Question

Problem 1. Find the pipe diameter D for the system shown in Figure 1. Water at 20°C is being pumped. (Hint: Write the energy equation between points 1 and 2 and solve it for D. However, the equation for D contains f which is a function of D. The pump head hP can be computed from the pump characteristic curve data. You can use Matlab or excel solver to solve for D). (22 - 2) = 45 m, L=0.50 km, e=0.9 mm, EK = 3.0, Q=1.2 m/s Pump characteristic curve is given below (hp is in meters and Q is in L/s): he =-8.036x10-6Q -0.0006429Q+159.8 (RⓇ = 0.9985)

Answer 1

Ans) Apply Bernoulli equation between point 1 and 2 respectively,

P1/$\gamma$ + V1^2/2g + Z1 + Hp = P2/$\gamma$ + V2^2/2g + Z2 + Hf + Hm

Since, both point 1 and 2 are open to atmosphere, pressure is only atmospheric,hence gauge pressure P1 = P2 = 0

Also, there is no velocity at surface , so V1= V2 = 0

Hp is pump head

Hf is head loss due to friction

Hm is minor loss

Hence, above equation reduces to,

Z1 + Hp = Z2 + Hf + Hm

Given , elevation, Z2 - Z1 = 45 m

=> Hp = 45 + Hf + Hm.......................................(1)

Also, frictional loss, Hf = 8 f L Q^2 / ($\pi$^2 g D^5)

where, f = Friction factor

L = Length of pipe = 0.5 km or 500 m

D = Pipe diameter

Since, friction factor is not known,solution is iterative ,so assume initial value as f = 0.02 to began iteration

=> Hf = 8(0.02)(500)(1.2)^2 / ($\pi$^2 x 9.81 x D^5)

=> Hf = 1.19/D^5

Also, Minor loss, Hm = $\sum$ k Q^2 / (2 g A^2)

Area of pipe = ($\pi$/4) D^2 = 0.785 D^2

=> Hm = 3(1.2)^2 / (2 x 9.81 x 0.785^2 x D^4)

=> Hm = 0.36/ D^4

According to question, pump characteristic equation,

Hp = -8.036 x 10^(-6) Q^2 - 0.0006429 Q + 159.8

Putting Q = 1.2 m^3/s or 1200 L/s

=>Hp = -8.036 x 10^(-6) (1200)^2 - 0.0006429 (1200) + 159.8

=> Hp = 147.45 m

Putting values in equation 1,

=> 147.45 = 45 + (1.19 / D^5) + (0.36/D^4)

=> 102.45 = (1.19 / D^5) + (0.36/D^4)

On solving above equation, D = 0.420 m

Iteration 1 :

Now,calculate improved value of friction factor using value of diameter calculated above

Flow velocity = Q / A = 1.2 / [($\pi$/4)(0.42)^2)] = 8.66 m/s

=> Reynold number, Re = V D / $\nu$

Kinematic viscosity of water = 10^(-6) m^2/s

=> Re = 8.66 x 0.42 / 10^(-6)

=> Re = 3.63 x 10^6

Roughness,e = 0.9 mm (given)

=> Relative roughness = e/D = 0.9 / 420 = 0.0021

According to Moody diagram, for Re = 3.63 x 10^6 and e/D = 0.0021, friction factor,f = 0.024

=> Hf = 8(0.024)(500)(1.2)^2 / ($\pi$^2 x 9.81 x D^5)

=> Hf = 1.43/D^5

Putting values in equation 1,

=> 147.45 = 45 + (1.43 / D^5) + (0.36/D^4)

=> 102.45 = (1.43 / D^5) + (0.36/D^4)

On solving above equation, D = 0.435 m

Iteration 2 :  

Now,calculate improved value of friction factor using value of diameter calculated above

Flow velocity = Q / A = 1.2 / [($\pi$/4)(0.435)^2)] = 8 m/s

=> Reynold number, Re = V D / $\nu$

=> Re = 8 x 0.435 / 10^(-6)

=> Re = 3.48 x 10^6

According to Moody diagram, for Re = 3.48 x 10^6 and e/D = 0.0021, friction factor,f = 0.024

Since, the friction factor comes out to be same as above iteration, Diameter of pipe will remain same as calculated in above iteration. Hence, D = 0.435 m