Question

Hey guys, I've got a energy question,

So the question is as follows:

Consider a river flowing toward a lake at an average velocity of 3m/s at a rate of 500m^3/s at a location 90m above the lake surface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location.

Please help! I've attached a diagram below, will give star and follow!

Answer 1

do you have the diagram attached

Answer 2

Yup, it's attached right there

Answer 3

$$
\Large{\rm \frac{Energy}{Unit ~Mass}= \frac{PE +KE}{mass} = \frac{mg + m\frac{v^2}{2}}{m}

}
$$

Answer 4

Ok, can you explain a bit more? I'm not quite getting it

Answer 5

$$
\Large{\rm
Energy~per~unit~Mass
= \frac{Total ~Energy }{Total~ mass}\\~\\
= \frac{PE +KE}{m} = \frac{mgh + m\frac{v^2}{2}}{m}\\
=\frac{m( gh + \frac{v^2}{2})} { m} =\frac{m( 9.81\cdot 90 + \frac{3^2}{2}) Joules} { m~ kg}
\\= \frac{m \cdot 887.4 ~J}{m ~~kg}
\\= \frac{\cancel m\cdot 887.4 ~J}{\cancel m ~kg}
\\= \frac{887.4 ~J}{1~kg}
\\= 887.4~\frac{J}{~kg} = 0.8874 ~ \frac{kJ}{kg}
}
$$

Answer 6

Ahh, sweet, thanks!

Answer 7

The power generation potential can be found by multiplying the mechanical energy by the mass flow rate .

Answer 8

$$\rm{\large{
mass = density \times volume \\
\frac{d}{dt}(m)= \frac{d}{dt}(\rho V) = \rho\cdot \frac{dV}{dt}
\\= \frac{1000~ kg}{m^3}\cdot 500 \frac{m^3}{s}
\\ = 500000 \frac{kg}{s}
\\~\\
Power~Potential = mass~flow ~rate \times mechanical~ energy
\\ = 500000 \frac{kg}{s} \cdot0.887 \frac {kJ}{kg}
= 444000 ~ \frac{kJ}{s}\\
= 444000 ~ kW
= 444~ MW ~ of ~power
}}
$$

Answer 9

This river provides 444 megawatts of power, assuming there is no loss of energy due to inefficiencies.

Answer 10

note how the units cancel nicely to give you energy per time (power)