formula.expert

Rooftop tank to wash-down nozzle

SPH4C Grade 12 College Physics · Hydraulic and Pneumatic Systems

An open storage tank on a barn roof keeps its water surface 12.0 m above a wash-down station in the yard, where a 19 mm bore nozzle hangs on a lever valve. Treat the water as 1000 kg/m³, the supply pipe as generously oversized (so friction can be neglected), and the tank as large enough that its level does not drop while the nozzle runs.

Given
  • h = 12 m — Water surface above the nozzle
  • ρ = 1000 kg/m³ — Water density
  • D = 19 mm — Nozzle bore
Determine
  1. (a)the static gauge pressure at the closed nozzle
  2. (b)the speed of the jet when the nozzle opens
  3. (c)the flow the nozzle passes
  4. (d)the rate at which the tank's height delivers energy to the jet
Step 1 of 4(a) · solve for Pressure

Closed valve first, because nothing is moving and the physics is pure statics: P = ρgh ≈ 118 kPa, about 17 psi on a gauge. Notice what is absent — the tank's width, the pipe's route, the volume stored. Height alone sets static pressure, which is the whole reason water towers are towers.

Rearranged for P
P=ρghP = \rho g h
Your values, in your units
P=(1,000 kg/m3) (9.80665 m/s2) (12 m)P = \left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 12\ \text{m} \right)
Answer
P=117.68 kPaP = 117.68\ \text{kPa}

Carried onward at full precision, not this rounded figure.

Open the Pressure Head (h = P/ρg) solver →

Step 2 of 4(b) · solve for Efflux speed

Open the valve and the same 12 m becomes speed instead of pressure: v = √(2gh) ≈ 15.3 m/s — exactly the speed of anything dropped 12 m, because it is the same energy exchange. The classic slip is forgetting the square root: doubling the height buys only 41% more speed, not double.

Rearranged for v
v=2ghv = \sqrt{2 g h}
Your values, in your units
v=2(9.80665 m/s2) (12 m)v = \sqrt{2 \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 12\ \text{m} \right)}
Answer
v=15.341 m/sv = 15.341\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Torricelli's Law (v = √(2gh)) solver →

Step 3 of 4(c) · solve for Flow rate

Speed through the 19 mm bore becomes flow: Q = Av ≈ 4.35 L/s, about 69 gpm. The area goes with the SQUARE of the bore — a 13 mm nozzle on the same tank passes less than half this — which is why nozzle size, not tank pressure, is the practical flow control on a gravity system.

Rearranged for Q
Q=πD24vQ = \tfrac{\pi D^{2}}{4} v
15.341 m/scarried from step 2
Your values, in your units
Q=π⋅(19 mm)24⋅(15.3414 m/s)Q = \tfrac{\pi \cdot \left( 19\ \text{mm} \right)^{2}}{4} \cdot \left( 15.3414\ \text{m/s} \right)
Answer
Q=260.98 L/minQ = 260.98\ \text{L/min}

Carried onward at full precision, not this rounded figure.

Open the Pipe Velocity from Flow and Diameter solver →

Step 4 of 4(d) · solve for Hydraulic power

P = ρgQh ≈ 512 W: every second, gravity hands the jet the energy of 4.35 kg of water dropped 12 m. Cross-check it as kinetic energy instead — ½ρQv² with v² = 2gh gives the identical 512 W, the two faces of the same conservation law. This is also the pump power a well would need to refill the tank as fast as the nozzle drains it.

Rearranged for P
P=ρgQhP = \rho g Q h
260.98 L/mincarried from step 3
Your values, in your units
P=(1,000 kg/m3) (9.80665 m/s2) (0.00434974 m3/s) (12 m)P = \left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 0.00434974\ \text{m}^{3}\text{/s} \right) \, \left( 12\ \text{m} \right)
Converted to base units
P=(1,000 kg/m3) (9.80665 m/s2) (260.984 L/min) (12 m)P = \left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 260.984\ \text{L/min} \right) \, \left( 12\ \text{m} \right)
Answer
P=511.88 WP = 511.88\ \text{W}

Carried onward at full precision, not this rounded figure.

Open the Hydraulic Power (P = ρgQh) solver →

The closed nozzle sees about 118 kPa of static pressure; opened, it throws a 15.3 m/s jet passing roughly 4.35 L/s, and the tank's 12 m of height delivers energy to that jet at about 512 W.

Why this order

This is the whole hydraulic-systems unit in one apparatus: the same 12 metres of height read four ways. Statics first — with the valve shut, elevation is pressure, P = ρgh, and nothing else about the installation matters. Then Torricelli converts the height to speed the instant flow starts, and it is worth saying out loud that v = √(2gh) is free fall in disguise: the water arrives at the nozzle at the speed it would have reached falling from the surface, because both are the same trade of mgh for ½mv². The order matters pedagogically because each quantity needs the previous one: flow is speed times the bore area, and power is the flow carrying its energy — computable either as ρgQh (potential energy spent per second) or ½ρQv² (kinetic energy delivered per second), and the fact that these agree to the last digit is not luck, it is Bernoulli's equation holding both ends of the pipe.

Three misconceptions do most of the damage in this unit. First, that a bigger tank means more pressure — it means more DURATION; only depth and density set pressure, which is why a skinny standpipe and a fat reservoir at the same height read identically. Second, dropped square roots: doubling h doubles pressure but only multiplies jet speed by √2, and mixing those scalings is the most common exam wreck. Third, ignoring what 'neglect friction' bought: a real 12 m gravity feed through a long small pipe delivers visibly less than 15.3 m/s, and the gap between the ideal jet and the measured one is precisely the friction loss the trade chains in this section spend their time computing. The ideal case is not wrong — it is the ceiling every real system is measured against.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.