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.

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.

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

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.

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.

To calculate the minimum volume of the reservoir tank required to hold the overflow of water, you need to consider the increase in volume of the pipe due to thermal expansion.

1. Begin by calculating the volume of the pipe:

Given:
Inside radius of the pipe, r = 9.10 * 10^-3 m
Length of the pipe, L = 74 m

The volume of a cylindrical pipe is given by V = πr^2L.

V = π * (9.10 * 10^-3 m)^2 * 74 m = 1.9354 * 10^-2 m^3

2. Calculate the change in volume of the water in the pipe due to the temperature increase from 20 to 75°C:

Given:
Thermal expansion coefficient of water, α = 4 * 10^-4 (1/°C)
Temperature change, ΔT = 75°C - 20°C = 55°C

The change in volume is given by ΔV = V * α * ΔT.

ΔV = 1.9354 * 10^-2 m^3 * 4 * 10^-4 (1/°C) * 55°C

ΔV = 0.042388 m^3

3. Find the minimum volume of the reservoir tank to hold the overflow of water:

Since the change in volume is the overflow that needs to be accommodated by the tank, the minimum volume of the reservoir tank should be equal to ΔV.

Minimum volume of the reservoir tank = ΔV = 0.042388 m^3

So, the minimum volume of the reservoir tank required to hold the overflow of water is 0.042388 m^3.