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Let's say we have a pressurized horizontal tube that has a hole along the way. What would be the flow rate through that hole, and what would the pressure be downstream of the hole?

Or if you prefer to work with numbers, let's say the following:

  • Inlet Pressure: 5 barg
  • Diameter of main stream: 1 inch
  • Initial Main Flow velocity: 1 m/s
  • Hole Diameter: 5 mm

Unknowns:

  1. Flow rate through hole
  2. Flow rate downstream of the hole
  3. The pressure downstream of the hole

I'm aware that conservation of mass can be used to determine one flow rate based on the other, but I'm not sure how to get the flow rate through the hole, and I'm also not sure how to then calculate the pressure drop, since the Bernoulli equation that I'm familiar with predicts a pressure increase due to the decrease in flow speed.

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  • $\begingroup$ I don't see what the problem is. The Bernoulli equation ought to be adequate for this problem. $\endgroup$
    – Chet Miller
    Jan 3, 2020 at 21:45
  • $\begingroup$ @Chet Bernoullis equation predicts that the pressure in the pipe just downstream of the hole increases due to the smaller velocity. That makes no sense to me. $\endgroup$
    – Hani
    Jan 4, 2020 at 8:55
  • $\begingroup$ Sorry about that, but, to decelerate the flow velocity, you need a higher pressure downstream. You actually need to apply the Bernoulli equation twice, once for the flow through the hole and the other time for the flow continuing down the pipe. $\endgroup$
    – Chet Miller
    Jan 4, 2020 at 13:32

3 Answers 3

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Solution

I had to use a hole diameter of 1mm instead of 5mm, or the pressure loss would be too big

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Edit: Found the solution to this, which is to re-derive the Bernoulli equation from energy conservation. The confusion in the original Bernoulli equation is due to the fact that mass is cancelled out, when in this problem it shouldn't be

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    $\begingroup$ The Bernoulli equation is derived from conservation of energy for an incompressible fluid flow. There is no such thing as an original Bernoulli equation (emphasis on original). Perhaps you instead mean SIMPLIFIED not original. To complete this now that you have an answer, folks might also appreciate when you would post a picture of your system, the analysis that you did, and the answer that you finally found. $\endgroup$
    – Jeffrey J Weimer
    Jan 5, 2020 at 0:13
  • $\begingroup$ Apologies for the choice of words. I'll write down a solution once I get back home. $\endgroup$
    – Hani
    Jan 5, 2020 at 0:17
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Bernoulli's equation has to be applied to the flow as a whole. It can't be applied to individual parts of the flow. If you split the flow into two subflows, the total of all the energies of the two subflows will be equal to that of the original flow. Since the tube is horizontal, we can ignore gravitational potential energy. Thus we have that

$m_0(v_0^2+\frac {p_0}{\rho}) = m_1(v_1^2+\frac {p_1}{\rho})+m_2(v_2^2+\frac {p_2}{\rho})$

where $v_0$ is the velocity before the hole, $v_1$ is the velocity in the tube after the hole, and $v_2$ is the velocity of the fluid escaping the hole, likewise for $p_0, p_1, p_2$, and $m_0, m_1, m_2$ are weights representing the relative masses of the flows.

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  • $\begingroup$ How can your equation be valid? Wouldn't you need to account for the different mass flows of each stream, as I did above earlier? $\endgroup$
    – Hani
    Jan 5, 2020 at 8:59
  • $\begingroup$ Of course you would. This answer is not correct. $\endgroup$
    – Chet Miller
    Jan 5, 2020 at 13:38
  • $\begingroup$ @Hni Good point. With one flow, the masses cancel out, but with two you have to take them into account. $\endgroup$
    – Acccumulation
    Jan 5, 2020 at 19:40

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