Page 358 - Advanced thermodynamics for engineers

16.2 FURTHER CONSIDERATIONS OF BASIC ENGINE CYCLES 347

Since the purpose of this book is to discuss thermodynamic principles rather than detailed engine
design, having pointed out a practical aspect of engine operation, we will assume that ivc is at bdc; this
makes the nominal and effective compression ratios equal.
It was shown that the thermal efficiency of an Otto cycle is defined by Eqn (3.16)
1
h Otto ¼ 1
r ðk 1Þ
where r is the compression ratio. It is apparent from this expression that higher values of compression
ratio, r, result in higher thermal efficiencies. However, differentiating Eqn (3.16) with respect to r gives
dh th ðk 1Þ
¼ ; (16.3)
dr r k
which indicates that there are diminishing returns with increasing compression ratio, and by a
compression ratio of 20:1 the increase of efficiency per unit increase in r is only 0.60%, or about 1%
improvement in power output (based on an instantaneous efficiency of 60%).
It is also necessary to introduce another parameter that is commonly used to assess the load on a
reciprocating engine: this is referred to as the mean effective pressure (mep), p. If the mep is evaluated
from the indicator diagram of an air-standard cycle (or any other p–V diagram) then it is called the
indicated mean effective pressure (imep). The mep is defined as the pressure, which if it operated over
the whole swept volume, would produce the same work output as the actual cycle. Let the mep be
denoted by p, then
I
pV s ¼ pdV (16.4)

Hence, for the Otto cycle, the imep is
H
pdV
p ¼ (16.5)
i
V s
The work output, H pdV, can be evaluated from the energy addition and the thermal efficiency.
The energy addition is equivalent to the product of the lower calorific value of the fuel ðQ Þ, which is
0
v
the negative of the heat of reaction (Q v ), and the mass of fuel, which is normally defined in terms of the
air–fuel ratio, ε. The quantity of energy added in the air-standard cycle is ðQ =εÞ, in kJ/kg of air. The
0
v
mass of air trapped in the cylinder is given by
p 1 V 1
m ¼ (16.6)
RT 1
0
Q p 1 V 1
v
which results in an energy addition of :
εRT 1
Hence the imep of an Otto cycle is
p 1 V 1 Q 0 v h th
p ¼ (16.7)
i
RT 1 ε V s
1 1

But V s ¼ V 1 1 and h ¼ 1 giving
th
r r k 1