Functional Testing

Before user tests were conducted, the system needed to be tested on its functional requirements. The system could be user tested when all the “must” functional requirements, specified in section 5.1.1, were fulfilled. When the system did not fulfil one of the requirements, the prototype had been adjusted accordingly. At the end of the functional test, the system fulfilled all the functional “must” requirements specified in section 5.1.1.

7.2.1 Discharge Time

One of the requirements was that the buffers should be able to empty within 2 hours. The time it takes for the water to flow out of the buffer can calculated theoretically. For this Bernoulli’s principle will be used [48].

Bernoulli’s principle is as follow: 𝑣"

2 + 𝑔 ⋅ 𝑧 +

𝑃

𝜌 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Herein v is the flow speed of a fluid through a point, g is the gravitational constant, which in this case will be 9.811₂₃, z is the height from the reference point in meters, P is the pressure and 𝜌 is the density of the fluid, water in this case.

For the rain barrel used in this system, this equation can be used to calculate the outflow time of the water. This is done by using Bernoulli’s equation for two points in this system, first at the surface of the water in the rain barrel, which is point 1 in Figure 31. The second point is located at the opening of the nozzle of barrel, which is point 2 in Figure 31. For ease of calculation, it is assumed that the rain barrel is a perfect cylinder and that the bottom of the nozzle pipe is located exactly 0.05 meters (5cm) above the bottom of the rain barrel. The initial water height is assumed to be 0.9 meters above the bottom of the rain barrel, or 0.85 meters above the bottom of the nozzle. The variable h is the height difference between the water surface in the rain barrel and the bottom of the nozzle. The initial value for this, h0, is thus also 0.85m. d1 is the diameter of the rain barrel, which is 0.57m. d2 is the diameter of the

Because Bernoulli’s equation results in a constant, this can be compared for the two points on the rain barrel. This leads to the following;

453 " + 𝑔𝑧1 + 85 9 = 4"3 " + 𝑔𝑧2 + 8" 9

Because z2 is the reference point, this is set to 0 whereas z1 is equal to h. Furthermore, because both points are in contact with air, there is no pressure difference, therefore this can be negated. Finally, the equations are divided by g, the gravitational constant to make them simpler. The equation then looks like this;

𝑣1"_{= 𝑣2}"_{+ 2𝑔ℎ}

It is safe to assume that v2 is much bigger than v1, because d2 is much smaller than d1. Thus; 𝑣1"_{= 2𝑔ℎ}

And

𝑣1 = 2𝑔ℎ

However, although it is much smaller than v1, v2 is not equal to 0. Instead;

𝑣2 = −𝑑ℎ

𝑑𝑡

The mass flow through the two points is equal (because the same amount of water will flow through these points, albeit at different speeds.) Mass flow rate is given as the flow speed of something multiplied by the area through which this flows. This means that the following holds true;

𝑣1𝐴1 = 𝑣2𝐴2.

With v1 and v2 being the flow speed at their respective points and A1 and A2 being the area of the pipe at their respective points. if the earlier calculated equations for speed are plugged into this equation, this leads to the following equation. R is the radius of the rain barrel and equal to half d1, r is the radius of the faucet and equal to half d2.

2𝑔ℎ ∙ 𝜋 ∙ 𝑟"₌−𝑑ℎ

𝑑𝑡 𝜋𝑅" After moving isolating dh/dt 𝑑ℎ

𝑑𝑡 = − 2𝑔

𝑟"

𝑅" ℎ

Because everything, except for h, in this equation is a constant, we define a placeholder constant k as follow;

𝑘 = − 2𝑔 𝑟"

𝑑ℎ

𝑑𝑡 = 𝑘 ℎ

By changing this equation around a little more and then integrating, this equation can be used to give a formula for the time it takes to empty the barrel. This integration goes from h0 to h1, with h0 being the initial water height and h1 being the final water height (when the barrel is empty) where t is time in seconds. 1 ℎ D5 DE 𝑑ℎ = 𝑘 𝑑𝑡 F G

Which turns into 2 ℎ1 − ℎ0 = 𝑘𝑡

Dividing this by k gives the equation which can be used to calculate the time it takes for the barrel to empty itself.

𝑡 = 2 ℎ1 − ℎ0

𝑘

k was defined as follow;

𝑘 = − 2𝑔 𝑟"

𝑅"

When filling in the values for these constants (g = 9.81m/s2_{, r = 0.005m and R = 0.285m), the value for k is}

found.

𝑘 = − 2 ∗ 9.81 0.005"

0.285"= −0.0013633

Now t, which is the time it takes to empty the barrel, can be calculated. The initial water height, h0 is 0.85 meters above the reference point. The final height h1 is 0cm above the reference point. Filling this into the earlier given equation together with k results in the following;

𝑡 = 2 − 0.85

−0.0013633= 1352.5 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 = 22.56 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

Therefore, in theory it should take 22.56 minutes for the barrel to empty. However, due to friction of the sensors and valves, the time it takes to empty completely the barrel would be longer but will most probably stay under two hours, fulfilling the requirement.