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).

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:

[tex]P = \rho g h_1[/tex]

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

[tex]\rho = 13600 kg/m^{3}[/tex]

[tex]g = 9.8 m/s^{2}[/tex]

So, replace in the equation:

[tex]P = (13600 kg/m^{3} )(9.8 m/s^{2})(800 mmHg)(\frac{1 mHg}{1000 mmHg})[/tex]

[tex]P = 106624.0 \frac{kg}{s^{2}} = 106624.0 Pa[/tex]

Transforming from Pa to mmHg

[tex]P = 106624.0 Pa (\frac{760 mmHg}{101325 Pa}) = 799.7 mmHg[/tex]

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

[tex]P = \rho g h_2 + P_atm[/tex]

Isolate [tex]P_atm[/tex]:

[tex]P_atm = P - \rho g h_2[/tex]

Calculating:

[tex]P_atm = 799.75 mmHg - (13600 kg/m^{3} )(9.8 m/s^{2})(25 mmHg)(\frac{1 mHg}{1000 mmHg})(\frac{760 mmHg}{101325 Pa} )[/tex]

[tex]P_atm = 774.75 mmHg[/tex]

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

[tex]P_gau = P - Patm[/tex]

[tex]P_gau = 799.75 mmHg - 774.75 mmHg[/tex]

[tex]P_gau = 24.99 mmHg[/tex]

End.