Pressure Drop Testing - Too Many Unknowns?

Hello there,

I am having a discussion with a co-worker about a pressure drop test. I have two different 'boxes' for which I don't know the pressure drop. What I want to know is the difference in pressure drop between the two 'boxes'.

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Knowns
1) P_A1 = P_A2 = P_A = Inlet Pressure for Setup 1 and 2 = Measured (Same for both setups)
2) F_A1 = F_B1 = F₁ = Flow Rate for Setup 1 = Measured (Same throughout setup 1)
3) F_A2 = F_B2 = F₂ = Flow Rate for Setup 2 = Measured (Same throughout setup 2)

Unknowns
1) P_B1 = Pressure at Outlet in Setup 1
2) P_B2 = Pressure at Outlet in Setup 2

What I want to find out
1) P_B1 - P_B2 = ΔP_B = ???

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The first step I took is to use the equation:

[itex]\frac{F_1}{F_2}[/itex] = [itex]\sqrt{\frac{P_{B1}}{P_{B2}}}[/itex] (Eq. 1)

And simplified it to:

[itex]∆P_B=P_{B1}-P_{B2}=P_{B1}-P_{B1}(\frac{F_2}{F_1})^2[/itex] (Eq. 2)

So, at this point, I have two unknowns (ΔP and P_B1) and only one equation. I am claiming that we have to measure and/or solve for either P_B1 (or P_B2) in order to find an answer. My co-worker is claiming that we do NOT have to find either of these values. He wants to use another Equation 1, or something along those lines.

However, I feel you can’t use Equation 1 to compare different points in a system. For instance, Equation 3 below is NOT a usable equation.

[itex]\frac{F_{A1}}{F_{B1}}[/itex] = [itex]\sqrt{\frac{P_{A1}}{P_{B1}}}[/itex] (Eq. 3)

Since F_A1 = F_B1; this is meaningless, right? Or am I missing something?



Thanks in advance!
Alan