1.

A venturimeter has a threat of diameter 0-06 m. The diameter of the horizontal pipeline where the venturimeter is inserted is 0-1 m. The pressure difference between the mainline and the throat is 0-32 of water. Calculate the rate of now of water in the pipeline. ?

Answer»

Solution :Rate of flow of water ` Q = a_1a_2sqrt((2(P_1+P_2))/(RHO ( a_1^(2) - a_2^(2))))`
` P_1- P_2 = HRHO g "" ` where` h= 0.32 m of " water ", rho = 10 ^(3) kg//m^(3)`
`a_1 = PI r_1^(2), ""r_1 = 0.05m "" a_1 =0.0025 pi m^(2)`
` a_2 = pi r_2^(2),"" r_2=0.03 m , "" a_2= 0.0009 pi m^(2)`
` "" Q= 0.0025 pi xx 0. 0009 pisqrt((2xx 0.32 xx 10 ^(3) xx 9.8)/(10^(3) xx pi (0.0025^(2) - 0.0009^(2))))`
` = 0.00762 m^(3)//s`


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