Question

NOTE: Make sure to use consistent units. If measurements are given in Sl units, report your answers in SI units. Pressures should be reported in psi for English units and in kPa for S bo on In the following system (Questions 1-4), 568 liters/min of Monoethanolamine at 20°C (SG 1.018, Kinematic Viscosity 2.4 x 10 m2/s) are flowing at Point 1. Monoethanolamine is used to remove sulfur and carbon dioxide from gas streams. At Point 1 and Point 2, the pipe is 5 in. schedule 80 steel pipe. In branch A and branch B, the pipe is 2.5 in. schedule 80 steel pipe. Both branches are 4.6 meters long. The valve shown in branch B is a half open gate valve. The elbows that are part of branch A and branch B are standard elbows. US 1. Which branch would you expect to experience a higher volume flow rate? Why? (2 Points) 2. If the flow in each branch is in the completely turbulent region, do you need to guess friction factor or can you find it directly? (1 Point) Assuming that flow through this system is completely turbulent (it's not, but you can assume it is for this question), find the pressure drop between Point 1 and Point 2.(10 Points) bau 3. 4. If the valve in branch B were to be completely closed, what would be the flow rate in each branch? (2 Points) ou need to pump 53 gallons/min of water at 110'F up from Tank 1 and into Tank 2, as shown in the gure below. The pipe is shown with a dashed line because the pipe size is unknown. Based on your fi available electrical power, you expect the pump to add 21.3 ft-lb/b of head to the fluid. Tank 2 23 ft Tank 1ーーーーゴPump- 33 ft 49 ft total magnitude of the losses between Tank 1 and Tank 2? (2 points) s. What is the 6. You plan t the only losses in the system (ignore elbows, ent used? (5 Points) o complete this project using schedule 40 steel pipe . Assuming that friction losses are rances, exits, etc.J, what size pipe should be

Answer 1

1. A higher volume flow rate will be in pipe A. As pipe B has half open gate valve, it will reduce the volume flow rate in the pipe B even though the diameters of both the pipes are same (2.5 in). Cross sectional area available to flow in pipe B is smaller than available in pipe A, which results in decrease in volume flow rate in pipe B.
2. If the flow in each branch is completely turbulent, then we can find the value of friction factor directly based on the Reynolds Number using following equation.

f = 0.0791 / (Re)^1/4

As per given conditions for branch A and B,

Total Discharge, Q = Q_A + Q_B

Discharge, Q = 150 gal/min = 0.00946353 m³/s

Q_A = 2 * Q_B ( Branch B has half open gate valve)

Q = (2 * Q_B) + Q_B

Q_B = Q/3 = 0.00946353/3 = 0.00315451 m³/s

Therefore, Q_A = 2 * 0.00315451 = 0.00630902 m³/s

Reynolds number in branch A, Re_A = ( V_A* D_A ) / ʋ

Where, V_A = Velocity of fluid in branch A = Q_A / Area

Discharge, Q_A = 0.00630902 m³/s

Diameter of branch A, D_A = 2.5 in = 0.0635 m

Area = (л/4) * D_A² = (л/4) * 0.0635² = 3.16692*10^-3

Kinematic viscosity, ʋ = 2.6*10^-4 ft²/s = 2.415*10^-5 m²/s

L = 15 ft = 4.572 m

Therefore, V_A = Q_A / Area = 0.00630902/3.16692*10^-3 = 1.992 m/s

Re_A = ( V_A* D_A ) / ʋ = (1.992 * 0.0635 ) / 2.415*10^-5 = 5238.1918

The flow is turbulent and friction factor for branch A will be,

f_A = 0.0791 / (5238.1918)^1/4 = 0.00929

f_A = 0.00929

Reynolds number in branch B, Re_B = ( V_B* D_B ) / ʋ

Where, V_B = Velocity of fluid in branch B = Q_B / Area

Discharge, Q_B = 0.00315451 m³/s

Diameter of branch B, D_B = 2.5 in = 0.0635 m

Area = (л/4) * D_B² = (л/4) * 0.0635² = 3.16692*10^-3

Therefore, V_B = Q_B / Area = 0.00315451 /3.16692*10^-3 = 0.996 m/s

Re_B = ( V_B* D_B ) / ʋ = (0.996 * 0.0635 ) / 2.415*10^-5 = 2619.095

The flow is turbulent and friction factor for branch B will be,

f_B = 0.0791 / (2619.095)^1/4 = 0.011

f_B = 0.011

3. Pressure drop between point 1 and 2. (∆P)

Total head loss between 1 and 2, H

H = major loss + minor loss = h_f + h_m

Major loss in branch A and B,

h_f = [f_A * L * V_A² ] / ( 2 * g * D_A ) + [f_B * L * V_B² ] / ( 2 * g * D_B )

h_f = [0.00929 * 4.572 * 1.992 ² ] / ( 2 * 9.81 * 0.0635 ) + [0.011 * 4.572   * 0.996 ² ] / ( 2 * 9.81 * 0.0635 )

h_f = 0.1753 m

Minor loss in branch A and B,

Minor loss coefficients:

Standard Elbow, K_e = 0.9

Half open gate valve, Kv = 5.6

h_m = [ ( 2 * K_e * V_A² ) / (2*g) ]_A + [ ( 2 * K_e + Kv ) * (V_B² / (2*g)) ]_B

h_m = [ ( 2 * 0.9 * 1.992 ² ) / (2*9.81) ]_A + [ ( 2 * 0.9 + 5.6 ) * (0.996² / (2*9.81))]_B

h_m = 0.7381 m

therefore,

Total head loss between 1 and 2,

H = h_f + h_m = 0.1753 + 0.7381 = 0.91345 m

H = 2.996 ft

Therefore, pressure drop between 1 and 2,

∆P = ρ * g * H = ( 1.018*1000) * 9.81 * 0.91345 = 9122.2415 Pa

∆P = 9122.2415 Pa = 9.122 kPa

∆P = 1.3230 psi

4. If the valve in branch B is completely closed then discharge in branch A will equal to total discharge coming from pipe 1

Q_A = 150 gal/min