The volume flow rate Q of a viscous liquid through a pipe depends on the radius r of the p… — JAMB UTME

The volume flow rate $Q$ of a viscous liquid through a pipe depends on the radius $r$ of the pipe, the coefficient of viscosity $\eta$ , and the pressure gradient $\frac{\Delta P}{L}$ according to the dimensional equation $Q = k r^x \eta^y \left(\frac{\Delta P}{L}\right)^z$ , where $k$ is a dimensionless constant. What is the value of the exponent $x$ ?

Answer: 4

Answer

The value of the exponent

x

is 4.

Applying the principle of dimensional homogeneity, the dimensions of volume flow rate

[Q] = L^3 T^{-1}

are equated to

[r]^x [\eta]^y \left[\frac{\Delta P}{L}\right]^z = L^x (M L^{-1} T^{-1})^y (M L^{-2} T^{-2})^z

. Equating powers yields

y + z = 0

for mass,

-y - 2z = -1

for time, and

x - y - 2z = 3

for length. Solving these simultaneous equations gives

z = 1

y = -1

, and

x = 4

Step-by-Step Solution

Identify the base dimensions of each physical quantity in the given equation.

Flow rate

[Q] = M^0 L^3 T^{-1}

, radius

[r] = L

, viscosity

[\eta] = M L^{-1} T^{-1}

, and pressure gradient

\left[\frac{\Delta P}{L}\right] = M L^{-2} T^{-2}

Expressing each quantity in terms of fundamental dimensions (

M

L

T

) is necessary for dimensional analysis.

Substitute dimensions into the power-law equation and collect powers of base dimensions.

M^0 L^3 T^{-1} = M^{y+z} L^{x-y-2z} T^{-y-2z}

The principle of dimensional homogeneity requires both sides of a physically valid equation to have identical dimensions.

Set up and solve linear equations for the exponents

x

y

, and

z

Solving

y + z = 0

-y - 2z = -1

, and

x - y - 2z = 3

yields

z = 1

y = -1

, and

x = 4

Equating powers of

M

T

, and

L

allows step-by-step determination of each unknown exponent.

Key Concept

Dimensions of Physical Quantities and Dimensional Analysis

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic