written 6.7 years ago by | modified 2.8 years ago by |
written 2.8 years ago by | • modified 2.8 years ago |
Given,
$μ = 1.0\ poise = 0.1\ Ns/m^2$
$s = 0.9\ \implies ρ = 900\ kg/m^3$
$L = 50m$
$D = 300mm \implies 0.03m$
$R = D/2 = 0.015m$
$Pressure\ Drop\ = 20KN/m^2$
(i) Mass flow rate (m)
Q = ūA
$ū = \frac{(P_1-P_2)D^2}{32ūL}$
$(P_1-P_2) = \frac{32ūμL}{D^2} = \frac{32\times0.1\timesū\times50}{0.03m^2}$
$\implies 20,000 = 5333.33ū$
$ū = 3.75m/s$
$A = \frac{π}{4}D^2 \implies \frac{π}{4} \times 0.03^2 \implies A = 0.00070685m^2$
$Q = 3.75 \times 0.00070685 = 0.002650m^3/s$
$m = Q\timesρ = 0.002650 \times 900 = 2.385 kg/sec = 143.1kg/min$
Ans.
(ii) Shear stress at pipe wall (τ)
$τ = \frac{-∂P }{∂x} \times \frac{R}{2} $
$\frac{-∂P }{∂x} = \frac{-(P_2-P_1)}{x_2-x_1} = \frac{20,000}{50} = 400N/m^2$
$τ = 400 \times \frac{0.015}{2} = 3N/m^2$
Ans.
(iii) Reynolds (R)
$R = \frac {ρvD}{μ} = \frac{900\times3.75\times0.03}{0.1} = 1012.5$
Ans.
(iv) Power required per 50m length (P)
$P = W \times hf $
$W = Q \times ρg \implies 0.002650 \times 900 \times 9.81 = 23.39 kg-m/s^2$
$hf = \frac{32μūL}{ρgD^2} \implies \frac{32 \times 0.1 \times 3.75 \times 50}{900 \times 9.81 \times (0.03)^{2}}$
$hf = 75.50 m$
$P = 23.3968 \times 75.50 = 1766.66 N$