# Some mathematics of the 1st rapid at Penrith

If you are into kayaking the Penrith Whitewater Stadium is worth visiting.  To an engineer, the first rapid looks like a classic ‘flow over a hump in a channel’ situation with flow being forced to critical depth and then forming a hydraulic jump downstream that becomes a ‘stopper’ for kayakers.

The first rapid at Penrith with and without water

The standard analysis, from any hydraulics textbook, is as follows (see for example the discussion here).

Specific energy $E$ can be written as.

$E = y + \frac{Q^2}{2gA^2}$     (1)

where, $y$ is the depth, $Q$ is the discharge and $A$ is the area of flow.  At critical depth, the specific energy is a minimum so for constant Q, we take the derivative of E and set it to zero.

$\frac{dE}{dy} = 1 - \frac{Q^2}{gA^3}\frac{dA}{dy} = 0$      (2)

Now, for a rectangular chute, $\frac{dA}{dy} = T$   (the top width).     (3)

Therefore:

$\frac{Q^2 T_c}{gA_c^3} = 1$      (4)

Where the subscript c designates critical conditions.

The flow cross-section at this first rapid is approximately rectangular so the area can be written as:

$A = y \times T$      (5)

Therefore, equation (4) can be written as:

$\frac{Q^2}{gy_c^3T^2} = 1$      (6)

or

$Q = \sqrt{ \left( g y^3T^2 \right)}$      (7)

We can use equation (7) to estimate the discharge.

Looking at Google Earth, the width of the chute is about 4 m and, from the picture above, the critical depth looks to be about 1 m.

$Q = \sqrt{ \left( 9.8 \times 1^3 \times 4^2 \right)} = 12.5 \, \mathrm{m^3s^{-1}}$

The stadium website also provides information on the flow. There were 5 pumps operating when the photo above was taken and each pump has a capacity of 2.8 m3s-1.  So the total discharge was about 14 m3s-1.  Pretty close to the calculated value.

The velocity, at critical depth, is about:

$\frac{Q}{A} \approx 14/(1 \times 4) = 3.5 \, \mathrm{ms^{-1}}$

When you are in the chute, the water seems very fast but you need to paddle faster than the flow. On my first time down the rapid, I didn’t have enough boat speed to get through the stopper which meant my inflatable kayak was sucked back into the supercritical flow and filled up with water.

We can also look at how much energy is being dissipated.  The equation for power of water flow is:

$P = \rho g Q h$

The height difference between the start and end of the course is 5.5 m.  Take discharge as 14 m3s-1 and $\rho$ = 1000 kg/m3 (the density of water). Therefore:

$P = 1000 \times 9.8 \times 14 \times 5.5 = 754.6 \mathrm{kW}$

The pumps are electrically driven and I don’t know what the ‘wire to water’ efficiency is but lets say 75%. Therefore, the electrical energy consumed is:

$750/0.75 = 1000 \mathrm{kW}$

Guessing the price of electricity at $0.15/kWh means the pumps would use about$150 worth of electricity per hour. If the pumps operate for 6 hours per day the cost per month would be $27,000. This newspaper article suggests electricity costs are$30,000 per month so we are in the right ball-park.

All this swirling around in rapids is going to heat the water up. The specific heat of water is 4.2 kJ/kg. This is the energy required to heat up 1 kg of water by 1 oC. Therefore, we need to know the total total mass of water on the course.

The course is 320 m long and let’s guess the average width as 8 m and depth as 1.5 m. Therefore, the total volume is 3840 m3. There are also large pools at the start and end. The top pool has a volume of about 1200 m3 and the bottom one about 1500 m3. This is based on measurements from Google Earth and assuming an average depth of 1.5 m. There is also a connection to a nearby lake but let’s ignore that. Therefore the total water volume being heated up is 3840 + 1200 + 1500 = 6540 m3 = 6.5 x 106 kg.

About 750 kW is being dissipated per hour which is 2.7 x 106 kJ.

Therefore the amount of energy per kg per hour is

$\frac{2.7 \times 10^6}{6.5 \times 10^6} = 0.42 \mathrm{kJ/kg}$

This is one tenth of the amount of energy required to heat water up by 1 oC. So we are only increasing the water temperature but about 0.1 oC per hour. Not really noticeable.

## One thought on “Some mathematics of the 1st rapid at Penrith”

1. Colin Caprani

Nice work! Always good to get approximate agreement too. The top photo shows a small wall maybe 2-3 m downstream of the drop to force the stopper to form. What effect does this have on the hydraulic jump analysis?