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

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.

              $r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

                          = $\frac{2 \times 1}{2(1) + 2(2)}$

                          = $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

                     $D_{eq} = 4 \times r_{H}$

                                    = $4 \times \frac{1}{3} ft$  

                                    = $\frac{4}{3}$ ft

(b)    Formula for velocity floe is as follows.

                         Q = VA

                     V = $\frac{Q}{A}$

                        = $\frac{48000}{2 \times 1}$ ft/min

                        = 24000 ft/min

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

         $R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

                   = $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$  (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

                   = 53742.66 hr/min  

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

                            = 3224559.6

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

              $\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

               $\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

                      = $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

                      = 6.238 $lb/ft^{2}$