Chapter 7, Problem 7.1P

A certain gasoline engine has an efficiency of 30%: that is, it converts into useful work 30% of the heat generated by burning a fuel.

(a) 1f the engine consumes 0.80 L/h of a gasoline with a heating value of $\text{3}.\text{25}\times \text{1}{0}^4\text{ kJ}/\text{L}$, how much power does it provide? Express the answer both in kW and horsepower.

(b) Suppose the fuel is changed to include 10% ethanol by volume. The heating value of ethanol is approximately $\text{2}.\text{34}\times \text{1}{0}^4\text{ kJ}/\text{L}$and volumes of gasoline and ethanol may be assumed additive. At what rate (L/h) does the fuel mixture have to be consumed to produce the same power as gasoline?

Interpretation Introduction

(a)

Interpretation:

The power provided by the given gasoline engine is to be calculated.

Concept introduction:

An engine is a type of mechanical machine that convert different type of energy into useful mechanical energy. Different types of fuel are burnt in engine to produce heat. The energy produced in the form of heat is converted into mechanical work.

Answer to Problem 7.1P

The power provided by given gasoline engine in $\text{kW}$ and $\text{hp}$ is $2.1666\text{kW}$ and $2.903\text{hp}$ respectively.

Explanation of Solution

The efficiency of gasoline engine is $30\%$.

The rate consumption of gasoline by the given engine is $0.80\text{L}/\text{h}$.

The heating value of gasoline is $3.25\times {10}^4\text{kJ}/\text{L}$.

The relation between the rate of fuel consumption and power of provided by fuel is given as,

$P=rh$

Where,

• $P$ represents the power of provided by fuel.
• $r$ represents the rate of fuel consumption.
• $h$ represents the heat value of fuel.

Substitute the value of rate of consumption of gasoline and power of engine in the above equation.

$\begin{array}{c}P=\left(3.25\times {10}^4\text{kJ}/\text{L}\right)\left(0.80\text{L}/\text{h}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\\ =7.2222\text{kJ}/\text{s}\end{array}$

The power provided by engine is represented as,

$P\text{'}=\gamma P$

Where,

• $\gamma$ represents the efficiency of engine.
• $P$ represent the power provided by fuel.

Substitute the value of $\gamma$ and $P$ in the above equation.

$\begin{array}{c}P\text{'}=\left(\frac{30\%}{100\%}\right)\left(7.2222\text{kJ}/\text{s}\right)\\ =\left(2.1666\text{kJ}/\text{s}\right)\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =2.1666\text{kW}\end{array}$

The value of actual power of engine in $\text{hp}$ is represented as,

$\begin{array}{c}P=\left(2.1666\text{kW}\right)\left(\frac{1.341\text{hp}}{1\text{KW}}\right)\\ =2.903\text{hp}\end{array}$

Therefore, the power provided by given gasoline engine in $\text{kW}$ and $\text{hp}$ is $2.1666\text{kW}$ and $2.903\text{hp}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The rate at which the given fuel mixture consumed to produce the same power as gasoline is to be calculated.

Concept introduction:

An engine is a type of mechanical machine that convert different type of energy into useful mechanical energy. Different types of fuel are burnt in engine to produce heat. The energy produced in the form of heat is converted into mechanical work.

Answer to Problem 7.1P

The rate at which the given fuel mixture consumed to produce the same power as gasoline is $0.823\text{L}/\text{h}$.

Explanation of Solution

The efficiency of gasoline engine is $30\%$.

The power provided by gasoline is $7.2222\text{kJ}/\text{s}$.

The heating value of gasoline is $3.25\times {10}^4\text{kJ}/\text{L}$.

The heating value of ethanol is $2.34\times {10}^4\text{kJ}/\text{L}$.

The volume percentage of ethanol in fuel mixture is $10\%$.

The volume percentage of gasoline in fuel mixture is represented as,

$\begin{array}{c}{V}_1\%=100\%-10\%\\ =90\%\end{array}$

Therefore, the volume percentage of gasoline in fuel mixture is $90\%$.

The heating value of fuel mixture is given as,

$h=\frac{h_1{V}_1\%+h_2{V}_2\%}{100\%}$

Where,

• $h_1$ represents the heating value of gasoline.
• $V_1\%$ represents the volume percentage of gasoline in the given mixture.
• $h_2$ represents the heating value of ethanol.
• $V_2\%$ represents the volume percentage of ethanol in the given mixture.

Substitute the value of $h_1$, $V_1\%$, $h_2$ and $V_2\%$ in the above equation.

$\begin{array}{c}h=\frac{\left(3.25\times {10}^4\text{kJ}/\text{L}\right)\left(90\%\right)+\left(2.34\times {10}^4\text{kJ}/\text{L}\right)\left(10\%\right)}{100\%}\\ =3.159\times {10}^4\text{kJ}/\text{L}\end{array}$

The relation between the rate of fuel consumption and power of provided by fuel is given as,

$P=rh$

Where,

• $P$ represents the power of provided by fuel.
• $r$ represents the rate of fuel consumption.
• $h$ represents the heating value of fuel.

Rearrange the above equation for the value of $r$.

$r=\frac{P}{h}$

Substitute the value of power generated by gasoline and heating value of fuel mixture in above equation.

$\begin{array}{c}r=\left(\frac{7.2222\text{kJ}/\text{s}}{3.159\times {10}^4\text{kJ}/\text{L}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\\ =0.823\text{L}/\text{h}\end{array}$

Therefore, the rate at which the given fuel mixture consumed to produce the same power as gasoline is $0.823\text{L}/\text{h}$.