Fuel consumption in a truck Let’s begin with an efficiency analysis of a gasoline engine. The engine in a large truck takes in 2500 J of heat and delivers 500 J of mechanical work per cycle. The heat is obtained by burning gasoline with heat of combustion Lc = 5.0 × 10^4 J/g. (a) What is the

Question

Fuel consumption in a truck

Let’s begin with an efficiency analysis of a gasoline engine. The engine in a large truck takes in 2500 J of heat and delivers 500 J of mechanical work per cycle. The heat is obtained by burning gasoline with heat of combustion [latex]L_c = 5.0 imes 10^4 J/g.[/latex] (a) What is the thermal efficiency of this engine? (b) How much heat is discarded in each cycle? (c) How much gasoline is burned during each cycle? (d) If the engine goes through 100 cycles per second, what is its power output in watts? In horsepower? (e) How much gasoline is burned per second? Per hour?

Accepted Answer

SET UP It’s often useful to make an energy-flow diagram for heatengine problems; Figure 16.2 shows what we draw. We are given that [latex]\mathrm{Q_H = 2500 J and W = 500 J.}[/latex]

SOLVE Part (a): The thermal efficiency is found from Equation 16.3:

[latex]\mathrm{e=\frac{W}{Q_H}=\frac{500 J}{2500 J}=0.20 =20 } [/latex]%.

Part (b): From Equation 16.2, the heat [latex]\mathrm{Q_C}[/latex] discarded per cycle is the difference between the heat absorbed [latex]\mathrm{(Q_H)}[/latex] and the work W done by the engine:

[latex]\mathrm{W=Q=Q_H+Q_C=\left|Q_H ight|-\left|Q_C ight| }. [/latex] (16.2)

[latex]\mathrm{W = Q_H + Q_C,} [/latex]

[latex]\mathrm{500 J = 2500 J + Q_C,}[/latex]

[latex]\mathrm{Q_C = -2000 J.} [/latex]

Thus, 2000 J of heat leaves the engine during each cycle.
Part (c): Let m be the mass of gasoline burned during each cycle; then [latex]\mathrm{Q_H}[/latex] is m times the heat of combustion:[latex]\mathrm{Q_H = mL_c} [/latex]. Thus,

[latex]\mathrm{m=\frac{Q_H}{L_c}=\frac{2500 J}{5.0 imes 10^4 J/g} =0.050 g. } [/latex]

Part (d): The power P (rate of doing work) is the work per cycle multiplied by the number of cycles per second:

P = (500 J/cycle)(100 cycles/s) = 50,000 W = 50 kW, or

P = (50,000 W) (1 hp/746 W) = 67 hp.

Part (e): The mass of gasoline burned per second is the mass burned per cycle multiplied by the number of cycles per second:

(0.050 g/cycle) (100 cycles/s) = 5.0 g/s.

The mass burned per hour is

[latex]\mathrm{(5.0 g/s)\left(\frac{3600 s}{1 h} ight)=18,000 g/h=18 kg/h. } [/latex]

REFLECT The efficiency from part (a) is a fairly typical figure for cars and trucks if W includes only the work actually delivered to the wheels.
The density of gasoline is about 0.70 g/cm³ ; the volume of fuel burned per hour is about 25,700 cm³ , 25.7 L, or 6.8 gallons of gasoline per hour. If the truck is traveling at 55 mi/h (88 km/h), that volume represents fuel consumption of about 8.1 miles per gallon (mpg) (3.4 km/L).
Each cylinder in the engine goes through one cycle for every two revolutions of the crankshaft. For a four-cylinder engine, 100 cycles per second corresponds to 50 crankshaft revolutions per second, or [latex]\mathrm{3000 rev/\min}[/latex] (typical for highway speeds).

Practice Problem: A gasoline engine with a thermal efficiency of 28% has a power output of 50 hp. How much heat must be supplied per second to the engine? Answer: [latex]\mathrm{Q_H = 133 kJ.}[/latex]

Question 16.1: Fuel consumption in a truck Let’s begin with an efficiency a...

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