1V, = 1961 J Since the compressor is single acting, Nw = N = 200 and B
= 6540 W
178
12.4
Minimum work for multistage compression Minimum work required for a two stage compressor with
complete intercooling is obtained when 8
X8;8, or in other words, + -F +
Now,
041
1
2
1
1 2 12 . vr v v T T = =
= 2.7
i.e., T2 = 2.7
x
288 = 778 K At constant pressure from 2 to 3, since pv = RT for a perfect gas, then
2
3
2
3
vv
TT
=
i.e., 1 765 778 1373 2 3 . v v = Therefore
1 765 1
12
3
2
2
1
3
2
2
4
3
4
v. v
vv
vv
vv
vv
1
2 = = p r p p T T T T where rp is the pressure ratio, p2/p 1. i.e. ( ) 1/ 34 = pTrT
and ( ) 1/ 21 = pTrT
B
T2
A
T4
1
2
3
s
4
T
p1
1
23
v
4
p
p2 T 3
T1
p2
p1
Heat rejected
Cooler
Heat supplied
Compressor Turbine
Net work output
Heater
4
1
2
3
1
23
v
4
p
v2 v1
p3 = p 2
110
Heat supplied, Q1 = c p(T 3 T2) Heat rejected, Q 2 = c v(T 4 T1) There is no heat flow in processes
1-2 and 3-4 since they are isentropic.
By substituting in the equation of thermal efficiency, i.e.,
1
12 Q QQ
=
and expressi
Various cylinder arrangements (in-line, Vee, radial, flat, etc.) are used but we shall analyse the simple
in-line arrangement.
Force in con-rod =
MECHANICAL ACTION (static, frictionless) Force on piston: F
r
c
g
b
Therefore: Instantaneous torque ( t ) =
T
Characteristics of power cycles
The working fluid is a condensable vapour which is in liquid phase during part of the cycle
The cycle consists of a succession of steady-flow processes, with each process carried out in a
separate component specifically des
Gross work output is Work output 4 to 1 + work output 1 to 2
Now, for an isothermal process, Q + W = 0 i.e. W4-1 = Q 4-1 = area under the line 4-1 on Fig (a) =
( s1 s4)
x T1 = 0.214
x
1073 = 229.6 kJ/kg
For an isentropic process from 1 to 2, W = ( u2 u1),
987 515 1. 30 54 049 5151. 30 88 += = 3247 kJ/kg of C 2H6
Example 10.10 When all the products of the gaseous hydrocarbon fuel C 2H6 are in the gaseous
phase, the enthalpy of combustion at 25 oC is 47590 kJ/kg. (i) Find: (a) The corresponding internal
ener
INTERNAL COMBUSTION (IC) ENGINES
An IC engine is one in which the heat transfer to the working fluid occurs within the engine itself,
usually by the combustion of fuel with the oxygen of air. In externalcombustion engines heat is
transferred to the workin
Carnot cycle
It consists of two reversible isothermal processes at T a and T b respectively, connected by two
reversible adiabatic (isentropic) processes.
When the working fluid is a condensable vapour, the two isothermal processes are easily obtained by
i.e., p2 = 57.2
x
1.01 = 57.8 bar
Then substituting,
578 69 931
3
.
T
= = 1112 K
Now the heat added at constant volume is equal to the heat added at constant pressure in this
example, therefore cv(T3 T2) = cp(T4 T3) i.e., 0.718(1112 931) = 1.005(T4 1112)
The DUAL cycle approximates to the processes within a modern high speed diesel engine.
In petrol engines, and large slow speed (< 500 RPM) diesels, the pressure rise because of combustion
is comparatively rapid (relative to piston speed) and the cycle can
Supplementary Problems
10.1 A sample of bituminous coal gave the following ultimate analysis by mass: C 81.9%; H 4.9%; O
6%; N 2.3%; and ash 4.9%. Calculate: (a) The stoichiometric A/F ratio (b) The analysis by volume of
the wet and dry products of combus
For the throttling process, 11-12, h11 = h 12 = 429 kJ/kg The enthalpy of the compressed liquid in the
feed line is approximately equal to that of saturated liquid at the same temperature. Thus h6 = hf at
10 bar = 763 kJ/kg and h5 = h12 = 429 kJ/kg For th
o
In the diesel or oil engine, the oil is sprayed under pressure into the compressed air at the end of the
compression stroke, and the combustion is spontaneous due to the high temperature of the air after
compression.
o
In a gas engine a mixture of gas a
( ) ( ) [ ] 1 11 1
1
+
=
pvp
p
rrr
r
(8.9)
Thus the thermal efficiency of a dual-combustion cycle depends not only on the compression ratio
but also on the relative amounts of heat supplied at constant volume and at constant pressure.
Equation (8.9) is
Both of these reactions can take place simultaneously in the same combustion process. The
proportions of the constituents adjust themselves to satisfy the equilibrium conditions and their
actual values depend on the particular pressure and temperature. Th
Solution The ideal cycle is shown below on a T-s diagram.
From Eq. (8.3)
) /( pr 1 1
1 =
14411
6 12 1 02
1
.) /(.
.
= = 0.401
The net work output of the cycle is given by the work output of the turbine minus the work input in
the compressor, i.e., Net wo
T1 1
23
s
4
T
D
P1
C
p4
1
2
3
v
4
p
p2
P3
102
1.
The pressure of the gas changes continuously from p 4 to p 1 during the isothermal heat supply and
from p2 to p3 during the isothermal heat rejection. But in practice it is much more convenient to
heat a ga
Pi
i
Mm
M m RTU vapH vap += 0000 ()()
=
+
32 3.41
44 3.15
18 1.26 .153145 2988.43987 = 43900 kJ/kg
Example 10.13 A combustible mixture of carbon monoxide and air which is 10% rich is compressed to
a pressure of 8.25 bar and a temperature of 282oC. The mix
Ratio of cylinder diameters For a single acting two stage reciprocating air compressor with complete
intercooling, the ratio of cylinder diameters is <- <+(F -*) where D1 = diameter of LP cylinder and
D2 = diameter of HP cylinder
Example 12.12 A two-stage
8.67 = 0.07 kg. (e) The water vapour in the gaseous products can be found from a hydrogen balance.
Basis: 1 kg of fuel H2O entering with air = 0.07 kg H2O in the fuel = 0.10 kg H2O from the combustion
of H 2 = 0.03 x 9 = 0.27 kg Total = 0.07 + 0.10 + 0.27
04
1
2
1
1 2 18 . v v T T = =
= 3.18
i.e., T2 = 3.18
x
T1 = 3.18 x (20 + 273) = 931 K From 2 to 3 the process is at constant volume, hence
2
3
2
3
TT
pp
= since
2
22
3
33
T pv
T pv
= and v3 = v2
i.e.,
2
2
2
3
3
69 931 p
T
pp
T
=
To find p2, use the equa
The feed pump delivers water to the boiler. The pressure inside a boiler is high and so the pressure of
the feed water has to be increased accordingly before it is made to enter the boiler. Generally, the
pressure of the feed water is 20% more than that i