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प्रश्न
Derive Poiseuille’s formula for the volume of a liquid flowing per second through a pipe under streamlined flow.
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उत्तर
Consider a liquid flowing steadily through a horizontal capillary tube. Let v = `("V"/"t")` be the volume of the liquid flowing out per second through a capillary tube. It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient `("P"/"l")`.
Then, `"v" ∝ η^"a""r"^"b"("P"/"l")^"c"`
v = `"k"η^"a""r"^"b"("P"/"l")^"c"` .............(1)
where, k is a dimensionless constant.
Therefore, [v] = `"volume"/"time" = ["L"^3"T"^-1], ["dP"/"dx"] = "Pressure"/"distance"`
`["Ml"^-2"T"^-2], [η] = ["M"^-1"T"^-1]` and `["r"] = ["L"]`
Substituting in equation (1)
`["L"^3"T"^-1] = ["ML"^-1"T"^-1]^"a" ["L"]^"b" ["ML"^-2"T"^-2]^"c"`
`"M"^0"L"^3"T"^-1 = "M"^("a" + "c") "L"^(-"a" + "b" - 2"c") "T"^(-"a" - 2"c")`
So, equating the powers of M, L, and T on both sides, we get
a + c = 0, −a + b −2c = 3, and −a −2c = −1
We have three unknowns a, b and c. We have three equations, on solving, we get
a = – 1, b = 4 and c = 1
Therefore, equation (1) becomes,
v = `"k"η^-1"r"^4("P"/"l")^1`
Experimentally, the value of k is shown to be `π/8`, we have
v = `(π"r"^4"P")/(8η"l")`
The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (vc).
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