Its very basic question. But there is no data all over internet about it.

Let's see how we calculate mass flow rate through pipe.

We take velocity of fluid and c/s area of pipe. Then we find volume flow rate using continuity equation. After that we multiply it by density So far so good.

$$\frac{mass}{sec} = density \times area \times velocity$$

But which density?

The liquid is saturated (partly water and partly vapour with some dryness fraction)

Now if we see the density of water and water vapour at saturation point (100 celcius), there is a lot much difference.

  • for water its around 1000

  • for vapour its around 0.598

Both will give different answers.

This question is still haunting me even after my engineering is completed. Over the years, I have seen numericals using them both without any reasonable explaination. Help me understand this please! Let's say dryness fraction is 0.6

  • 1
    $\begingroup$ specify the dryness fraction and continue... $\endgroup$ – Solar Mike Mar 27 at 7:25
  • 1
    $\begingroup$ This is known as multi-phase flow. You will need to define the enthalpy/dryness of the saturated steam, since saturated water could be 100% liquid or gas or anywhere in between. But now that you have it defined you can research it more fully. This is graduate level engineering. $\endgroup$ – Tiger Guy Mar 27 at 7:52

If you interested in the mass rate of a partially full pipe at steady state , then things are very simple.

enter image description here

The total mass would be the sum of the mass rate

  • in the liquid cross-section
  • in the vapour cross-section

So total mass rate :

$$\dot{m}_{total} = \dot{m}_{liquid} + \dot{m}_{vapour}$$ $$\dot{m}_{total} = \rho_{liquid}\times A_{liquid} \times v_{liquid} + \rho_{vapour}\times A_{vapour} \times v_{vapour}$$

Just for comparison purposes, lets assume that the pipe is half full. In that case $A_{liquid} = A_{vapour}=\frac{\pi}{2} r^2 $. (Also, in most cases the velocity of the vapour and the liquid should not be all that different but lets not make that assumption here).

In that case you see that the mass rate is predominantly from the liquid. i.e.

$$\dot{m}_{total} = (\rho_{l}\times v_{l}+\rho_{v}\times v_v) \times A_{liquid}$$

$$\dot{m}_{total} = (1000\times v_{l}+0.6\times v_v) \times A_{liquid}$$

So you see even if you calculated only the mass rate of the liquid, you would only incur less than 0.1% error in the mass rate.

Other cases

Of course if the flow is not steady state, and you've got gas bubbles and pockets (see below), that is different ball game.

enter image description here

However, again if you could find a characteristic time that the flow can be considered periodic, you can define a control volume and again derive the same equation. If you are meticulous enough you should be able to derive the fluctuation of the mass rate within the characteristic time.

  • $\begingroup$ Very much thanks for such good answer.You said "if you calculated only the mass rate of the liquid, you would only incur less than 0.1% error in the mass rate" this is the statement I was looking for! Over the years I always felt something missing and now it became clear. $\endgroup$ – Niranjan Wagh Mar 27 at 10:45

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