Question

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2 ft cross section. . a hydraulic ughness factor for the duct, is 0.00015 ft. Determine: raulic radius; and equivalent diameter b. velocity of flow c. Reynolds number d. pressure drop per 100 ft of flow I nine line of constant diameter of

Answer 1

Step-by-step explanation:

The given data is as follows.

Air is at
`75^(o)F` and 14.6 psia.

`\varepsilon` = 0.00015 ft, Flow rate, (Q) = 48000
`ft^(3)/m`

(a) Formula to calculate hydraulic radius
`(r_(H))` is as follows.

`r_(H) = \frac{\text{free flow area}}{\text{wet perimeter}}`

=
`(2 * 1)/(2(1) + 2(2))`

=
`(1)/(3)` ft

Formula for equivalent diameter is as follows.

`D_(eq) = 4 * r_(H)`

=
`4 * (1)/(3) ft`

=
`(4)/(3)` ft

(b) Formula for velocity floe is as follows.

Q = VA

V =
`(Q)/(A)`

=
`(48000)/(2 * 1)` ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

`R_(e)` =
`(D * V * \rho)/(\mu)`

=
`((4)/(3) * 24000 * 0.0744)/(0.0443)` (as
`\rho = 0.0744 lb/ft^(3)` and
`\mu` = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So,
`53742.66 hr/min * (60 min)/(1 hr)`

= 3224559.6

(d) Formula to calculate pressure drop
`(\Delta P)` is as follows.

`(\Delta P)/(L) = (4f \rho V^(2))/(2Dg_(c))`

Putting the given values into the above formula as follows.

`(\Delta P)/(L) = (4f \rho V^(2))/(2Dg_(c))`

=
`\frac{4 * 0.00015 * 100 * 0.0744 * (24000)^(2)}{2 * (4)/(3) * {4}{3}}`

= 6.238
`lb/ft^(2)`