 20 to 75°C, what must be the minimum volume of the reservoir tank to hold the overflow of water?

Question

Many hot-water heating systems have a reservoir tank connected directly to the pipeline, so as to allow for expansion when the water becomes hot. The heating system of a house has 74 m of copper pipe whose inside radius is 9.10 10-3 m. When the water and pipe are heated from 20 to 75°C, what must be the minimum volume of the reservoir tank to hold the overflow of water? 
m3

I know the volume of the pipe is 
pi R^2 L = 1.93*10^-2 m^2 = 19.3 liters

I calculated how much that volume of water increases when being heated from 20 to 75 C. I used the thermal expansion coefficient of water. It varies from 20 to 75 C.

delta V = V*4*10^-4*(75 - 20) 
= 0.43 liters

My finnally answe has to be in m^3 so I divided .43l liters by 10^-3 m^3 but I can't come up with right answer for this problem.

Damon · Answer

change in V/V = beta *(Tf-Ti)
= 4*10^-4 (55) = 220*10^-4 = 2.2 * 10^-2

pi r^2 L = pi * (9.1*10^-3)^2 m^2 * 75m
= 1.95*10^-2 METERS CUBED NOT liters

so
delta V = 1.95*10^-2 * 2.2*10^-2 = 4.29 * 10^-4 meters cubed

delta v/V can be liters per liter or meters^3 per meter^3 or thimbles per thimble but in this case everything is in meters.

Happy · Answer

my computer home work is tell me .000429 is not correct I have reworked the problem but it keep tell me the answer is not correct

Damon · Answer

Well, I assumed your thermal expansion coef for water was correct. Otherwise I did it all independently from your work. However water is non-linear and even changes sign just above freezing (why ice freezes at the surface of the pond) so if it is a little off it is because it is hard to define that coef. If it is a lot off, either we are both crazy or the computer homework is wrong.

Tom · Answer

Try this one.
you seem to have forgotten about the coeficient for copper pipe. Remember volume gained (reservoir) equals volume lost (pipe)

deltaV =beta_water x V_i x deltaT
this gives the change in V for water

then
deltaV = beta_copper x V_i x deltaT
this gives the change in V for Copper

delta_T and V_i are the same for each equation. Then SUBRACT the lower number from the higher number. I had a similar problem and got the correct answer.

hope it helps.