Question

Water at 60.0°F (p =62.4 lb/ft2) is flowing through a pipe at steady state. A differential manometer attached to the pipe the height of the manometer fluid is 48 inches. measures a pressure drop of 3.55 lb/in2 when (17 pts.) What is the specific gravity (60.0°F/60.0°F) of the manometer fluid? а. b. (6 pts.) What is the pressure drop in millimeters of mercury (SG=13.6)?

Answer 1

Step-by-step explanation:

The given data is as follows.

Temperature =
`10^(o)F`

Density (
`\rho`) = 62.4
`lb/ft^(2)`

Pressure drop = 3.55
`lb/in^(2)`

= 511.6
`lb/ft^(2)` (as 1
`lb/in^(2) = 144 lb/ft^(2)`)

Height of manometer = 48 inch = 4 ft (as 1 inch = 0.0833)

g = 9.81
`m/s^(2)`

= 32.18
`ft/s^(2)` (as 1
`m/s^(2)m = 3.280 ft/s^(2)`)

(a) Specific gravity of monometric fluid
`(S_(G))` will be calculated as follows.

`S_(G) = (\rho_(m))/(\rho_(soln))`

where,
`\rho_(m)` = density of manometric fluid

`\rho_(soln)` = density of standard fluid (water) = 1000 g/ml

So,
`S_(G) = (\rho_(m))/(1000)`

`\Delta P = \rho_(m) * g * h_(m)`

where,
`h_(m)` = height of manometric fluid

`\rho_(m)` = density of manometric fluid

Now, putting the given values into the above formula as follows.

`\Delta P = \rho_(m) * g * h_(m)`

511.6
`lb/ft^(2)` =
`\rho_(m) * 32.18ft/s^(2) * 4 ft`

`\rho_(m)` = 3.97
`lb/ft^(3)`

Therefore, calculate the value of
`S_(G)` as follows.

`S_(G) = (\rho_(m))/(1000)`

=
`(3.97 lb/ft^(3))/(1000)`

=
`3.97 * 10^(-3)`

And, for mercury
`\Delta P = \rho_(m)gh`

`\Delta P` =
`849 lb/ft^(3) * 32.18 ft/s^(2) * 4 ft`

= 109283.28
`lb/ft^(2)`

= 39246.99 mm Hg (1
`lb/ft^(2)` = 0.3591 mm Hg)

Thus, we can conclude that the pressure drop in millimeters of mercury is 39246.99 mm Hg.