Question

In a hydroelectric dam, water falls 25 m and then spins a turbine to generate electricity.

a. What is ∆UG of 1.0 kg of water?

b. Suppose the dam is 80% efficient at converting the water’s potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 50 MW of electricity? This is a typical value for a small hydroelectric dam.

Short answer

(a) As a result, the potential energy of the water descending from the dam's top changes $245J$.

(b) As a result, the required amount of water per second is$2.55\times {10}^5\mathrm{kg}$.

Step-by-step solution

Introduction

The following is the formula for calculating the change in gravitational potential energy:

$\Delta U_G=mgh$

we have$h$= height and $m$=mass.

The formula for calculating the power is:

$P=\frac{W}{t}$

we have, $W$= total work done, $t$is the time and$P$= power.

The amount of water needed is calculated as follows:

$\text{Amount of water required}=\frac{\text{Total work to be done}}{\text{Work done by}1\mathrm{kg}\text{of water}}$

Explanation (a)

Recall the expression for potential energy change.

$\Delta U_{\mathrm{G}}=mgh$

Putting $1\mathrm{kg}$for $m,9.8\mathrm{m}/{\mathrm{s}}^2$for $g$and $25\mathrm{m}$for $h$,

$\Delta U_{\mathrm{G}}=(1\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(25\mathrm{m})$

$=245\mathrm{J}$

As a result, the potential energy of the water descending from the dam's top changes $245\mathrm{J}$.

Explanation (b)

Remember the formula for calculating power:

$P=\frac{W}{t}$

To find the value of, rearrange the expression $W$,

$W=P\times t$

The amount of energy that must be generated is enormous $50\mathrm{MW}$.

Putting $50\mathrm{MW}$for $P$and $1\mathrm{s}$for $t$.

$W=\left(50\mathrm{MW}\times \frac{{10}^6\mathrm{W}}{1\mathrm{MW}}\right)(1\mathrm{s})$

$=5.0\times {10}^7\mathrm{J}$

As a result, $5.0\times {10}^7\mathrm{J}$the dam requires electricity every second. Since the dam has been built, $80\%$the turbine's efficiency is limited to the amount of energy it can convert $80\%$It converts the potential energy of falling water into electrical energy. The work done by the water is equal to the change in potential energy of the falling water.

The amount of effort that can be done per kilogramme of water is,

$W_{eff}=(245\mathrm{J})\left(\frac{80}{100}\right)\mathrm{J}/\mathrm{kg}$

$=196\mathrm{J}/\mathrm{kg}$

Recall the expression to calculate the amount of water required,

$\text{Amount of water required}=\frac{\text{Total work to be done}}{\text{Work done by}1\mathrm{kg}\text{of water}}$

Putting $W$for Total work to be done and $W_{eff}$for Work done by $1\mathrm{kg}$of water.

Amount of water required $=\frac{\mathrm{W}}{E_{eff}}$.

Putting $5.0\times {10}^7\mathrm{J}$for $W$and $196\mathrm{J}/\mathrm{kg}$for $W_{df\text{'}}$,

$\text{Amount of water required}=\frac{5.0\times {10}^7\mathrm{J}}{196\mathrm{J}/\mathrm{kg}}$

$=2.55\times {10}^5\mathrm{kg}$

As a result, the required amount of water per second is $2.55\times {10}^5\mathrm{kg}$.