I met an inventor named Bob at the Long Beach FRC robotic regional competition who wanted my opinions on how to best stabilize single nozzle version of the Zapata racing Flyboard which was inspired by the JetLev. Another newer version of this is called the Jetovator. All three are summarized in this Gismag article.
I wanted to refresh myself on the physics of the thrust that drives the jet pack.
We know that F = d(mv)/dt which under steady speed conditions and equal input and output pressures leads to the thrust (T) equation (ref NASA thrust site)
T = F = m_dot*(v_e – v_i)
where m_dot= time rate of change of water mass flowing through a hose and
v_e = water exit speed , v_i = water input speed along the same direction.
The mass continuity equations yields
m_dot = A_hose*rho*(v_hose) = A_nozzle*rho*v_e
where rho is the density of water and A_hose is the area of the hose, v_hose is the water flow rate in the hose and A_nozzle is the area of the thruster nozzle.
Since v_i = 0 for a stationary jet ski pumping water vertically to the flyboard then thrust equation for T in terms of v_e gives
T = A_nozzle*rho*V_e^2
If the hose and nozzle areas are different the relationship between v_hose and v_e is
v_e = v_hose*A_hose/A_nozzle.
Substituting into the T equation for V_e give T in terms of v_hose.
T = A_hose*rho*v_hose*v_e = (A_hose*v_hose)^2*rho/A_nozzle
or in terms of hose flow rate q = A_hose*v_hose
T = q^2 *rho/A_nozzle
Other data we will need:
A_hose = 1/12 sqft (approx 4 in hose diameter),
g = 32.2 ft/s/s,
rho = 1.94 slug/cft
Under steady state flight the total thrust must equal to the weight. It is assumed that the water weight is supported by the thrust since it puts a downward force on the hose directly beneath the vertical water column. The thrust must also support the dry weight of the hose extending above the water. (Note: the hose has viscosity drag due to the high-speed water flowing through it. The reaction force adds lift to the side of the hose. but is cancelled by the opposing downward force on the bottom of the hose.)
T = W_platform + W_hose_dry + W_water
W_hose _dry= 85 lb*H/100ft (4 in standard fire hose)
and H is the length of the hose above the water.
Substituting the weights into the T equation above we can rearrange terms and solve for q
q = sqrt((W_platform + W_hose_dry + W_water)/ (rho/A_nozzle ))
Comparison of Bob’s Test numbers with Theoretical calculation:
Bob built a test platform which weighed 250 lbs with him aboard. At a height of 30 feet the exit flow rates were around 1000 gpm and the jet ski ran at 7100 rpm. At 5 ft , the ski jet rpm lowered to 4500 rpm.
The rpm ratio he saw was 7200/4500 = 1. 6 between heights. He said he had two nozzles with a diameter of 1.7 inches (A_nozzle =2* .0158 sqft= .0316 sqft) .
The total length of these hoses are typically 50 ft when you account for the portion of the hose in the water.
So lets get started with the calculation based upon my equations:
The dry weight of the hose at 30ft and the water weight are:
W_hose_30ft = 85*30/100 = 26 lbs , W_water_30ft = 156 lbs
W_hose_5 ft = 4.25 lbs , W_water_5ft = 26 lbs
Lets calculate the required flow rate q at 30 feet and 5 ft and see how it compares with Bobs.
q_30 = 2.65 cfs ( 1189 gpm ) , q_5 = 2.13 cfs (958 gpm)
The flow ratio = (q_30/q_5) = 2.65/2.13 = 1.24
The calculated flow rate of 1189 gpm compares within 19% of the observed 1000 gpm .I would expect a little better agreement… I am not sure to what accuracy Bob estimated his observation or how it was measured.
RPM ratio :
Normally, pump affinity laws state that flow rates are directly proportional to rpm’s and we could stop here but as you see the flow ratio is 1.24 compared to the observed rpm ratio of 1.6. This is because there is static pump pressure and friction losses as well as dynamic pump pressure. So it is more accurate to compare the total pump pressure at each height and use a pump affinity rule that says the rpm is proportional to the sqrt of total pump pressure.
To calculate the rpm ratio of the pump we need the pump head or pump pressure p_p. Once we have that on the assumption of a centrifugal pump we know that the rpm’s are proportional to the sqrt(p_p) assuming the same pump efficiency. Using Bernoulli’s total pressure relationship we know that the total pressure at the pump must be higher than the total pressure at the nozzle by the added pressure of the water weight (p_h) and the friction back pressure (p_fl).
p_p = p_f l + p_T + p_h
To compute p_fl I assume that the total hose length is 50 feet and doesn’t change with how much hose is above water.
I found an internet equation (ref) that gives the hose friction pressure FL in terms of lbs per sqin.
FL _lbs_per_in2= C*(q_hose_gpm/100)^2*L_ft/100 where
C = Friction loss Coefficient = .2 for a 4 in fire hose.
To keep units the same we can write p_fl in terms of q (cft/s)
p_fl =144*FL = C’*q^2*50ft with C ‘ = C*(448.8)^2*144/(10^6)= 5.8
p_T = T/A_nozzle is the pressure on the nozzle area due to the thrust and is the total pressure at that point
p_h = W_water/A_hose is the pressure due to the water height
Pressure …………………… H_ft = 30 ft……………………..H_ft = 5 ft
Recall that assuming the same pump efficiency, the rpm’s are proportional to the sqrt of the total pump pressure. So
rpm_ratio = sqrt(p_p_30/p_p_5) = sqrt(26548/11037) = sqrt(2.4) = 1.6
Amazing! This is matches the observed rpm ratio of 1.6.