Elementary Principles of Chemical Processes, Binder Ready Version
4th Edition
ISBN: 9781118431221
Author: Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher: WILEY

#### Concept explainers

Textbook Question
100%
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 3 . 25 × 1 0 4  kJ / 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 2 . 34 × 1 0 4  kJ / 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?

Expert Solution
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.

The power provided by given gasoline engine in kW and hp is 2.1666kW and 2.903hp respectively.

### Explanation of Solution

The efficiency of gasoline engine is 30%.

The rate consumption of gasoline by the given engine is 0.80L/h.

The heating value of gasoline is 3.25×104kJ/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.

P=(3.25×104kJ/L)(0.80L/h)(1h3600s)=7.2222kJ/s

The power provided by engine is represented as,

P'=γP

Where,

• γ represents the efficiency of engine.
• P represent the power provided by fuel.

Substitute the value of γ and P in the above equation.

P'=(30%100%)(7.2222kJ/s)=(2.1666kJ/s)(1kW1kJ/s)=2.1666kW

The value of actual power of engine in hp is represented as,

P=(2.1666kW)(1.341hp1KW)=2.903hp

Therefore, the power provided by given gasoline engine in kW and hp is 2.1666kW and 2.903hp respectively.

Expert Solution
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.

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

### Explanation of Solution

The efficiency of gasoline engine is 30%.

The power provided by gasoline is 7.2222kJ/s.

The heating value of gasoline is 3.25×104kJ/L.

The heating value of ethanol is 2.34×104kJ/L.

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

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

V1%=100%10%=90%

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

The heating value of fuel mixture is given as,

h=h1V1%+h2V2%100%

Where,

• h1 represents the heating value of gasoline.
• V1% represents the volume percentage of gasoline in the given mixture.
• h2 represents the heating value of ethanol.
• V2% represents the volume percentage of ethanol in the given mixture.

Substitute the value of h1, V1%, h2 and V2% in the above equation.

h=(3.25×104kJ/L)(90%)+(2.34×104kJ/L)(10%)100%=3.159×104kJ/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 heating value of fuel.

Rearrange the above equation for the value of r.

r=Ph

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

r=(7.2222kJ/s3.159×104kJ/L)(3600s1h)=0.823L/h

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

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