Question 22 on the second afternoon of 2016
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The approach to solving this problem is to consider the pump’s head as being entirely used to overcome the pressure loss in the piping system, but no such information is provided in the question Excuse me, the experts in the group – how was this determined?
V=Q/S=(100/3600)/(3.14*0.075*0.075)=1.57 m/s; the flow rate is given in meters per second. The simplified formula is V=353.678*Q/D2=(353.678*100)/(150*150)=1.57 m/s. Here, the flow rate is in cubic meters per hour, and the diameter D is in millimeters
H=80-0.001Q2=80-0.001*100*100=70m. Hf=H=70m. Hf=λ*(L/D)*V2/2g=λ*（3500/0.15）*（1.57*1.57）/(2*9.81）=2931.4λ; λ=0.0239
After connection in parallel, the Hf value of the two tubes is the same. V1/V2 = sqrt(((4500/0.08) / (3500/0.15))) = 1.55; therefore V2 = 0.645 * V1. Q = Q1 + Q2 = (πD1²*V1/4 + πD2²*V2/4) = 3.14*0.15*0.15*V1/4 + 3.14*0.08*0.08*0.645*V1/4 = 0.0209*V1
H = 80 – 0.001Q2 = 80 – 0.001*(0.0209V1*3600)2 = 80 – 5.66*V1*V1.
Hf = λ*(L/D)*(V1)2/2g = 0.0239*(3500/0.15)*(V1*V1)/(2*9.81) = 28.4*V1*V1.
Since H = Hf, we have 80 – 5.66*V1*V1 = 28.4*V1*V1; therefore, V1 = 1.53 m/s.
Q = 0.0209V1*3600 = 0.0209*1.53*3600 = 115 m³/h
Hf = λ*(L/D)*(V1)2/2g = 0.0239*(3500/0.15)*(V1*V1)/(2*9.81) = 28.4*V1*V1.
Since H = Hf, we have 80 – 5.66*V1*V1 = 28.4*V1*V1; therefore, V1 = 1.53 m/s.
Q = 0.0209V1*3600 = 0.0209*1.53*3600 = 115 m³/h
The questions for registered chemical engineers are quite difficult
The things I learned are returned to the teacher if not used for a long time