# Bernoulli's Equation for flow of gas and changing area

My task is to determine the size of piping needed in a system. In this system, hydrogen gas flows horizontally in a $\frac{3}{4}\,\mathrm{inch}$ pipe and into a vertical pipe of a larger diameter. My constraint is that the velocity of the hydrogen cannot exceed $1\,\mathrm{m/s}$ once it expands and goes up the vertical tube.

|
| (2)
|
|__________________ (1) <- Hydrogen Gas


The pressure in the horizontal tubing at 1 is $2\,\mathrm{psig}$ and the pressure in the vertical tubing 2 is $0\,\mathrm{psig}$ (open to atmosphere).

The 2 equations I attempted to use were:

1. Bernoulli's equation:$$\frac{P_1}{\rho} + \frac{1}{2} V_1^2 + gh_1 = \frac{P_2}{\rho} + \frac{1}{2} V_2^2 + gh_2$$

2. equation of continuity:$$\rho_1A_1V_1 = \rho_2A_2V_2$$

Using Bernoulli's equation, I receive a very large negative root or a velocity of about ~550m/s in section 1 which seems very ridiculous. Is there a better suited equation for this application? The goal is to determine the size of piping needed for section 2.

• Psig is psi gauge, it is in reference to atmospheric pressure. Because the tube reads 2 psig, it is 2 psi above atmospheric pressure. – AkzoNorman Sep 24 '15 at 14:35
• @AkzoNorman: Bernoulli's equation is only applicable to imcompressible fluids, not gases. Have a look at this: en.wikipedia.org/wiki/… – Gert Sep 24 '15 at 15:55
• @Gert Most gases at low Mach numbers can be modeled as incompressible. Moving at 1 m/s is most decidedly low Mach and so it is not an issue to model this as an incompressible fluid (and note -- fluid means liquid, gas, or plasma). – tpg2114 Sep 24 '15 at 15:56
• @AkzoNorman Have you ever heard of choked nozzles? Your very large velocity would indicate that the incoming pipe has to have supersonic flow to meet your maximum velocity in the much larger pipe. This is pretty common in experimental setups. – tpg2114 Sep 24 '15 at 15:58
• I have heard of choked nozzles but have not done any real work or calculations with them. Is the large number I am getting the actual velocity needed for 1 m/s maximum velocity in the larger pipe or am I applying the wrong equation to this scenario? (I used Bernoulli's still because it was a very low velocity) – AkzoNorman Sep 24 '15 at 16:06

$$u_2 = \frac{q_2}{A} = \frac{(p_1 - p_2)R^2}{8 \mu_{av} L} \frac{z_2(p_1+p_2)}{2z_{av}p_2}$$
where $\frac{z_2(p_1+p_2)}{2z_{av}p_2}$ is a correction factor that accounts for the gradual expansion of the gas when going from the high pressure inlet to the low pressure outlet. See this link for the compressibility correction factor derivation: http://oaktrust.library.tamu.edu/bitstream/handle/1969.1/ETD-TAMU-2010-12-8739/LING-DISSERTATION.pdf. view pages 26 through 34 (of 243).
Fig. 2 shows the modulus of the current velocity (left), the maximum output velocity depending on time and the distribution $$p+\frac {1}{2}\rho \vec {v}^2$$ at the last moment. It can be seen that the Bernoulli integral is not conserved due to the large density and pressure gradient in the thin pipe.