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Two mercury manometers, one open-end and the other sealed-end, are attached to an air duct. The reading on the open-end manometer is 25 [mm] and that on the sealed-end manometer is 800 [mm]. Determine the absolute pressure in the duct, the gauge pressure in the duct, and the atmospheric pressure, all in (mm Hg).

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Answer:

Pressure in duct = 799.75 mmHg

Atmospheric pressure = 774.75 mmHg

Gauge pressure = 24.99 mmHg

Step-by-step explanation:

First of all, it is needed to set a pressure balance (taking in account that diameter of manometer is constant) in the interface between the air of the duct and the fluid mercury.

From the balance in the sealed-end manometer, we have the pressure of air duct as:


P = \rho g h_1

We have that ρ is density of mercury and g is the gravity


\rho = 13600 kg/m^(3)


g = 9.8 m/s^(2)

So, replace in the equation:


P = (13600 kg/m^(3) )(9.8 m/s^(2))(800 mmHg)((1 mHg)/(1000 mmHg))


P = 106624.0 (kg)/(s^(2)) = 106624.0 Pa

Transforming from Pa to mmHg


P =  106624.0 Pa ((760 mmHg)/(101325 Pa)) = 799.7 mmHg

From the balance in the open-end manometer, we have the pressure of air duct as:


P = \rho g h_2 + P_atm

Isolate
P_atm:


P_atm = P - \rho g h_2

Calculating:


P_atm = 799.75 mmHg - (13600 kg/m^(3) )(9.8 m/s^(2))(25 mmHg)((1 mHg)/(1000 mmHg))((760 mmHg)/(101325 Pa) )


P_atm = 774.75 mmHg

Finally, gauge pressure is the difference between duct pressure and atmospheric pressure, so:


P_gau = P - Patm


P_gau = 799.75 mmHg - 774.75 mmHg


P_gau = 24.99 mmHg

End.

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