1.

Calculate the rate of flow (volume per second) of a liquid through a capillary tube of diameter 0.20 xx 10^(-3) mand length 1m, eta = 3.00 xx 10^(-5) Nm^(-2)sand pressure gradient = 10 atm.

Answer»

SOLUTION :`R=(0.2 xx 10^(-3))/2 = 0.1 xx 10^(-3) m`
`=10^(-4) m, p=10 ATM = 10 xx 1.01 xx 10^(5) Pa`
`=1.01 xx 10^(6) Pa`
We have,
`eta = (pi PR^(4)t)/(8VI)`
`THEREFORE` rate of flow `=V/t = (pipr^(4))/(8eta l)`
`=((22//7) xx (1.01 xx 10^(6))(10^(-4))^(4))/(8 xx (3.0 xx 10^(-3))xx 1)`
`=1.32 xx 10^(-8) m^(5) s^(-1)`


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