A calculation problem from 2010

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15. The methanol feed is pumped from a storage tank to the reactor; the flow rate of methanol is 90 m3/h, the inner diameter of the pipe is 150 mm, the equivalent length is 80 m, and the friction coefficient λ = 0.018. The reactor in the factory has been modified, resulting in a doubling of its production capacity. There is an urgent need to add a pipeline capable of transporting the same volume of methanol. It is planned to use two pipelines currently available in the warehouse, with inner diameters of 200 mm and 100 mm respectively, connected in series. If the friction coefficients of the DN200 and DN100 pipelines are 0.02 and 0.016 respectively, then what should be the equivalent length (in meters) of the DN200 pipeline required to maintain the same pressure loss as that of the new and old pipelines? (A)51 (B)61 (C)71 (D)81 Solution: u=4Q/3.14d²=4×(90/3600)/(3.14×0.15²)=1.415 (m/s) ∑hf=(λLe/d)u²/2g=0.018×80×1.415²/(0.15×2×9.81)=0.98 (m) u₁=4Q/3.14d₁²=4×(90/3600)/(3.14×0.22)=0.7958 (m/s) u₂=4Q/3.14d₂²=4×(90/3600)/(3.14×0.12)=3.183 (m/s) ∑hf=(λ₁Le₁/d₁)u₁²/2g+(λ₂Le₂/d₂)u₂²/2g=0.98 (m) Le₁+Le₂=800.02×Le₁×0.7958²/(0.2×2×9.81)+ 0.016×(80-Le₁)×3.183²/(0.1×2×9.81)= 0.98 Le₁≈71 (m) Answer: C I would like to ask fellow enthusiasts why Le₁+Le₂=80? ? How did this equation come about? What is the principle? Thank you!