Question : Using method of dimension, derive an expression for rate of flow (V) of a liquid through a pipe of radius (r) under a pressure gradient (P/l). Given that V also depends on coefficient of viscosity (η) of the liquid.
Doubt by Shruti
Solution :
Volume flows out through a pipe per unit time is called rate of flow. It is generally denoted by
Q = V/t
Q = V/t
Dimensional Formula [M0L3T-1]
Rate of flow (V) = [M0L3T-1]
Radius (r) = [L]
Pressure Gradient : Pressure / Length
Pressure = Force / Area
=[MLT-2]/[L2]
=[ML-1T-2]
Pressure Gradient : Pressure / Length
[P/l] = [ML-1T-2]/[L]
[P/l] = [ML-2T-2]
Length = [L]
Coefficient of viscosity (η) = [ML-1T-1]Pressure = Force / Area
=[MLT-2]/[L2]
=[ML-1T-2]
Pressure Gradient : Pressure / Length
[P/l] = [ML-1T-2]/[L]
[P/l] = [ML-2T-2]
Length = [L]
V=kra(P/l)bηc — (1)
where k is a dimensionless constant
[M0L3T-1]=[L]a×[ML-2T-2]b×[ML-1T-1]c
[M0L3T-1]=[Mb+c La-2b-c T-2b-c]
b+c = 0 — (2)
a-2b-c=3 — (3)
-2b-c=-1 — (4)
Solving equation (2) and (4)
[M0L3T-1]=[L]a×[ML-2T-2]b×[ML-1T-1]c
[M0L3T-1]=[Mb+c La-2b-c T-2b-c]
b+c = 0 — (2)
a-2b-c=3 — (3)
-2b-c=-1 — (4)
Solving equation (2) and (4)
b+c = 0
-2b-c=-1
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-b+0=-1
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-b=-1
b=1
b=1
putting in equation (2)
1+c=0
c=-1
1+c=0
c=-1
putting the values of b and c in equation (3)
a-2(1)-(-1)=3
a-2+1=3
a-1=3
a=3+1
a=4
a-2+1=3
a-1=3
a=3+1
a=4
Substituting these values of 'a', 'b' and 'c' in equation (1)
V=kr4(P/l)1η-1
V=kr4(P/l)1η-1
V=kr4P/lη
V=kPr4/lη
where k is dimensionless constant, it value is π/8 which can be determined by other methods. so the final equation will look like V=πPr4/8lη
