Respuesta :
Explanation:
The given data is as follows.
Air is at [tex]75^{o}F[/tex] and 14.6 psia.
[tex]\varepsilon[/tex] = 0.00015 ft, Â Â Flow rate, (Q) = 48000 [tex]ft^{3}/m[/tex]
(a) Â Formula to calculate hydraulic radius [tex](r_{H})[/tex] is as follows.
       [tex]r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}[/tex]
             = [tex]\frac{2 \times 1}{2(1) + 2(2)}[/tex]
             = [tex]\frac{1}{3}[/tex] ft
Formula for equivalent diameter is as follows.
           [tex]D_{eq} = 4 \times r_{H}[/tex]
                  = [tex]4 \times \frac{1}{3} ft[/tex] Â
                  = [tex]\frac{4}{3}[/tex] ft
(b) Â Â Formula for velocity floe is as follows.
             Q = VA
           V = [tex]\frac{Q}{A}[/tex]
            = [tex]\frac{48000}{2 \times 1}[/tex] ft/min
            = 24000 ft/min
(c) Â Formula to calculate Reynold's number is as follows.
     [tex]R_{e}[/tex] = [tex]\frac{D \times V \times \rho}{\mu}[/tex]
          = [tex]\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}[/tex]  (as [tex]\rho = 0.0744 lb/ft^{3}[/tex] and [tex]\mu[/tex] = 0.0443 lb/ft. hr)
          = 53742.66 hr/min Â
As 1 hr = 10 min. So, [tex]53742.66 hr/min \times \frac{60 min}{1 hr}[/tex]
              = 3224559.6
(d) Â Formula to calculate pressure drop [tex](\Delta P)[/tex] is as follows.
       [tex]\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}[/tex]
Putting the given values into the above formula as follows.
        [tex]\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}[/tex]
           = [tex]\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}[/tex]
           = 6.238 [tex]lb/ft^{2}[/tex]
