Q1: Consider the following manometer: When flow through a pipe is suddenly restr

Question

Q1:

Consider the following manometer: When flow through a pipe is suddenly restricted, as in this diagram, there is a pressure drop due to this increase in velocity. In this diagram, let us assume that there is a fluid with density d_1 = 1.36 g/cm^3 flowing through this system, and that we can measure the pressure drop with this manometer that contains mercury, with a density d_2 = 13.55 g/cm^3. Let us also assume that h_1 = 8.0m and that h_2 = 5.2m. You can also assume that the atmospheric pressure for the day is 745 mm Hg. Calculate the pressure drop, P_A - P_B, in Pa.

Explanation / Answer

Using the equation ,pressure (p)=height (h)*density (d)*acceleration due to gravity(g)

d1=1.36 g/cm3=1.36 *10^-3 kg/10^-6 m3=1360kg/m3

d2=13.55 g/cm3=13.55 *10^-3 kg/10^-6 m3=13550 kg/m3

patm=745 mmhg *101.325 kpa/760mmhg=99.325kpa

pressure at point A=pA=-(pressure at point2 applied by fluid with density d1)=-(h1*d1*g)

pressure at point B=pB=-(pressure at point 4 ) =p(atmospheric)-(pressure of mercury at height h2)=patm - h2*d2*g

pA-pB=-h1*d1*g-patm+h2*d2*g=-(8.0m)*(1360 kg/m^3)*(9.8 m/s2)-99.325kpa+(5.2m)*(13550 kg/m3)*(9.8 m/s2)=-106624kpa-99.325kpa+690508kpa=583784.675 kpa

The pressure drop=583784.675 kpa

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