Question

The volume V of liquid that flows through a pipe in time t is given by the equation

V /t = Pi(the symbol of Pi) P r^4 /8CL

Where P is a pressure difference between the end of the pipe of radius r and the length L. The constant C depends on the frictional effects of the liquid.

Determine the base unit of C

Answer 1

Answer on Question #75214, Physics / Mechanics | Relativity

The volume $V$ of liquid that flows through a pipe in time $t$ is given by the equation

$$\frac{V}{t} = \frac{\pi P r^4}{8 C l}$$

where $P$ is a pressure difference between the end of the pipe of radius $r$ and the length $L$. The constant $C$ depends on the frictional effects of the liquid.

Determine the base unit of C.

Solution:

V units: $\mathrm{m}^3$

V / t units: $\mathrm{m}^3 \cdot \mathrm{s}^{-1}$

Pressure units (allow use of $P = F/A$): $\mathrm{kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-2} / \mathrm{m}^2 = \mathrm{kg} \cdot \mathrm{s}^{-2} / \mathrm{m}$

Clear substitution of units for $P, r^4, l$

$$C = \frac{\pi P r^4 t}{8 V l}$$ $$C = \frac{\mathrm{kg} \cdot \mathrm{s}^{-2} \cdot \mathrm{m}^4 \cdot \mathrm{s}}{\mathrm{m} \cdot \mathrm{m}^3 \cdot \mathrm{m}} = \frac{\mathrm{kg} \cdot \mathrm{s}^{-1}}{\mathrm{m}} = \mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}$$

Answer: $\mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}$

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