|Nov2-12, 02:27 PM||#1|
Heat Transfer in thermos
Im trying to figure out how to calculate the pressure necessary to create a 30 degree celsius temperature drop in water, by using a small pressurized cylinder of air within the water. The volume of the water is .00075 m^3 and the volume of the canister is .00025 m^3. I calculated the amount of energy lost by the water to be 94.05 KJ. But im not sure how to calculate the corresponding pressure necessary in the pressurized air container to cause this temperature change in the water.
These are the values i was using:
P1,air= 3bar (not sure if its enough)
I wasn't sure what values had to be assumed and but i assumed:
t=180seconds (for the air to escape the canister)
ρair= 1.204 kg/m^3
Any help is appreciated
|Nov3-12, 11:04 AM||#2|
If the water, as you have calculated, would 'lose' 94.05 kJ to drop in temperature by 30 degrees C, then something else has to 'gain' 94.05 kJ of energy.
You could start by asking yourself how much air ( mass ) at what temperature could cool 0.75kg of water by from 293 to 263 degrees C. Subsequently, you could calculate the pressure of that mass of air within the canister.
Also, are you considering that a release of air to atmosphere will cause the pressure of air within the canister to drop and thus lower the temperature of air within the canister? It seems that you have a T1,air=293K.
Note that if so there will be less mass of air within the canister to 'gain' 94.05kJ of energy.
|Nov4-12, 11:06 AM||#3|
I used Q=mCΔT to find the mass of air required to cool the water (both at room temperature).
However the air will be escaping at a certain flow rate. Does this mean i have to assume a flow rate and time to reach P2,air = atm ?
Also, the material and surface area of the container would impact the rate of heat transfer. This means i would have to use Fourier's law but i'm not sure to incorporate it into the equation.
|Nov5-12, 02:57 AM||#4|
Heat Transfer in thermos
If you are saying that air in the cannister is escaping at some flow rate, then the figure 94.05KJ has to be associated with some time term.
(mcΔT/t)water = (mcΔT/t)air
Thereby, there will be a time in which water has to cool by 30degress and the same time must be used to calculate the flow rate of air.
There are two unknowns, the time for heat transfer and the final temp of air.
as you said assuming t = 150s
0.75*4180*30/150 = m * 0.0003 * ΔT/t
After temperature is calculated pressure can be calculted.
m/t on RHS will give mass flow rate of air
It seems it has to be a numerical method with trial and error.
taking m = 0.0003kg will give very large values of ΔT which are not practical.
So I think there must be some more constriants to the problem.
|control volume, heat transfer, ideal gas law, thermo|
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