Thermodynamics-Two Insulated Tank System

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In the discussion about a thermodynamic problem involving two insulated tanks filled with air, participants analyze the effects of equalizing mass and pressure after opening a connecting valve. The initial conditions include one tank at 1 bar and the other at 10 bar, both at 300 K. Calculations show that the mass in tank one is 18.24 kg and in tank two is 182.4 kg, leading to a final equal mass of 100.32 kg in each tank. Key points include the need to consider pressure equalization rather than mass and the importance of using energy balance and heat capacity equations to determine the final temperatures. The conversation emphasizes the complexity of the calculations and the assumptions required for accurate results.
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Homework Statement



Two large, well-insulated tanks of 2m diameter and 5m in height are filled with air. One tank is initially at 1bar and 300oK and the other is at 10bar at 300oK. Determine the final temperature in both tanks if a valve connecting the two tanks is opened and the mass equalized (mf1=mf2) very quickly then closed again. Assume the enthalpy of the inlet and exit air for each tank is constant at 300.19kJ/kg. (Hint: One tank is ~198oK higher than the other after the valve shuts)

Homework Equations



PV=mRT
H=U+PV=U+RT


The Attempt at a Solution



I've used PV=mRT for both tanks to find the mass in tank 1 is 18.24kg while the mass in tank 2 is 182.4 kg. This gives a collective mass of 200.64kg. As stated in the problem statement the final mass in both tanks is equal which would then be 100.32 kg. From there I am not quite sure how to incorporate enthalpy into finding the final temperature in each tank?
 
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mpn17 said:

Homework Statement



Two large, well-insulated tanks of 2m diameter and 5m in height are filled with air. One tank is initially at 1bar and 300oK and the other is at 10bar at 300oK. Determine the final temperature in both tanks if a valve connecting the two tanks is opened and the mass equalized (mf1=mf2) very quickly then closed again. Assume the enthalpy of the inlet and exit air for each tank is constant at 300.19kJ/kg. (Hint: One tank is ~198oK higher than the other after the valve shuts)

Homework Equations



PV=mRT
H=U+PV=U+RT


The Attempt at a Solution



I've used PV=mRT for both tanks to find the mass in tank 1 is 18.24kg while the mass in tank 2 is 182.4 kg. This gives a collective mass of 200.64kg. As stated in the problem statement the final mass in both tanks is equal which would then be 100.32 kg. From there I am not quite sure how to incorporate enthalpy into finding the final temperature in each tank?

(1) It is not the mass, but the pressure in both tanks that will be equal at the end of the operation. Remember that the temperature will be different in each of the tanks.

(2) I cannot see how to do this calculation without assuming that air is (to a good approximation) made up of diatomic molecules, and introducing the heat capacity equation.
 
my professor mentioned being able to obtain the mass flow rate from the problem statement and then using an energy balance to then find the final temperature in each tank. He also said the pressure is not equal in both tanks.
 
mpn17 said:
my professor mentioned being able to obtain the mass flow rate from the problem statement and then using an energy balance to then find the final temperature in each tank. He also said the pressure is not equal in both tanks.

An interesting idea, and I am not convinced that it is a correct one. A pressure difference is a plausible driving force for transfer of material to equalize pressures while the tap is open; a mass difference is not.
 
H=U+PV
Given P,V,H
U only depends on the temperature assuming Ideal Gas Model
U=Cv(T2-T1)
((H-PV)/Cv)+T1=T2
Cv=Specific Temp of air (volume)
 
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