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Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-15: Discussion question: 327. If the mass flow rate of steam is M=100 t/h, what should be the inner diameter D of the pipeline? If the density of the steam is ρ=38 kg/m3, what is the average flow velocity of the steam in the pipe? (5 points are given only to those who provide the correct answer; additional points are awarded for a correct calculation process or detailed explanation.)
s=(M*1000/38 /3600)/*1000000=23.28m/s
s=(M*1000/38 /3600)/*1000000=23.28m/s
s = (m*1000/38 / 3600) = (100*1000/38/3600) = 23.28 m/s; the average speed is 23.08 m/s.
s=(M*1000/38 /3600)/*1000000=23.28m/s
s = (m*1000/38 / 3600) = (100*1000/38/3600) = 23.28 m/s; the average speed is 23.08 m/s.
s=(M*1000kg/h/38kg/m3 /3600s)/*1000∧-2m2=23.28m/s
If the mass flow rate of steam is M=100 t/h, the selected inner diameter of the pipe is D=200 mm. If the density of the steam is ρ=38 kg/m3, what is the average flow velocity of the steam in the pipe? Solution: The relationship between mass flow rate and volume flow rate is: mass flow rate = density multiplied by volume flow rate (M = ρQ). Therefore, Q = M/ρ = 100*1000/38 = 2631.58 m3/h = 0.731 m3/s. The cross-sectional area of the pipe, S, is equal to 3.14*(D/2)2 = 3.14*0.01 = 0.0314 m2 ; The average flow velocity of the steam in the pipe is u = Q/S = 0.731/0.0314 = 23.28 m/s. Answer: The average flow velocity of steam in the pipeline is 23.28 m/s.
s = (m*1000/38 / 3600) = (100*1000/38/3600) = 23.28 m/s; the average speed is 23.28 m/s.
The relationship between mass flow rate M and volume flow rate Q is M = ρQ; therefore, Q = M/ρ. Hence, Q = 100 / ρ = 2631 kg/m3. The cross-sectional area of the pipe, S, is given by S = π/4 · D2 = π/4 (0.2)2 = 0.0314 m2. Since Q = SU, the average velocity of the steam, U, is calculated as U = Q/S = 0.731 / 0.0341 = 23.3 m/s
s=(M*1000/38 /3600)/*1000000=23.28m/s