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Calculate the rate of flow of glycerine ...

Calculate the rate of flow of glycerine of density `1.25xx10^(3) kg m^(-3)` through the conical section of a pipe, if the radii of its ends are`0.1` m and `0.04` m and the pressure drop across its length is `10 Nm^(-2)`.

Text Solution

Verified by Experts

The correct Answer is:
6.53

Using Bernoulli.s theorem for horizontal flow, `p_1+1/2 rhov_1^2=p_2+1/2rhov_2^2`
or `v_2^2-v_1^2=(2(p_1-p_2))/(rho)`
`=(2xx10Nm^(-2))/(1.25 xx10^3 "kgm"^(-3))=16 xx10^(-3)`
According to equation of continuity, `a_1v_1 = a_2 v_2`
`:. v_1/v_2=(a_2)/(a_1)=(pir_2^2)/(pir_1^2)=r_(2)^2/r_(1)^2=((0.04)^2)/((0.1)^2)=16xx10^(-2)`
`v_1 = 16xx10^(-2) v_2`
Hence `v_2^(2) -(16 xx10^(-2)v_2)^(2)=16 xx10^3`
or `v_2^2[1-(256)/10000]=16xx10^(-3)`
or `v_2^2xx(9744)/(10000) = 16 xx10^(-3)`
or `v_2^(2) = (160)/9744 ` or `v_2 = 0.13 "ms"^(-1)`
Rate flow of glycerine `= a_(2) v_(2) = 3.14 xx (0.04)^2 xx 0.13 = 6.53xx10^(-4) m^(3)s^(-1)`.
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