1. You are planning to build a mini hydro power plant. A river near your home is w = 4.0 m wide and h = 1.0 m deep. You measure the water flow to be v = 1.50 m/s over the brink of a waterfall H = 10.0 m high. The company that makes the turbine advertises e = 25% efficiency in converting the potential energy of the water into electric energy. Derive an expression for the generated power in terms of the above variables and any physics constants. How many 100watts bulbs will this power handle?
4 x 1 x 1.5 = 6 cubic meters of water per second = 6000 liters of water = 6000 kilos of water
amount of potential energy = m*g*h
E = 6000 * 9.81 * 10 = 588600 joules
25% efficiency => 147150 joules of energy
1 Watt = 1 joule / second
147150 / 100 = 1471.5 bulbs
this is wrong^
To derive an expression for the generated power of the mini hydro power plant, you need to consider a few factors: the volume flow rate of the water, the potential energy of the falling water, and the efficiency of the turbine.
First, let's find the volume flow rate of the water. The volume flow rate (Q) can be calculated by multiplying the cross-sectional area of the river (A) by the velocity of the water (v):
Q = A * v
Given that the river is 4.0 m wide and 1.0 m deep, the cross-sectional area would be:
A = width * depth = 4.0 m * 1.0 m = 4.0 m^2
So, the volume flow rate can be expressed as:
Q = 4.0 m^2 * 1.50 m/s = 6.0 m^3/s
Next, let's calculate the potential energy of the falling water. The potential energy (PE) is given by the equation:
PE = mass * gravity * height
Since the mass is equal to the density of water (ρ) multiplied by the volume (V), and the volume is equal to the cross-sectional area (A) multiplied by the height (H), we can rewrite the equation as:
PE = ρ * A * H * g
The density of water (ρ) is approximately 1000 kg/m^3, and the acceleration due to gravity (g) is approximately 9.8 m/s^2. Substituting these values and the given measurements, we get:
PE = 1000 kg/m^3 * 4.0 m^2 * 10.0 m * 9.8 m/s^2
= 392,000 J
Now, let's calculate the generated power. The generated power (P_gen) is equal to the potential energy of the water multiplied by the efficiency of the turbine (e), divided by time (t):
P_gen = (PE * e) / t
Since the problem doesn't specify a time interval, we'll assume a standard value of 1 second, so:
P_gen = PE * e
Substituting the values we calculated earlier, the expression for the generated power becomes:
P_gen = 392,000 J * 0.25
= 98,000 J
Finally, to determine how many 100-watt bulbs this power can handle, we just need to divide the generated power by the power consumption of a single bulb:
Number of bulbs = P_gen / Power per bulb
Number of bulbs = 98,000 J / 100 W
= 980 bulbs
Therefore, this mini hydro power plant will be able to handle 980 100-watt bulbs.