Question

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series and a pump between them. The first pipe is 20 m long and has a 6-cm diameter, while the second pipe is 35 m long and has a 3-cm diameter. The water level in the reservoir is 30 m above the centerline of the pipe. The pipe entrance is sharp-edged, and losses associated with the connection of the pump are negligible. Neglecting the effect of the kinetic energy correction factor, determine the required pumping head and the minimum pumping power to maintain the indicated flow rate.

Answer 1

Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

Answer 2

The required pumping head is; 1309.18 m

The minimum pumping power to maintain the indicated flow rate is; 228.86 kW

We are given;

Discharge rate; Q = 18 L/s = 0.018 m³/s

Diameter of first pipe; d₁ = 6 cm = 0.06 m

Length of first pipe; l = 20 m

Diameter o second pipe; d₂ = 3 cm = 0.03 m

Length of second pipe; L = 35 m

• Formula for the Bernoulli energy equation here is;

$\frac{P_{1}}{\gamma} + \frac{V_{1}^{2}}{2g} + z_{1} = \frac{P_{2}}{\gamma} + \frac{V_{2}^{2}}{2g} + z_{2} + h_{i} - h_{pump}$

At initial conditions, V₁ = 0 and z₂ = 0; P₁ = P₂. Thus, we now have;

$z_{1} = \frac{V_{2}^{2}}{2g} + h_{i} - h_{pump}$

Making $h_{pump}$ the subject gives;$h_{pump} = \frac{V_{2}^{2}}{2g} + h_{i} - z_{1}$

• Now, for pipe flow in Pipe 1, formula for velocity V₁ is;

V₁  = Q/A₁

Thus; V₁ = 0.018/(π × 0.06²/4)

V₁ = 6.366 m/s

Now, Reynolds number is calculated from;

R_e = ρVd/μ

where;

R_e is Reynolds number

ρ is density of water at 15°C = 999.1 kg/m³

μ is dynamic viscosity of water at 15°C = 1.138 × 10⁻³ kg/m·s

Thus;

R_e = (999.1 × 6.366 × 0.06)/(1.138 × 10⁻³)

R_e ≈ 335339

Relative pipe roughness is gotten from the formula;

Relative pipe roughness = ε/d

where ε is roughness of cast iron pipes = 0.00026 m.

Thus;

Relative pipe roughness = 0.00026/0.06 = 0.0043

To get the head for pipe 1, we will use the formula;

$h_{1} = \frac{V_{1}^{2}}{2g}(k_{L} + \frac{fL}{d})$

where;

$k_{L}$ is loss coefficient for a sharp-edged entrance = 0.5

f is friction factor

L is length of pipe

d is diameter

From moody's chart attached, at R_e ≈ 335339 and relative pipe roughness of 0.0043, we have;

friction factor; f = 0.0294

Thus;

$h_{1} = \frac{6.366^{2}}{2*9.8}(0.05 + \frac{0.0294*20}{0.06})$

h₁ = 21.3 m

• For Pipe 2;

formula for velocity V₁ is;

V₂  = Q/A₂

Thus; V₂ = 0.018/(π × 0.03²/4)

V₂ = 25.46 m/s

Reynolds number; R_e = (999.1 × 25.46× 0.03)/(1.138 × 10⁻³)

R_e ≈ 6705573

Relative pipe roughness = 0.00026/0.03 = 0.0087

From moody's chart attached, at R_e ≈ 6705573 and relative pipe roughness of 0.0087, we have;

friction factor; f = 0.033

Thus head in pipe 2 is;

$h_{2} = \frac{25.46^{2}}{2*9.8}(0.05 + \frac{0.033*35}{0.03})$

h₂ = 1289.81 m

Total head; h_i = h₁ + h₂

h_i = 21.3 m +  1289.81 m

h_i = 1311.11 m

Recall that our pumping head formula gotten earlier is;

$h_{pump} = \frac{V_{2}^{2}}{2g} + h_{i} - z_{1}$

Thus;

$h_{pump} = \frac{25.46^{2}}{2*9.8} + 1311.11} - 35$

$h_{pump} = 1309.18 m$

Minimum pumping power is gotten from the formula;

P = Q * ρ * g * h_pump

P = 0.018 * 991 * 9.8 * 1309.18

P = 228,860.9 W

P = 228.86 kW

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