Physics of a Fly-Board: Lifting a Rider and Calculating Pump Power
Analyzing the physics of a fly-board by calculating the thrust required to lift the rider and the power needed for steady flow.
Problem Statement
A fly-board is a device that uses high-velocity water to produce thrust, allowing a rider to hover above the ground. The device consists of:
- A small platform (the board) connected by a thick hose to a powerful water pump.
- Water is forced up the hose and exits downward through nozzles, generating thrust that lifts the rider.
We make the following assumptions:
- The total mass of the rider, board, and hose portion is $M = 100\,\text{kg}$.
- Water is treated as an ideal fluid with density $\rho = 1000\,\text{kg/m}^3$.
- The hose is vertical with an inner diameter $d = 0.11\,\text{m}$, so $r = \frac{d}{2} = 0.055\,\text{m}$.
- Water enters and exits the board at the same speed $v$, with the flow reversed by $180^\circ$ inside the board.
- The acceleration of gravity is $g = 9.81\,\text{m/s}^2$.
Tasks
- (a) Determine the velocity $v$ of water needed to lift the rider in steady hover.
- (b) Calculate the minimum power $P$ required for the pump to sustain steady flow when the rider is hovering at a height $h = 7.0\,\text{m}$ above the pump’s water inlet.
Solution
(a) Velocity of Water for Steady Hover
We want to find the flow speed $v$ that provides enough thrust to balance the weight $Mg$ of the rider and board.
1. Mass Flow Rate
Let $\dot{m}$ be the mass flow rate of water through the hose:
\[\frac{dm}{dt} = \rho \, A \, v = \rho \,\pi\,r^2\,v,\]where $A = \pi r^2$ is the hose cross-sectional area.
2. Force on the Water (Momentum Change)
Within the board, water velocity is reversed from upward ($+v$) to downward ($-v$). Thus, the net velocity change per unit mass is
\[\Delta v = v_{\text{out}} - v_{\text{in}} = (-v) - (+v) = -\,2v.\]The rate of change of the water’s momentum (force on the water by the board) is
\[F_{w,M} = \frac{dm}{dt} \,\Delta v = (\rho \,\pi\,r^2\,v)\,(-2v) = -\,2\,\pi\,\rho\,r^2\,v^2.\]By Newton’s 3rd Law, the force on the rider+board from the water is the opposite:
\[F_{M,w} = -\,F_{w,M} = 2\,\pi\,\rho\,r^2\,v^2.\]3. Force Balance for Hover
To hover, the upward thrust $F_{M,w}$ must equal the weight $Mg$:
\[F_{M,w} = M\,g \quad \Longrightarrow \quad 2\,\pi\,\rho\,r^2\,v^2 = M\,g.\]Solving for $v$:
\[v = \sqrt{\frac{M\,g}{2\,\pi\,\rho\,r^2}}.\](b) Minimum Pump Power
We now compute the power $P$ that the pump must deliver to keep the rider at height $h$ above the inlet.
1. Energy per Unit Mass
Each kilogram of water gains:
- Gravitational potential energy $g\,h$,
- Kinetic energy $\tfrac12\,v^2$.
Hence, per unit mass of water, the energy is
\[g\,h + \frac12\,v^2.\]2. Power = Energy per Time
The pump supplies this energy to a mass flow rate $\dot{m}$:
\[P_{\text{pump}} = \dot{m} \Bigl( g\,h + \tfrac12\,v^2 \Bigr) = \rho\,\pi\,r^2\,v \Bigl( g\,h + \tfrac12\,v^2 \Bigr).\]In expanded form, you might also see:
\[P_{\text{pump}} = \frac12\,\pi\,\rho\,r^2\,v \Bigl( 2\,g\,h + v^2 \Bigr).\]Final Calculations
Plug in the known values:
- $M = 100\,\text{kg}$
- $\rho = 1000\,\text{kg/m}^3$
- $r = 0.055\,\text{m}$
- $g = 9.81\,\text{m/s}^2$
- $h = 7.0\,\text{m}$
- Water speed $v$:
- Pump Power $P_{\text{pump}}$:
First, note:
\[v^2 \approx (7.2)^2 \approx 51.8, \quad 2\,g\,h = 2 \times 9.81 \times 7.0 \approx 137.3.\]Then:
\[\dot{m} = \rho\,\pi\,r^2\,v \approx 1000\,\pi\,(0.055)^2 \times 7.2 \approx 68\,\text{kg/s}.\]So,
\[P_{\text{pump}} = \dot{m} \Bigl(g\,h + \tfrac12\,v^2\Bigr) \approx 68 \Bigl(9.81 \times 7.0 + \tfrac12 \times 51.8\Bigr).\]Inside the parentheses:
\[g\,h = 9.81 \times 7.0 = 68.7, \quad \tfrac12\,v^2 = 25.9, \quad \text{sum} = 68.7 + 25.9 = 94.6.\]Hence,
\[P_{\text{pump}} \approx 68 \times 94.6 \approx 6430\,\text{J/s} = 6.43\,\text{kW}.\]Rounding slightly, we get about $6.4\text{–}6.5\,\text{kW}$.
Final Results
- Velocity of Water:
- Minimum Pump Power:
Conclusion
By reversing the water flow from upward to downward at about $7.2\,\text{m/s}$, the fly-board produces enough thrust to lift a $100\,\text{kg}$ rider and board. In an ideal frictionless scenario, maintaining that flow at $7\,\text{m}$ height requires about $6.4\,\text{kW}$ of pump power. Real systems will require more power to account for losses, but this analysis highlights the fundamental physics of momentum change and energy demands in a fly-board system.