모두CBT
수량 20m3/h20m^3/h20m3/h를 양수하는데 필요한 펌프의 구경은? (단, 양수펌프 내 유속은 2m/s2m/s2m/s로 한다.)
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Q=20 m3/h=203600=0.00556 m3/sQ = 20\ \text{m}^3/\text{h} = \dfrac{20}{3600} = 0.00556\ \text{m}^3/\text{s}Q=20 m3/h=360020=0.00556 m3/sQ=Av ⇒ A=Qv=0.005562=0.00278 m2Q = A v \;\Rightarrow\; A = \dfrac{Q}{v} = \dfrac{0.00556}{2} = 0.00278\ \text{m}^2Q=Av⇒A=vQ=20.00556=0.00278 m2d=4Aπ=4×0.00278π≈0.0595 m=59.5 mmd = \sqrt{\dfrac{4A}{\pi}} = \sqrt{\dfrac{4 \times 0.00278}{\pi}} \approx 0.0595\ \text{m} = 59.5\ \text{mm}d=π4A=π4×0.00278≈0.0595 m=59.5 mm → 약 60mm
Q=20 m3/h=203600=0.00556 m3/sQ = 20\ \text{m}^3/\text{h} = \dfrac{20}{3600} = 0.00556\ \text{m}^3/\text{s}Q=20 m3/h=360020=0.00556 m3/s
Q=Av ⇒ A=Qv=0.005562=0.00278 m2Q = A v \;\Rightarrow\; A = \dfrac{Q}{v} = \dfrac{0.00556}{2} = 0.00278\ \text{m}^2Q=Av⇒A=vQ=20.00556=0.00278 m2
d=4Aπ=4×0.00278π≈0.0595 m=59.5 mmd = \sqrt{\dfrac{4A}{\pi}} = \sqrt{\dfrac{4 \times 0.00278}{\pi}} \approx 0.0595\ \text{m} = 59.5\ \text{mm}d=π4A=π4×0.00278≈0.0595 m=59.5 mm → 약 60mm
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