NBPhO 2026

Part i) (1.5 points)

Solution 1 (direct from the problem’s drag scales). The problem gives the laminar drag per unit area as $F_L \sim \mu v/D$ (the velocity gradient at the wall is of order $v/D$). The turbulent drag per unit area is $F_T \sim \rho v^2$ (dynamic pressure carried by the vortices). The dimensionless ratio is

$$R = \frac{F_T}{F_L} \sim \frac{\rho v^2}{\mu v/D} = \frac{\rho v D}{\mu}.$$

Grading (Solution 1)

• $R = F_T/F_L \sim \rho v^2/[\mu\,du/dr]$: 0.5 pts
• $du/dr \sim v/D$: 0.5 pts
• $R = \rho v D/\mu$: 0.5 pts
• Includes numerical factor $> 10$: −0.2 pts

Solution 2 (dimensional analysis). Look for $R = D^d \rho^r \mu^m v^u$ to be dimensionless. Mass, length, and time give three equations:

$$\begin{cases} r + m = 0,\\ d - 3r - m + u = 0,\\ -m - u = 0. \end{cases}$$

Solving: $m = -r$, $u = r$, $d = r$. So $R = (Dv\rho/\mu)^r$. This seemingly leaves one degree of freedom, but $\mu$ appears only in the laminar drag (denominator of $F_T/F_L$), so $m = -1$ and hence $r = 1$:

$$R = \frac{\rho v D}{\mu}.$$

Grading (Solution 2)

• Setting up $R = D^d \rho^r \mu^m v^u$ as dimensionless: 0.3 pts
• Correct system of equations from mass, length, time dimensions: 0.5 pts
• Solving the system to obtain $R = (Dv\rho/\mu)^r$: 0.4 pts
• Correctly fixing $r = 1$ by noting $\mu$ enters with power $-1$ (or by appeal to the standard Reynolds-number form): 0.3 pts

Part ii) (0.5 points)

The hose has cross-section area $\pi d^2/4$, so by continuity the mean flow speed is

$$v = \frac{Q}{\pi d^2/4} = \frac{4Q}{\pi d^2} \approx 3.1\,\tfrac{\text{m}}{\text{s}}.$$

The Reynolds number is then

$$R = \frac{\rho v d}{\mu} = \frac{4\rho Q}{\pi d \mu} \approx 37000.$$

Since $37000 \gg 2500$, the flow is turbulent.

Grading

• Computing the mean flow speed $v$ from $Q$ via continuity: 0.2 pts
• Correctly evaluating $R \approx 37000$: 0.2 pts
• Concluding turbulent flow on the basis of the numerical comparison $R \gg 2500$: 0.1 pts

Part iii) (3 points)

The pump’s power $P$ goes into three reservoirs: gravitational potential energy of the lifted water, work against drag in the hose, and kinetic energy of the exiting jet. Per unit volume, these are $\rho gh$, $\Delta p_\text{drag}$, and $\rho v_\text{exit}^2/2$. The power balance is therefore

$$P = Q\left(\rho gh + \Delta p_\text{drag} + \tfrac{1}{2}\rho v_\text{exit}^2\right).$$

Since the flow is turbulent (part ii), the drag scales as $v^2$: $\Delta p_\text{drag} = kv^2$, where $k$ is a constant for this hose.

Calibration (no nozzle). With no nozzle, $v_\text{exit} = v_0$ (the hose-exit speed from part ii), $v_0 \approx 3.14\,\text{m/s}$). Then

$$P = Q\left(\rho gh + kv_0^2 + \tfrac{1}{2}\rho v_0^2\right).$$

Numerically:

$$\rho gh = 1.96\times 10^5\,\text{Pa}, \quad \tfrac{1}{2}\rho v_0^2 = 4.93\times 10^3\,\text{Pa},$$ $$P/Q - \rho gh - \tfrac{1}{2}\rho v_0^2 = kv_0^2 \approx 3.99\times 10^5\,\text{Pa},$$ $$k = \frac{3.99\times 10^5\,\text{Pa}}{v_0^2} \approx 4.04\times 10^4\,\frac{\text{Pa}\cdot\text{s}^2}{\text{m}^2}.$$

With nozzle. The exit area is $f$ times the hose area; by mass continuity, $v'_\text{exit} = v'/f$, where $v'$ is the (new) hose flow speed. Substituting into the power balance:

$$P = \frac{\pi d^2}{4}v'\left(\rho gh + kv'^2 + \frac{\rho v'^2}{2f^2}\right).$$

This is a cubic in $v'$.

Dimensionless form. Introduce $x = v'/v_0$. Each of the three reservoirs in the unblocked case contributes a fraction of the total power:

$$a = \frac{\rho gh}{\rho gh + kv_0^2 + \tfrac{1}{2}\rho v_0^2}, \quad b = \frac{kv_0^2}{\rho gh + kv_0^2 + \tfrac{1}{2}\rho v_0^2},$$

and $c = 1 - a - b$. Numerically: $a \approx 0.327$, $b \approx 0.665$, $c \approx 0.0082$. The power balance becomes

$$x\left[a + \left(b + \frac{c}{f^2}\right)x^2\right] = 1.$$

With $f = 0.15$, $c/f^2 \approx 0.364$, so $b + c/f^2 \approx 1.029$.

Iteration. Starting from $x_0 = 1$ on the right-hand side:

$$x^3 + 0.318 x = 0.972 \implies x = (0.972 - 0.318x)^{1/3}.$$ $$x_1 = (0.972 - 0.318)^{1/3} = 0.868,$$ $$x_2 = (0.972 - 0.276)^{1/3} = 0.886,$$ $$x_3 = (0.972 - 0.282)^{1/3} = 0.884,$$ $$x_4 = (0.972 - 0.281)^{1/3} = 0.884.$$

Converged: $x \approx 0.884$.

Results.

$$Q'/Q = v'/v_0 = x \approx \boxed{0.88}, \qquad \text{so } Q' \approx 22.1.$$

The exit speed is $v'_\text{exit} = v'/f = xv_0/f$, and the dirty-surface excess pressure scales as $\rho v_\text{exit}'^2/2$:

$$\frac{p'_\text{excess}}{p_\text{excess}} = \frac{v_\text{exit}'^2}{v_0^2} = \frac{x^2}{f^2} = \frac{0.884^2}{0.15^2} \approx \boxed{35}.$$

Grading

• Power balance: gravity term $\rho gh$: 0.2 pts
• Power balance: drag term $\Delta p_\text{drag}$: 0.2 pts
• Power balance: exit kinetic energy term $\rho v_\text{exit}^2/2$: 0.2 pts
• Identifying $\Delta p_\text{drag} = kv^2$ from the turbulent-regime scaling: 0.3 pts
• Calibrating $k$ from the unblocked case: $k \approx 4\times 10^4\,\text{Pa}\cdot\text{s}^2/\text{m}^2$: 0.3 pts
• Mass continuity giving $Q_m = Q'_m$: 0.3 pts
• $v'_\text{exit} = v'/f$: 0.2 pts
• Updated power balance with the kinetic-energy term boosted by $1/f^2$: 0.5 pts
• Volumetric flow rate ratio $Q'/Q \approx 0.88$ within tolerance ($\pm 0.02$): 0.4 pts
• Formula for excess-pressure ratio $(v'_\text{exit}/v_0)^2$: 0.2 pts
• Excess-pressure ratio $\approx 35$ within tolerance ($\pm 1$): 0.5 pts
• If accuracy $> 1\,\%$: −0.1 pts