The Amazon wall socket

If this explanation's motivation is not quite clear, please feel welcome to ask in the comments to expand further!

Consider some fluid element $dV$ (that is, an infinitesimally small volume element of fluid).

The mass of this element, if its density $\rho$ is homogenous, is $\rho\, dV$ , and say this element (and all other elements in consideration) have some velocity $v_0$ . Then, the kinetic energy $dT$ of this element is: $dT=\frac{1}{2}d(mv^2)=\frac{1}{2}(\rho\, dV)(v_0)^2$ This happens as $v_0$ is constant (that is, $dv_0=0$ ), as defined.

Note that, in our case, we can consider the volume element to be of a cylindrical shape. Note also that the radius of our aperture is constant and hence $dr_0=0$ . So, we have: $dV=d(\pi \,r_0^2h)=\pi r_0^2\,dh$

Putting this back into our original expression gives us: $dT=\frac{1}{2}(\rho\pi r_0^2\,dh)(v_0)^2=\frac{1}{2}(\rho\pi r_0^2v_0^2)dh$

Dividing through by $dt$ , we get: $\frac{dT}{dt}=P=\frac{1}{2}(\rho\pi r_0^2v_0^2)\frac{dh}{dt}$

We know that $\frac{dh}{dt}=v_0$ , so, putting this back in, we get: $P=\frac{1}{2}\rho\pi r_0^2v_0^3$

The units check out, which is fantastic, and so, we simply solve for the case, plugging in our given values $\rho=1000\text{ kg/m}^3$ , $r_0=.05\text{ m}$ , $v_0=1\text{ m/s}$ , $E=5\text{ W-hrs}$ , and $C_{\text{eff}}=.5$ : $\frac{E_\text{total}}{C_\text{eff}P}=\frac{2E}{C_\text{eff}\,\rho\pi r_0^2v_0^3}=\frac{2(5)}{(.5)(1000)(.05)^2(1)^3\pi}$

Hence: $\frac{E_\text{total}}{C_\text{eff}P}\approx 2.54648\text{ hrs}$

Cheers. And, again, if any clarifications are necessary, please comment.

We can think of the water as a cylinder of length $h$ and radius $r$ , so the volume of water moving past a point per second is $\pi r^2 h$ , which is equal to $\pi\times 0.05^2$ . Hence the mass of this water is $\frac{5\pi}{2}$ . $\text{Kinetic energy}, E_K=\frac{1}{2}mv^2$ , $v=1$ , so $E_K=\frac{5\pi}{4}$ . We only have $50\%$ efficiency, so the kinetic energy transferred to the battery per second is $\frac{5\pi}{8}$ . The battery is rated at $5~Wh=5\times 3600~J$ , so the battery will take $(5\times 3600)\div\frac{5\pi}{8}~s$ to charge, which is equal to $5\div\frac{5\pi}{8}~h=2.55~h$ .

Electric Energy=Kinetic Energyx50% 5W-h= (1/2 mv^2) W=J/s mfr=pvA=(1000kg/m^3)(1m/s)(0.05^2pi)=2.5pi kg/s mfr=m/T 5W-h= 1/2 (mfr x T) x v x 50% 5W-h=1/2 (2.5pi kg/s x T) (1m/s)^2 x 50% 8/pi W-h/j/s= T T=2.55h

The mass flow rate is given by $\rho v \pi r^2$ and has units of kg/s. The kinetic energy, or $0.5 m v^2$ is now a unit of power, or $0.5 \rho v \pi r^2 v^2$ and has units of watts. Multiply the power by 0.5 to account for the efficiency and divide $5$ Wh into the result to find the hours needed to charge the battery.

Consider a cross-section of the aperture.

In 1 second, the volume of water flows through it would be a cylinder of radius 5 cm and length 100 cm, means that it has a volume of 5 5 pi*100 = 7853.98 cm^3

Therefore, the mass of that volume is 7.85 kg

Its kinetic energy is 0.5 m v^2 = 3.93 J

=> Its power is 3.93 J/s or 3.93 W

Now, as one-half of the energy is converted to electricity, we have the HydroBee's power is 3.93/2 or 1.96 W

=> The time it takes: 5 W.h / 1.96 W = 2.54 hours

Let area of cross section of pipe be 'A' and velocity of water through it be 'v'.

Given,

A= 3.14 * (0.05)^2 = 0.00785 sq. m, v = 1 m/s.

Kinetic Energy (K) which is transferred for charging = 0.5 * 0.5 * m * ( v^2).

Mass, m is calculated as follows density = mass/volume

1000 = mt/(A * v * t) .......... from eqn. of continuity, t is time m = 7.85t Kg.

So, K = 1.9625t J

And this K is equivalent to the completely charged battery of rating 5 W-h.

1.9625t = 5

t = 2.547 h (Ans)

I did it by working out the mass passing the turbine per second with the velocity to get the KE delivered by the mass of water per second. The mass per second is density X Area x Velocity. Then with 1/2mv^2 the KE "arriving" per second gives the power. Then knowing the 5Wh needed you can covert this to an energy required (x2 due to the efficiency of the system) and work out how long you'd need the water power running for to get the energy.