Question

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).

Answer 1

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.