Respuesta :
Explanation:
The given data is as follows.
Temperature = [tex]10^{o}F[/tex]
Density ([tex]\rho[/tex]) = 62.4 [tex]lb/ft^{2}[/tex]
Pressure drop = 3.55 [tex]lb/in^{2}[/tex]
= 511.6 [tex]lb/ft^{2}[/tex] (as 1 [tex]lb/in^{2} = 144 lb/ft^{2}[/tex])
Height of manometer = 48 inch = 4 ft (as 1 inch = 0.0833)
g = 9.81 [tex]m/s^{2}[/tex]
= 32.18 [tex]ft/s^{2}[/tex] (as 1 [tex]m/s^{2}m = 3.280 ft/s^{2}[/tex])
(a) Specific gravity of monometric fluid [tex](S_{G})[/tex] will be calculated as follows.
[tex]S_{G} = \frac{\rho_{m}}{\rho_{soln}}[/tex]
where, [tex]\rho_{m}[/tex] = density of manometric fluid
[tex]\rho_{soln}[/tex] = density of standard fluid (water) = 1000 g/ml
So, [tex]S_{G} = \frac{\rho_{m}}{1000}[/tex]
[tex]\Delta P = \rho_{m} \times g \times h_{m}[/tex]
where, [tex]h_{m}[/tex] = height of manometric fluid
[tex]\rho_{m}[/tex] = density of manometric fluid
Now, putting the given values into the above formula as follows.
[tex]\Delta P = \rho_{m} \times g \times h_{m}[/tex]
511.6 [tex]lb/ft^{2}[/tex] = [tex]\rho_{m} \times 32.18ft/s^{2} \times 4 ft[/tex]
[tex]\rho_{m}[/tex] = 3.97 [tex]lb/ft^{3}[/tex]
Therefore, calculate the value of [tex]S_{G}[/tex] as follows.
[tex]S_{G} = \frac{\rho_{m}}{1000}[/tex]
= [tex]\frac{3.97 lb/ft^{3}}{1000}[/tex]
= [tex]3.97 \times 10^{-3}[/tex]
And, for mercury [tex]\Delta P = \rho_{m}gh[/tex]
[tex]\Delta P[/tex] = [tex]849 lb/ft^{3} \times 32.18 ft/s^{2} \times 4 ft[/tex]
= 109283.28 [tex]lb/ft^{2}[/tex]
= 39246.99 mm Hg (1 [tex]lb/ft^{2}[/tex] = 0.3591 mm Hg)
Thus, we can conclude that the pressure drop in millimeters of mercury is 39246.99 mm Hg.
