Chapter 3 – What Causes Surge?
Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Chapter 3 – Problem 3.7
3.7 Resistance to flow K is defined as the pressure headloss ∆H across the device divided by the square of the flowrate Q, all parameters in standard English or SI units. K = ∆H/Q2.
a. Consider a 3m long 1000 mm internal diameter pipe section. Compute the resistance(s) of this pipe section if its friction is defined by a Hazen-William roughness coefficient of 140 for three different flowrates of 0.5 m3/s, 1.5 m3/s, and 2.5 m3/s.
b. Compute the resistance(s) for the same pipe section and the same set of flowrates if the pipe friction is defined by a constant Darcy-Weisbach friction factor of 0.012.
c. Compute the resistance of a 200 mm globe valve with a fixed minor loss coefficient m of 10. What is the (pressure) headloss across the valve at a flow velocity of 2 m/s?
d. Compute the resistance of a 200 mm butterfly valve with a fixed minor loss coefficient m of 0.2. What is the (pressure) headloss across the valve at a flow velocity of 2 m/s?
e. Compute the resistance of a 2000 mm butterfly valve with a fixed minor loss coefficient m of 0.2. What is the (pressure) headloss across the valve at a flow velocity of 2 m/s?
Solution:
a. Consider a 3m long 1000 mm internal diameter pipe section. Compute the resistance(s) of this pipe section if its friction is defined by a Hazen-William roughness coefficient of 140 for three different flowrates of 0.5 m3/s, 1.5 m3/s, and 2.5 m3/s.
Length of pipe section L = 3m, diameter D = 1m, and Hazen-William roughness coefficient C = 140.
For Q = 0.5 m3/s, ∆H = (10.67 L Q1.852) / (C1.852 D4.87) = 0.00094m, K = 0.00094/0.52 = 0.00376
For Q = 1.5 m3/s, ∆H = (10.67 L Q1.852) / (C1.852 D4.87) = 0.0072m, K = 0.0072/1.52 = 0.003196
For Q = 2.5 m3/s, ∆H = (10.67 L Q1.852) / (C1.852 D4.87) = 0.0185m, K = 0.0185/2.52 = 0.00296
b. Compute the resistance(s) for the same pipe section and same set of flowrates if the pipe friction is defined by a constant Darcy-Weisbach friction factor of 0.012.
Length of pipe section L = 3m, diameter D = 1m, and Darcy-Weisbach friction factor f = 0.012.
For Q = 0.5 m3/s, ∆H = (f L Q2) / (2 g (π/4)2 D5) = 0.00074m, K = 0.00074/0.52 = 0.00298
For Q = 1.5 m3/s, ∆H = (f L Q2) / (2 g (π/4)2 D5) = 0.0067m, K = 0.0067/1.52 = 0.00298
For Q = 2.5 m3/s, ∆H = (f L Q2) / (2 g (π/4)2 D5) = 0.0186m, K = 0.0186 /2.52 = 0.00298
c. Compute the resistance of a 200 mm globe valve with a fixed minor loss coefficient m of 10. What is the pressure headloss across the valve for a flow velocity of 2 m/s?
Pressure headloss across the valve is given by ∆H = m V2/2g = m Q2 / (2g ((π/4)D2)2)
Valve resistance K = ∆H / Q2 = m / (2g ((π/4)D2)2) = 516.418
Flow velocity of 2 m/s implies a flowrate of ((π/4)0.22)2)*2 = 0.0628 m3/s and ∆H = K Q2 = 2.038m
d. Compute the resistance of a 200 mm butterfly valve with a fixed minor loss coefficient m of 0.2. What is the pressure headloss across the valve for a flow velocity of 2 m/s?
Pressure headloss across the valve is given by ∆H = m V2/2g = m Q2 / (2g ((π/4)D2)2)
Valve resistance K = ∆H / Q2 = m / (2g ((π/4)D2)2) = 10.328
Flow velocity of 2 m/s implies a flowrate of ((π/4)0.22)2)*2 = 0.0628 m3/s and ∆H = K Q2 = 0.041m
e. Compute the resistance of a 2000 mm butterfly valve with a fixed minor loss coefficient m of 0.2. What is the pressure headloss across the valve for a flow velocity of 2 m/s?
Pressure headloss across the valve is given by ∆H = m V2/2g = m Q2 / (2g ((π/4)D2)2)
Valve resistance K = ∆H / Q2 = m / (2g ((π/4)D2)2) = 0.001033
Flow velocity of 2 m/s implies a flowrate of ((π/4)0.22)2)*2 = 6.283 m3/s and ∆H = K Q2 = 0.041m
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