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Difficulty: HardPhysical Quantities, Units and Dimensions
The volume flow rate QQ of a viscous liquid flowing through a pipe of radius rr under a pressure gradient ΔPl\frac{\Delta P}{l} is modeled by the equation:
Q=kηxry(ΔPl)zQ = k \eta^x r^y \left(\frac{\Delta P}{l}\right)^z
where η\eta is the coefficient of dynamic viscosity and kk is a dimensionless constant. What are the values of the exponents xx, yy, and zz respectively?
  1. x=−1,y=4,z=1x = -1, y = 4, z = 1Answer
  2. B
    x=1,y=4,z=−1x = 1, y = 4, z = -1
  3. C
    x=−1,y=3,z=1x = -1, y = 3, z = 1
  4. D
    x=1,y=2,z=1x = 1, y = 2, z = 1

Answer

x=−1x = -1, y=4y = 4, and z=1z = 1
The dimensional representation of volume flow rate is [L3T−1][L^3 T^{-1}], dynamic viscosity is [ML−1T−1][M L^{-1} T^{-1}], radius is [L][L], and pressure gradient is [ML−2T−2][M L^{-2} T^{-2}]. Equating the powers of MM, LL, and TT gives x+z=0x + z = 0, −x−2z=−1-x - 2z = -1, and −x+y−2z=3-x + y - 2z = 3. Solving these simultaneously yields x=−1x = -1, y=4y = 4, and z=1z = 1.

Step-by-Step Solution

1
Determine the fundamental dimensions of each physical quantity
Volume flow rate Q=VolumeTime=[L3T−1]Q = \frac{\text{Volume}}{\text{Time}} = [L^3 T^{-1}];
Dynamic viscosity η=[ML−1T−1]\eta = [M L^{-1} T^{-1}];
Radius r=[L]r = [L];
Pressure gradient ΔPl=PressureLength=[ML−1T−2][L]=[ML−2T−2]\frac{\Delta P}{l} = \frac{\text{Pressure}}{\text{Length}} = \frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}].
Correct base dimensions are required for dimensional analysis.
2
Set up the dimensional equation by substituting the base dimensions into the formula
[L3T−1]=[ML−1T−1]x[L]y[ML−2T−2]z=Mx+zL−x+y−2zT−x−2z[L^3 T^{-1}] = [M L^{-1} T^{-1}]^x [L]^y [M L^{-2} T^{-2}]^z = M^{x+z} L^{-x + y - 2z} T^{-x - 2z}.
The principle of dimensional homogeneity requires both sides of the equation to have matching exponents for MM, LL, and TT.
3
Equate exponents for MM, TT, and LL to form algebraic equations
For MM: x+z=0  ⟹  z=−xx + z = 0 \implies z = -x
For TT: −x−2z=−1-x - 2z = -1
For LL: −x+y−2z=3-x + y - 2z = 3.
This creates a linear system of equations for the exponents xx, yy, and zz.
4
Solve the system of linear equations
Substituting z=−xz = -x into the TT equation: −x−2(−x)=−1  ⟹  x=−1-x - 2(-x) = -1 \implies x = -1.
Hence z=−(−1)=1z = -(-1) = 1.
Substituting x=−1x = -1 and z=1z = 1 into the LL equation: −(−1)+y−2(1)=3  ⟹  1+y−2=3  ⟹  y=4-(-1) + y - 2(1) = 3 \implies 1 + y - 2 = 3 \implies y = 4.
Yields the unique set of exponents x=−1,y=4,z=1x = -1, y = 4, z = 1.

Key Concept

Dimensional Analysis and Homogeneity
Estimated Time:2m 0s
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Topics

Physical Quantities, Units and Dimensions

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