The Simple Physics of Energy Use by Unknown

Author:Unknown
Language: eng
Format: epub

V

γ –1

1

η

1

= 1 –

= 1 –

(9.18)

V 2

rγ –1

The thermal efficiency improves as the compression ratio is increased. A car these

days might have a compression ratio of 10, which implies a thermal efficiency of

60%. In practice, it is not that good. There are various losses and inefficiencies

that all combine to reduce the overall efficiency.1 The intake and compression

stage is not perfect; there are pressure losses as the air gets sucked through the air

filter and the inlet manifold, and the inlet valve might not close perfectly. These

are called volumetric losses, and are of the order of 15%. The mixture might not

be optimal, although, with electronic fuel injection, where the exact amount of

fuel is carefully regulated according to the air mass sucked into the engine, this

is less of a problem. We will allow for the possibility of a 10% loss due to im-

perfect combustion. There is also friction in the engine bearings and from the

piston sliding up and down in the cylinder, which could potentially lose another

12%. Accessories such as the alternator, fuel pump, power steering pumps and

air conditioning cause further losses. Let us assume that these only result in a

5% loss in power, although this might be a serious underestimation with all the

things added to the average modern car. Finally, there is potential loss of another

5% from badly fitting piston rings and exhaust blowback. The combination of all

these power losses reduce the efficiency from 60% to about 36%.

1 The analysis is based on aircraft pis-

ton engines (Busch, 2014). Automobile

0.6

engines could have lower volumetric losses

× 0.85 × 0.90 × 0.88 × 0.95 × 0.95 ∼ 0.36

and better-adjusted fuel/air mixtures, but

have greater losses from cooling and more

A characteristic of internal combustion engines is that the turning force, or torque,

power-robbing accessories such as air con-

that they produce, turning the crankshaft, increases with angular velocity or rpm

ditioning.

136

Ground Transportation: Road and Rail

FIAT

2012 MY Engine Performance Rating

1.4L Multiair - Fiat 500

140

140

Torque

120

120

Power

100

100

101 bhp @ 6500 rpm

133 Nm @ 4000 rpm

(Nm)

80

80

er (bhp)w

oeque

60

60

rected T

rected po

Cor

Cor

40

40

20

20

0

0

1000

2000

3000

4000

5000

6000

7000

Fig. 9.7 Torque and Power Curves

Engine Speed (rpm)

for a Petrol Engine (Fiat Chrysler

Net engine output obtained and corrected

in accordance with SAE J1349 March 2008

Automobiles).

until it reaches a maximum before dropping off. Power is torque multiplied by

angular velocity, so the maximum power is reached at a higher angular velocity or

rpm than the maximum torque. For the Fiat engine shown in Figure 9.7, the max-

imum torque is at 4000 rpm, although in practice the torque is in an acceptable

range of about 2000–5000 rpm. To match the engine’s angular velocity (rpm),

which gives acceptable torque to the angular velocity of the wheels as the car ac-

celerates, requires different gears, which allow for changing the ratio of these two

angular velocities. From Figure 9.7, the power output in the normal engine op-

erating range of 2000–3000 rpm is only about 40–50% of the peak power, so the

overall efficiency is less than 20%.

How does the diesel engine compare?

Otto and diesel cycles

137

Q1

P

2

3

4

Q2

1

V

V2

V3

V1 = V4

1-2 Isentropic compression

2-3 Isobaric addition of heat

3-4 Isentropic expansion

Fig. 9.8 Pressure Against Volume for a

4-1 Heat removal at constant volume

Diesel Engine.

The four stages are shown in the pressure–volume diagram of Figure 9.

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