In the following series of questions:

"Under ordinary operation a steady flow of helium gas of 60 g/s passes from a large storage manifold through an expansion engine as shown in Figure. There are certain times when the engine must be shut down for short intervals, but not the inlet flow rate, however, flow rate can be decreased at such times and thus helium stream must be diverted. It is proposed to employ adjacent system that consists a large insulated tank (tank C) of 6 m3 capacity with a safety valve venting to the atmosphere through the plant ducting system.

Figure 1

The emergency diverting system is operated in the following manner. If the expansion engine must be shut down (or operates improperly), valve B is shut and A is opened, letting helium into tank. When the pressure in C raises to 3.5 bar, safety valve D operates, venting gas. Initially tank C is evacuated to very low pressures. We shall assume that when gas enters in this tank, it is well mixed but has negligible heat transfer with the tank walls. Assume that helium is an ideal gas with a constant heat capacity.

(a). Write the suitable final form of energy balance equation for the process occurring in control volume C. How long will flow enter tank C before the pressure reaches to 3.5 bar and the safety valve opens. What is the temperature and mass of helium in this condition?"

For this part using the First Law of Thermodynamics for a Transient Process, I obtained that: $ u_2 = h_i $ where $ u_2 $ = Internal Energy of tank at when P = 3.bar. And hence, $ T_2 = 566.667K $, and $ m_2 = 1.7841Kg $.

In the next part,

(b) If the safety valve on tank C should operate, we would like to maintain a constant mass of gas in this tank equal to the mass at the time the safety tripped. The mass flow rate into the tank is, as noted above, 60 g/s, the flow rate from the safety valve is expressed as, $$ \dot m(g/s)= \frac{KAP}{T} $$ Where, K is a constant, A is the throat area of the valve, P and T are pressure and temperature in tank C. To keep the tank mass constant, the valve flow area A will be varied. What are the pressure and temperature of the tank C 10 s after the safety valve opens? Write the suitable final energy equation for this process. Write the expression for Area “A “to keep the mass of the tank C constant and find change in A with time.

Again using the Transient flow equation we get that,

$$ \frac{dU_{cv}}{dt} = \dot m_ih_i - \dot m_e h_e $$ and since $\dot m_e = \dot m_i = \dot m$, and $ U_{cv} = mC_{v}T_{cv}$, $ h_{e} = C_{p}T_{cv} $ after simplification and rearranging:

$$ \int_{T2}^{T2'} \frac{dT_{cv}}{T_i - T_{cv}} = \int_0^{10} \frac {\dot m}{m} \frac{C_p}{C_v}dt $$

which on solving gave $$ T_2' = 469.4086K, P_2 = 289.9287kPa $$

The following part has:

(d) Find the total entropy change of the gas entering in tank, tank, gas leaving to surroundings, and the universe during the time between the opening of the valve D and a time 10 s later Write a suitable equation for finding entropy change for the above mentioned cases. The vented gas mixes with an infinite amount of air exterior to the tank and cools to 300 K and 1 bar. Assume the entropy change of mixing is negligible.

I used the Second Law for a control volume for adiabatic conditions to get,

$$ \frac{dS_{cv}}{dt} = \dot m_i s_i - \dot m_e s_e + \dot S_{gen} $$

However I have no idea how to proceed further as $ s_e $ is a function of T and I can't get a formulation for the same.

  • $\begingroup$ Yes you can. By doing that, you are assuming that the gas in the tank is well-mixed. $\endgroup$ Commented May 1, 2020 at 11:29
  • $\begingroup$ @ChetMiller actually that did not work out. Instead wrote U as function of $T_2'$, and $m_eh_e$ as well, and integrated wrt it. Now the problem is that I need to find the entropy change of the surroundings for (a) and then for (b) and for that I tried using 2nd Law for a control Volume transient process. And found that I need the value of $s_i$ for it, and have no way to calculate that. What do you suggest? $\endgroup$ Commented May 1, 2020 at 12:25

1 Answer 1


Actually, I would have written $$m\frac{du}{dt}=\dot{m}(h_e-h)$$or$$mC_v\frac{dT}{dt}=\dot{m}C_p(T_{in}-T)$$I guess that's the same thing you ended up with.

In terms of the entropy change, you've already calculated the entropy change of the gas in the tank, right? And the entropy change of the gas that has exited the tank is determined by its cumulative number of moles, and its final temperature and pressure of 300 K and 1 bar. To get the entropy change of the surrounding air, you need to determine the amount of heat transferred between the exit helium and the surrounding air. Do you know how to calculate the average temperature of the exit gas? And, based on that, do you know how to calculated the heat transferred?


For the average temperature of the helium exiting the tank during the 10 seconds, I get $$\bar{T}=T_{in}+(T_1-T_{in})\left[\frac{1-e^{-x}}{x}\right]$$where $$x=\frac{m\gamma t}{\dot{m}}$$This comes out to 513.5 K. The gas then cools to 300 K. The amount of heat transferred from this 150 moles of He to the surrounding air is $$Q=150 C_p(513.5-300)$$The heat transfer causes an increase in the entropy of the surrounding air by Q/300.

  • $\begingroup$ No the entropy part is not working out, I am getting an integral of $ln(T_i + (T_2 - t_i)e^{-\frac{t \gamma \dot m}{m}} $ from 0 to 10 $\endgroup$ Commented May 1, 2020 at 16:02
  • $\begingroup$ If looks to me like a math issue. I'll add some to my answer a little later. For the temperature at time t, I get $$T=T_{in}+(T_1-T_{in})e^{-\frac{\gamma \dot{m}t}{m}}$$ where Tin is 340 and T1 is 567. This should match your 469. Do we agree so far? $\endgroup$ Commented May 1, 2020 at 16:14
  • $\begingroup$ See my EDIT in the Answer. $\endgroup$ Commented May 1, 2020 at 18:04

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