Question

.35m Smooth Pipe - 5си Q Open Jet e- 60m T 10 M Water K 45m The smooth pipe flow given in the below figure is driven by pressurized air above the water surface in the tank. What gage pressure pz is needed to provide a 20°C water flow rate Q= 50 m3/h? The pipe outlet is open to atmosphere. Use explicit Haaland formulation to find the friction factor. Take viscosity of water as u=1x10-3kg/ms, density p=1000kg/m3 and gravity as 9.81m/s2. Assume that width of the tank is much larger than the diameter of the pipe. Minor Loss Type KL Sharp Entrance 0.50 90° Elbow 0.95

Answer 1

Q = 50 m3/h or 0.01388888 m3/s

d = 5 cm or 0.05 m

V = Q / A = 0.01388888 / (π/4) 0.052 = 7.07355 m/s

By Haaland equation

1/sqrt(f) = - 1.8 log10 [{(e/d) / 3.7}1.11 + 6.9 / Re]

Re = ρ V d / μ = 1000 * 7.07355 * 0.05 / 1 * 10-3 = 3.53677 * 105

e/d = 0 (for smooth pipe)

1/sqrt(f) = - 1.8 log10 [ 0 + 6.9 / 3.53677 * 105 ]

sqrt(f) = 0.1179584

f = 0.34345

hf = f L V2 / 2 g d

hf = 0.34345 * 140 * 7.073552 / 2 * 9.81 * 0.05

hf = 2452.4344 m

Sharp entrance

h1 = k V2 / 2 g

h1 = 0.5 * 7.073552  / 2 * 9.81

h1 =1.2751 m

2 900 elbow

h2 = n k V2 / 2 g

h2 = 2 * 0.95 * 7.073552  / 2 * 9.81

h2 = 4.84539 m

HL = hf + h1 + h2

HL = 2452.4344 + 1.2751 + 4.84539

HL = 2458.55 m

b is 35m Snooth Pipe d=5cm Q Open Jet e 60M 1 10M а Water K 45m

By bernoulli equation

Pa + 0.5 ρ Va2/2g + ρ g za = Pb + 0.5 ρ Vb2/2g + ρ g zb + ρ g HL

Va = sqrt (2gh) = sqrt (2 * 9.81 * 10) = 14.00714 m/s

za = 0

Pb = 101325 N/m2

zb = 60 m

Pa + 0.5 * 1000 * 14.007142/ 2 * 9.81 + 0 = 101325 + 0.5 * 1000 * 7.073552 / 2 * 9.81 + 1000 * 9.81 * 60 + 1000 * 9.81 * 2458.55

Pa = 24804575.604 N/m2

P1 + Pw = Pa

Pw = ρ g h = 1000 * 9.81 * 10 = 98100‬ N/m2

P1 = 24804575.604 - 98100‬

P1 = 24706475.6 N/m2 atm

P1 = 24706475.6 - 101325

P1 = 24605150.6 N/m2 gauge