The pressure at the mix point is equal to the pressure drop in the 4" pipe from the mix point to the open tank. Add the two flow rates together and calculate the pressure drop in that pipe.

In physics, Pascal’s principle states that given a fluid in a totally enclosed system, a change in pressure at one point in the fluid is transmitted to all points in the fluid, as well as to the enclosing walls. In other words, if you have a fluid enclosed in a pipe (with no air bubbles) and change the pressure in the fluid at one end of the pipe, the pressure changes all throughout the pipe to match.

Use the equivalent length method on engineering toolbox to estimate the pressure loss from point P1 to end of pipe:

engineering toolbox

Note that if you have pulsating flow from the positive displacement pump, it will be more difficult to estimate the pressure loss.

Let's say the discharge pressure of fluid 1 is 120 psi and the discharge at fluid 2 is about 30 psi. Will there be an issue of the fluid pumped from P2, since the discharge pressure of P2 is small compared to dicharge pressure at P1? Will there be a proper mix?

You're doing this wrong.

Start from the end point and work backwards.

In your example this is 0 barg.

Then do as don 1980 says and calculate pressure drop to your common point pm.then calculate back to your pump discharge.

Pd pumps pump a near constant flow with only the pressure which is needed. Your case is not valid.

If indeed type p2 pump is limited in pressure then for flow to occur pm needs to be lower than p2 max pressure.

This sounds like a theory test?

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