I have stuck with the engineering calculation for my work

Suppose that we have a underground pipeline with constant flow of liquid, which connected with closed underground vessel through the valve (vessel are full of same liquid).

In pipeline we know pressure P1, temperature T1, in vessel we know pressure P2, temperature T2, volume of liquid in vessel V0 (vessel is full). P1 > P2, T1 > T2. Pipeline and vessel are made of steel.

Let's assume that we opened the valve. Pressure in vessel will become equal to pressure in a pipeline, due to this, a little amount of liquid is added to the vessel, with the temperature T1 > T2. Now we have volume of liquid in a vessel V1 = V0 + Vadded.

Then we closed the valve. Due to temperature difference, small amount of added volume of liquid will start to cool down. Volume of added liquid will decrease slightly. When the cooling process is over, the pressure in the vessel will be set at the level of P3, which is P1 > P3 > P2. Volume of a liquid in a vessel V2 = V0 + Vadded - Vcooled.

The question is - how to calculate P3 if we know all of initial parameters (P1, P2, T1, T2, all geometric parameters, all liquid properties)?

P.S. In my calculations, I got that final pressure P3 is independent of the initial pressure in a vessel P2. And this result really confused me, so I want to ask.

Full description of analyzed system in numbers: Type of fluid in the pipeline and in the vessel - oil, density = 850 $kg/m^3$, coefficient of thermal expansion b = $8.31 * 10^{-4}$ $K^{-1}$, oil compressibility k = $7*10^{-10}$ $Pa^{-1}$. The Pipeline and vessel are made of steel, bulk modulus = $2 * 10^{11}$ Pa. Pressure in the pipeline P1 = 2 MPa, temperature of fluid in the pipeline T1 = 286 K. Initial pressure in the vessel P2 = -0.1 MPa, initial temperature of fluid in the vessel T2 = 280 K. Diameter of the pipeline = 1 meter, diameter of the vessel = 1 meters, length of pipeline = infinity, length of the vessel = 15 kilometers.

Sample image of system: enter image description here

My analysis: I have divided the process into three stages:

  1. Adding a volume of "warm" liquid to the vessel
  2. (after closing the valve) Reducing the volume of added "warm" liquid due to cooling,
  3. Reducing the pressure in the vessel due to volume reduction of "additional" liquid.

For each stage, I obtained the following ratios:

  1. Additional volume of "warm" liquid: $$\Delta V_{warm} = V_0*k*(P_1 - P_2)$$

  2. Reducing the volume of added "warm" liquid due to cooling: $$\Delta V_{cooling} = \Delta V_{warm}*b*(T_1 - T_2)$$

  3. Reducing the pressure in the vessel due to volume reduction: $$(P_3 - P_1)*k*\Delta V_{warm} = - \Delta V_{cooling}$$ then: $$P_3 = \frac{-\Delta V_{cooling} + \Delta V_{warm} * k * P_1}{\Delta V_{warm} * k }$$ then after substitution: $$P_3 = \frac{b*(T_2 - T_1) + k*P_1}{ k }$$ According to the ratio I obtained, $P_3$ does not depend on $P_2$, so I thought I was wrong.

  • $\begingroup$ This is a very imprecise description of what you are trying to analyze. Is the vessel of constant volume, or does it change also? Let's see your analysis so far. Is the total volume of fluid really changing, or is it just the density? Is the same fluid in the vessel as In the pipeline? What are the numbers, like the pressures, temperatures, bulk modulus, coefficient of thermal expansion, etc? $\endgroup$ Commented Jun 9, 2021 at 18:03
  • $\begingroup$ corrected the first message, added information $\endgroup$
    – 4lk4st
    Commented Jun 10, 2021 at 16:08
  • $\begingroup$ Did you really mean an initial value of P2 is -0.1 MPa? And did you really mean 15 km? $\endgroup$ Commented Jun 10, 2021 at 16:18
  • $\begingroup$ At first I wanted to write 0 MPa, but my pressure sensors just don't show a vacuum, and in fact there is a really small vacuum, about in between -0,1 and -0,3 MPa. And yes, in reality vessel is just a long pipeline with closed gate valve at the end, from 1 to 15 km $\endgroup$
    – 4lk4st
    Commented Jun 10, 2021 at 16:23
  • $\begingroup$ In other words, the vessel is not filled 100% with liquid? And, that 15 km length?? $\endgroup$ Commented Jun 10, 2021 at 16:25

1 Answer 1


Let's assume that the volume V of the vessel is constant (neglecting expansion of the vessel itself). Let $\rho_0$ be the density of the liquid at T2 and P2, and, at arbitrary P and T, the density is $$\rho=\rho_0\exp{\left[\frac{(P-P_2)}{B}-\alpha(T-T_2)\right]}$$So the mass of liquid 2 in the tank is always $M_2=\rho_0V$. After the pressure rises to P1, the volume of liquid 2 in the tank is $$V_2(P_1,T_2)=\frac{M}{\rho_0}\exp{\left[-\frac{(P_1-P_2)}{B}\right]}=V\exp{\left[-\frac{(P_1-P_2)}{B}\right]}$$ So the volume of fluid 1 which enters the tank is $$V\left(1-\exp{\left[-\frac{(P_1-P_2)}{B}\right]}\right)$$ and the mass of fluid 1 in the tank is $$M_1= V\left(1-\exp{\left[-\frac{(P_1-P_2)}{B}\right]}\right)\rho_0\exp{\left[\frac{(P_1-P_2)}{B}-\alpha(T_1-T_2)\right]}$$ So the total mass in the vessel is now:
$$V\rho_0+V\rho_0\left(1-\exp{\left[-\frac{(P_1-P_2)}{B}\right]}\right)\exp{\left[\frac{(P_1-P_2)}{B}-\alpha(T_1-T_2)\right]}$$In the final thermally equilibrated state, this is also equal to $$V\rho_0\exp{\left[\frac{(P_3-P_2)}{B}\right]}$$So setting these two expressions equal and linearizing with respect to the pressure and temperature changes, we have finally: $$P_3=P_1$$ The temperature difference effect is, of course, second order (since the only thing that changes temperature is the tiny amount of fluid that enters from the pipeline).


If we mathematically manipulate our final equation just a little bit, we obtain: $$\left[1-\exp{\left(-\frac{(P_1-P_2)}{B}\right)}\right][1-\exp{(-\alpha(T_1-T_2)}]=1-\exp{\left(\frac{P_3-P_1}{B}\right)}$$Expanding both sides of this equation in Taylor series and retaining only the lowest order terms on both sides yields: $$P_3=P_1-\alpha(P_1-P_2)(T_1-T_2)$$Evaluating this expression for the parameters that have be specified in the problem statement indicates that P3 is 0.01 MPa lower than P1.

  • $\begingroup$ Chet, honestly said, in practice P3 = P1 only in initial time. After 4-8 hours it is dramatically falls down. Perhaps this is because the temperatures in the pipeline and the vessel are always different? In my system, the vessel is located at a distance from the pipeline and its temperature is always lower than the temperature in the pipeline. $\endgroup$
    – 4lk4st
    Commented Jun 10, 2021 at 23:58
  • $\begingroup$ Is the final pressure closer to P2 or P1? $\endgroup$ Commented Jun 11, 2021 at 0:05
  • $\begingroup$ Based on the data you provided, the mass of liquid that enters the tank when the valve is opened is only 0.14% of the mass of liquid in the tank to begin with. And this is the only liquid that cools down. The rest of the liquid is already at T2. $\endgroup$ Commented Jun 11, 2021 at 2:13
  • $\begingroup$ $P_3$ is much closer to $P_2$. Yes, 0.14% of mass of liquid - i think it is correct. It turns out that this 0.14% cools down to T2 and all the pressure that the additional volume created falls down dramatically. $\endgroup$
    – 4lk4st
    Commented Jun 11, 2021 at 7:15
  • $\begingroup$ See the ADDENDUM for the analytic solution for the difference between P1 and P3. It shows that after the value is closed, the pressure should only drop by about 0.01 MPa. $\endgroup$ Commented Jun 11, 2021 at 11:51

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