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Water flows through a horizontal tube as shown in figure. If the difference of height of water column in the vertical tubes in 2cm and the areas of corss-section at A and B are 4 cm^(2) respectively. Find the rate of flow of water across any section. . |
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Answer» `p_(A)+(1)/(2)rho v_(A)^(2)+rhogh_(A)=p_(B)+(1)/(2)rhov_(B)^(2)+rho gh_(B) (h_(A)=h_(B))` `or, (1)/(2)rho v_(B)^(2)-(1)/(2)rho v_(A)^(2)=rhog (h_(A)=h_(B))` or `v_(B)^(2)-v_(A)^(2)=2g (h_(A)=h_(B))` `=2xx10xx0.02` or `v_(B)^(2) -v_(A)^(2) = 0.4 m^(2)//s^(2)`....(i) `v_(A)A_(A)=v_(B)A_(B)` or, `4v_(A)=2v_(B)` `:. v_(B)=2v_(A)` ...(ii) SOLVING EQS, (i) and (ii), we get `v_(A)=0.363 m//s` VOLUME Flow rate`=v_(A)A_(A)` `=(0.365)(4xx10^(-4))` `=1.46xx10^(-4) m^(3)//s=146cm^(3)//s`. |
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