Water flows (v = approx 0) over a dam at the rate of 660 kg/s and falls vertically 86 m before striking the turbine blades. Calculate the rate at which mechanical energy is transferred to the turbine blades, assuming 53% efficiency.
Express your answer using two significant figures. P = W ??
Power out = (efficiency)*(mass flow rate)*(potential energy loss per mass)
= 0.53*(660 kg/s)*(9.8 m/s^2)*(86 m)
The answer will be in Watts
Well, well, well, looks like we've got a hydroelectric situation here! Let's get quacking on the math, shall we?
First, let's find the potential energy of the falling water. The potential energy (PE) can be calculated using the formula PE = mgh, where m is the mass of the water, g is the acceleration due to gravity, and h is the height fallen.
PE = (660 kg/s)(9.8 m/s^2)(86 m) = 557208 J/s
Now, let's account for the efficiency of the turbine. Since it's mentioned to be 53% efficient, we need to multiply the potential energy by 0.53 to find the actual mechanical energy transferred to the turbine blades.
Mechanical energy transferred = (PE)(0.53) = 557208 J/s * 0.53 ≈ 295609 J/s
So, the rate at which mechanical energy is transferred to the turbine blades is approximately 295609 J/s. And that, my friend, is the power of hydroelectric humor!
To calculate the rate at which mechanical energy is transferred to the turbine blades, we need to find the power output.
Power (P) is given by the formula:
P = W / t
Where:
P is the power,
W is the work done,
t is the time taken.
The work done (W) can be calculated as:
W = mgh
Where:
m is the mass of water per second (660 kg/s),
g is the acceleration due to gravity (9.8 m/s^2),
h is the height through which the water falls (86 m).
Let's plug in the given values and calculate W:
W = 660 kg/s * 9.8 m/s^2 * 86 m
W = 560088 Joules per second
Now, we need to find the power output by using the efficiency (53%). Power output is the product of the efficiency and the work done:
Power output = Efficiency * W
Power output = 0.53 * 560088 Joules per second
Now, we can round the result to two significant figures:
Power output = 297000 Joules per second
P = 300000 Joules per second (to two significant figures)
To calculate the rate at which mechanical energy is transferred to the turbine blades, we can use the formula:
P = mghη,
where
P is the power (rate of energy transfer),
m is the mass flow rate of water,
g is the acceleration due to gravity,
h is the height of the dam,
and η is the efficiency.
Given values:
m = 660 kg/s,
g = 9.8 m/s²,
h = 86 m,
η = 53% = 0.53.
Substituting these values into the formula:
P = (660 kg/s)(9.8 m/s²)(86 m)(0.53).
Calculating this expression:
P ≈ 2,472,504 W.
Therefore, the rate at which mechanical energy is transferred to the turbine blades is approximately 2,472,504 Watts or 2.5 MW (megawatts) when rounded to two significant figures.