1. **Problem Statement:** Find a formula for the flow rate $\alpha$ of a fluid through a pipe in terms of fluid viscosity $\eta$, pressure gradient per unit length $\frac{\Delta P}{L}$, and pipe radius $r$ using dimensional analysis. 2. **Given Units:** - Viscosity $\eta$: N·s/m$^2$ (which is equivalent to kg/(m·s)) - Pressure gradient $\frac{\Delta P}{L}$: pressure per length, units N/m$^3$ or kg/(m$^2$·s$^2$) - Radius $r$: meters (m) - Flow rate $\alpha$: volume per time, units m$^3$/s 3. **Step 1: Identify dimensions of each quantity:** - Viscosity $\eta$: $[M][L]^{-1}[T]^{-1}$ - Pressure gradient $\frac{\Delta P}{L}$: Pressure is force per area $[M][L]^{-1}[T]^{-2}$, so gradient per length is $[M][L]^{-2}[T]^{-2}$ - Radius $r$: $[L]$ - Flow rate $\alpha$: volume per time $[L]^3[T]^{-1}$ 4. **Step 2: Assume a formula:** $$\alpha = k \eta^a \left(\frac{\Delta P}{L}\right)^b r^c$$ where $k$ is a dimensionless constant. 5. **Step 3: Write dimensions of right side:** $$[M]^a [L]^{-a} [T]^{-a} \times [M]^b [L]^{-2b} [T]^{-2b} \times [L]^c = [M]^{a+b} [L]^{-a - 2b + c} [T]^{-a - 2b}$$ 6. **Step 4: Equate dimensions to flow rate $[L]^3 [T]^{-1}$:** - For mass $M$: $a + b = 0$ - For length $L$: $-a - 2b + c = 3$ - For time $T$: $-a - 2b = -1$ 7. **Step 5: Solve the system:** From $a + b = 0$, we get $a = -b$. Substitute into time equation: $$-(-b) - 2b = -1 \Rightarrow b - 2b = -1 \Rightarrow -b = -1 \Rightarrow b = 1$$ Then $a = -1$. Substitute $a$ and $b$ into length equation: $$-(-1) - 2(1) + c = 3 \Rightarrow 1 - 2 + c = 3 \Rightarrow c = 4$$ 8. **Step 6: Final formula:** $$\alpha = k \eta^{-1} \left(\frac{\Delta P}{L}\right)^1 r^4 = k \frac{r^4}{\eta} \frac{\Delta P}{L}$$ 9. **Interpretation:** The flow rate is proportional to the pressure gradient and the fourth power of the radius, and inversely proportional to the viscosity. **Answer:** $$\boxed{\alpha = k \frac{r^4}{\eta} \frac{\Delta P}{L}}$$ This matches the known Hagen-Poiseuille law form (with $k$ depending on pipe geometry).