4. Problem 4 a) Determine the Delta p = pA- pB across the restriction of the pip

Question

4. Problem 4 a) Determine the Delta p = pA- pB across the restriction of the pipe in the figure below, knowing that the manometer is filled with water (p water= 1000 kg/m^3) and the fluid flowing in the duct is oil (poil = 912 kg/m^3). Additional data are listed here: ? zA = 1 m ? z1= 0.5 m ? z2= 1.5 m ? zB=2 m b) How could the resulting relation for Delta p be simplified if the fluid in the duct was air at standard conditions and the fluid in the manometer was mercury (p mercury = 13534 kg/m^3)?

Explanation / Answer

Pa + rho(oil)*g*(za-z1) + rho(water)*g*z1 - rho(water)*g*z2 - rho(oil)*g*(zb-z2) = Pb

Pa-Pb = rho(water)*g*z2 + rho(oil)*g*(zb-z2)- rho(oil)*g*(za-z1) - rho(water)*g*z1

Pa-Pb = rho(water)*g*(z2- z1) + rho(oil)*g*(zb-za + z1-z2)

Pa-Pb = 1000*10*(1) + 912*10*(1 -1 )

Pa-Pb = 10000 Pa

Now

Pa-Pb = rho(mer)*g*z2 + rho(air)*g*(zb-z2)- rho(air)*g*(za-z1) - rho(mer)*g*z1

Since

Rho(mercury) > Rho(air)

Rho(air) = 0

Pa-Pb = rho(mer)*g*(z2-z1)

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