Temperature after closing the valve

In summary: Delta m)\ln{m_2}$$or$$\frac{m_2}{m_1}\gamma(\Delta m)$$$$T_2=\frac{371,3}{448}K$$See my corrected version.
  • #1
hilfethermo
4
0
I have the following task and I am not getting anywhere:

A steel tank with 15 m³ volume, well insulated from the outside, contains air at a temperature of 288 K and a pressure of 1 bar. It is connected to a compressed air line via a valve that is initially blocked. which constantly supplies air at a temperature of 310 K and a pressure of 5 bar. The valve is opened so that air flows into the tank. As soon as the pressure in the tank reaches 3 bar the valve is closed again.

Air can be considered as an ideal gas with constant specific heat capacities: R = 287 J/kgK; κ = 1.4
The steel tank has a mass of 1300 kg and a specific heat capacity of cT = 448 J/kgK. The mass and heat capacity of its insulation can be neglected.

What is the temperature in the tank immediately after closing the valve if the air supply is so fast that there is no heat exchange between the air and the tank wall?My approach:
I tried to calculate with isentropic change of state but get wrong result, the solution is T2 = 371,3 K. Can help me further.
 
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  • #2
Are you familiar with the open system version of the 1st law of thermodynamics?
 
  • #3
Chestermiller said:
Are you familiar with the open system version of the 1st law of thermodynamics?
Yes I am familiar with it.

I am thinking for the problem above it will look like this:
dU = h * dm, with dU = m * cv * dT

But here T2 and m2 are unknown to solve the task. Or did I use the wrong equation?
 
  • #4
hilfethermo said:
Yes I am familiar with it.

I am thinking for the problem above it will look like this:
dU = h * dm, with dU = m * cv * dT

But here T2 and m2 are unknown to solve the task. Or did I use the wrong equation?
Very close. dU=mdu+udm=hdm

and ##du=C_vdT##
$$mC_VdT+udm=u_0dm+Pvdm$$or$$mC_vdT=C_v(T_0-T)dm+RT_0dm$$or$$C_VdT=(C_pT_0-C_vT)d\ln{m}$$or$$\frac{dT}{\gamma T_0-T}=d\ln{m}$$where ##T_0=350\ K## and R=8.314 J/mole.K. Also, $$m=\frac{PV}{RT}$$
 
Last edited:
  • #5
Chestermiller said:
Very close. dU=mdu+udm=hdm

and ##du=C_vdT##
Thank you, now we can replace h with Cp * T, but how should I calculate dm ?
 
  • #6
I meant To is 310, not 350.
 
  • #7
Chestermiller said:
I meant To is 310, not 350.
Thank you for your help. Appreciate it
 
  • #8
hilfethermo said:
Thank you for your help. Appreciate it
See my corrected version.
 
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  • #9
This problem can also be solved using the closed system version of the 1st law of thermodynamics it we take as our system the moles of gas initially in the tank ##m_1## plus the number of moles allowed to enter from the compressed air line ##\Delta m=(m_2-m_1)##, where ##m_2## is the final number of moles in the tank, with
$$m_1=\frac{P_1V}{RT_1}$$
$$m_2=\frac{P_2V}{RT_2}$$where P1 = 1 bar and P2 = 3 bars
From the first law of thermodynamics, we have $$U_2-U_1=W$$where W is the work done by gas in the compressed air line in forcing $\Delta m$ moles of gas ahead of it into the tank. In this equation, $$U_1=m_1C_V(T_1-T_{ref})+(\Delta m)C_V(T_0-T_{ref})$$
$$U_2=m_2C_V(T_2-T_{ref})$$and $$W=(\Delta m)RT_0$$where ##T_{ref}## is a reference temperature of zero internal energy datum which drops out of the calculation. Combining the previous 4 equations, we have $$m_2C_VT_2-m_1C_VT_1=(\Delta m)C_pT_0$$or$$m_2T_2-m_1T_1=(m_2-m_1)\gamma T_0$$
 

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