💧 Impact of a Jet
A virtual F1-16 apparatus: a vertical water jet strikes a deflector on a weighted stem. Load the pan with masses, open the valve until the pan floats back to its datum — momentum balanced against gravity — and verify the momentum equation for flat (90°), 120° and hemispherical (180°) targets. The hemisphere should need exactly half the dynamic force of the flat plate.
Objective
To investigate the reaction force produced by the change in momentum of a jet striking flat and curved surfaces, and compare measured forces with the momentum equation.
Theory
v = Q/A • Fy = ρQv(1−cosθ) = ρA·v²(1−cosθ) • balance: W = mg = Fy • slope of W vs v²: S = ρA(1−cosθ)
Deflection factors: flat plate θ = 90° → (1−cos90°) = 1; conical 120° → 1.5; hemispherical cup 180° → 2. Plotting the applied weight against v² should give straight lines through the origin whose slopes are in the ratio 1 : 1.5 : 2.
Equipment
F1-10 hydraulics bench; F1-16 impact-of-a-jet apparatus (clear cylinder, nozzle, interchangeable deflectors on a spring-supported stem with weight pan, datum level gauge); stopwatch.
Nozzle d = 8 mm Masses 50 → 500 g in 50 g steps Collect ≥ 60 s per flow
Procedure — perform it here
Observations & Computations
Record with the pan balanced. W = mg; Fy = ρQv(1−cosθ) from the measured flow; the last column compares them.
| No. | Deflector | m (g) | W (N) | Vol (L) | t (s) | Q (L/s) | v (m/s) | v² (m²/s²) | Fy (N) | W/Fy |
|---|---|---|---|---|---|---|---|---|---|---|
| No readings yet — balance the pan, collect, then Record. | ||||||||||
W vs v² — slopes against theory
Discussion
Does this verify the momentum equation?
Yes — quantitatively. The fitted W–v² slopes reproduce ρA(1−cosθ) within a few percent, and the three deflectors fall in the predicted 1 : 1.5 : 2 ratio. The residual gap is systematic: the jet slows slightly (gravity, air entrainment and splash) between nozzle and target, so slightly more flow is needed than the ideal theory says.
Would results differ if the deflector were closer to the nozzle?
Yes. The jet decelerates under gravity as it rises (v² = v₀² − 2gs), so a closer target intercepts a faster jet and the measured force would sit closer to the ideal prediction.
Why does the hemispherical cup double the force?
It reverses the jet completely (θ = 180°), so the momentum change per unit mass is 2v instead of v for the flat plate, which merely destroys the vertical component. That doubling is the basis of the Pelton bucket's shape.
Precautions: level the cylinder; check the pan swings freely and set the datum gauge with no weights; wait for the pan to settle before recording; collect for at least 60 seconds.