Chapter 6, Problem 102QAP

To determine

(a)

The average output power per hydroelectric plant.

Answer to Problem 102QAP

The average output power per hydroelectric plant is $7.77\text{MW/dam}$.

Explanation of Solution

Given:

  $\begin{array}{l}E=282\times {10}^9\text{kWh}\\ numberofdams=4138\\ Elecrticalenergy=90\%\\ h=50.0\text{m}\\ \text{1}\text{.00}\text{kWh=3}\text{.60}\times {\text{10}}^6\text{J}\end{array}$

Concept Used:

Theory of power for transfer of energy.

Calculation:

According to theory of power for transfer of energy:

  $\begin{array}{l}P=\frac{W(Amountofworkdone)}{t(Time)}\\ P=\frac{282\times {10}^9\text{kWh}}{(1\text{yr})(4138\text{dams})}\times \frac{3.60\times {10}^6\text{J}}{1.00\text{kWh}}\times \frac{1\text{yr}}{365.25\text{days}}\times \frac{1\text{day}}{24\text{hr}}\times \frac{1\text{hr}}{3600\sec }\\ P=7.77\times {10}^6\text{W/dam}\times \frac{1\text{MW}}{{10}^6\text{W}}\\ P=7.77\text{MW/dam}\end{array}$

Conclusion:

The average output power per hydroelectric plant is $7.77\text{MW/dam}$.

To determine

(b)

The total mass of water flowed over the dams during $2006$.

Answer to Problem 102QAP

The total mass of water flowed over the dams during $2006$ is $2.30\times {10}^{15}\text{kg}$.

Explanation of Solution

Given:

  $\begin{array}{l}E=282\times {10}^9\text{kWh}\\ numberofdams=4138\\ Elecrticalenergy=90\%\\ h=50.0\text{m}\\ g\text{=9}\text{.80}{\text{m/sec}}^2\\ \text{1}\text{.00}\text{kWh=3}\text{.60}\times {\text{10}}^6\text{J}\end{array}$

Concept Used:

Interpreting potential energy.

Calculation:

Potential energy $U$ associated with the gravity during its motion:

  $U_{grav}=mgh$

As per given condition each plant is efficient of converting $90\%$ mechanical energy to electrical energy.

  $\begin{array}{l}{U}_{grav}=282\times {10}^9\times \frac{100\%mechanicalenergy}{90\%electricalenergy}\\ U_{grav}=3.13\times {10}^{11}\text{kWh}\\ m=\frac{U_{grav}}{gh}\end{array}$

On replacing the values, we get

  $\begin{array}{l}g=9.8{\text{m/sec}}^{\text{2}}\\ h=50.0\text{m}\end{array}$

  $\begin{array}{l}{U}_{grav}=3.13\times {10}^{11}\text{kWh}\\ \text{1}\text{.00}\text{kWh=3}\text{.60}\times {\text{10}}^6\text{J}\\ m=\frac{3.13\times {10}^{11}\text{kWh}}{9.8\times 50.0}\times \frac{\text{3}\text{.60}\times {\text{10}}^6\text{J}}{\text{1}\text{.00}\text{kWh}}\\ m=2.30\times {10}^{15}\text{kg}\end{array}$

Conclusion:

The total mass of water flowed over the dams during $2006$ is $2.30\times {10}^{15}\text{kg}$.

To determine

(c)

The average mass of water per dam and average volume of water per dam that provides the mechanical energy to generate the electricity.

Answer to Problem 102QAP

The average mass of water per dam is $5.56\times {10}^{11}\text{kg/dam}$ and average volume of water per dam that provides the mechanical energy to generate the electricity is $5.56\times {10}^8{\text{m}}^{\text{3}}\text{/dam}$.

Explanation of Solution

Given:

  $\begin{array}{l}E=282\times {10}^9\text{kWh}\\ numberofdams=4138\\ Elecrticalenergy=90\%\\ h=50.0\text{m}\\ g\text{=9}\text{.80}{\text{m/sec}}^2\\ \text{1}\text{.00}\text{kWh=3}\text{.60}\times {\text{10}}^6\text{J}\\ \text{Density=1000}{\text{kg/m}}^3\end{array}$

Concept Used:

Interpreting potential energy.

.

Calculation:

Potential energy $U$ associated with the gravity during its motion:

  $U_{grav}=mgh$

As per given condition each plant is efficient of converting $90\%$ mechanical energy to electrical energy.

  $\begin{array}{l}{U}_{grav}=282\times {10}^9\times \frac{100\%mechanicalenergy}{90\%electricalenergy}\\ U_{grav}=3.13\times {10}^{11}\text{kWh}\\ m=\frac{U_{grav}}{gh}\end{array}$

On replacing the values, we get

  $\begin{array}{l}g=9.8{\text{m/sec}}^{\text{2}}\\ h=50.0\text{m}\end{array}$

  $\begin{array}{l}{U}_{grav}=3.13\times {10}^{11}\text{kWh}\\ \text{1}\text{.00}\text{kWh=3}\text{.60}\times {\text{10}}^6\text{J}\\ m=\frac{3.13\times {10}^{11}\text{kWh}}{9.8\times 50.0}\times \frac{\text{3}\text{.60}\times {\text{10}}^6\text{J}}{\text{1}\text{.00}\text{kWh}}\\ m=2.30\times {10}^{15}\text{kg}\end{array}$

  $Averagemassofwaterperdam=\frac{Massofwater}{No.ofdams}$

On replacing the value, we get

  $\begin{array}{l}Averagemassofwaterperdam=\frac{2.30\times {10}^{15}\text{kg}}{4138dams}\\ Averagemassofwaterperdam=5.56\times {10}^{11}\text{kg/dam}\end{array}$

The average volume of water per dam that provides the mechanical energy to generate the electricity is:

  $\begin{array}{l}Volume=\frac{Averagemassofwaterperdam}{Densityofwater}\\ Volume=\frac{5.56\times {10}^{11}}{{10}^3}=5.56\times {10}^8{\text{m}}^{\text{3}}\text{/dam}\end{array}$

Conclusion:

The average mass of water per dam is $5.56\times {10}^{11}\text{kg/dam}$ and average volume of water per dam that provides the mechanical energy to generate the electricity is $5.56\times {10}^8{\text{m}}^{\text{3}}\text{/dam}$.

To determine

(d)

The gallons that the gasoline for the $4138$ dams save.

Answer to Problem 102QAP

The gallons that the gasoline for the $4138$ dams save is $22.6\times {10}^9\text{gallons}$.

Explanation of Solution

Given:

  $\begin{array}{l}E=282\times {10}^9\text{kWh}\\ numberofdams=4138\\ Elecrticalenergy=90\%\\ h=50.0\text{m}\\ g\text{=9}\text{.80}{\text{m/sec}}^2\\ \text{1}\text{.00}\text{kWh=3}\text{.60}\times {\text{10}}^6\text{J}\\ \text{Density=1000}{\text{kg/m}}^3\end{array}$

A gallon of gasoline contains $45.0\times {10}^6\text{J}$ of energy.

Concept Used:

Interpreting potential energy.

Calculation:

We know that energy is given as:

  $\begin{array}{l}E=282\times {10}^9\text{kWh}\\ \text{1}\text{.00}\text{kWh=3}\text{.60}\times {\text{10}}^6\text{J}\end{array}$

A gallon of gasoline contains $45.0\times {10}^6\text{J}$ of energy

So, the gallons that the gasoline for the $4138$ dams save is:

  $\begin{array}{l}=282\times {10}^9\text{kWh}\times \frac{\text{3}\text{.60}\times {\text{10}}^6\text{J}}{\text{1}\text{.00}\text{kWh}}\times \frac{1\text{gal}}{45.0\times {10}^6\text{J}}\\ =22.6\times {10}^9\text{gal}\end{array}$

Conclusion:

The gallons that the gasoline for the $4138$ dams save is $22.6\times {10}^9\text{gallons}$.