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 +

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

1, = 1702 Nm (a) Mean effective pressure is 8G HIJK LIMN OPJIKN QIRSTN C
.B)
= 180700 N/m 2 = 1.807 bar (b) Since the compressor is single acting, Nw = N and power required to
drive the compressor is C
9 = 14183 W 12.3 Multistage compression It is not adv

. T = = 408 K
Now the heat supplied, Q 1, is given by Q1 = cv(T3 T2) + cp(T4 T3) or Q1 = 2 cv(T3 T2) since in
this example the heat added at constant volume is equal to the heat added at constant pressure.
Therefore Q1 = 2
x
0.718
x (1112 931) = 260 kJ/kg

Otto Cycle
The Otto cycle is a model of the real cycle that assumes heat addition at top dead center. The Otto
cycle consists of four internally reversible cycles, that describe the process of an engine. Figure 2.1,
shows the p-v and T-s diagram for the O

Wet volumetric analysis (a) The amount-of-substance in the products is: 6 (CO 2) + 7 (H 2O) + 1.9 (O
2) + 42.89 (N 2) = 57.79 kmol Now the mole fraction of a constituent equals the volume fraction. The
wet volumetric analysis if all the water vapour is pr

crank angle
torque (Nm)
r = 50 mm c = 150 mm F = 1000N
F = 1000N constant
F = 1000N up to 30 then dropping pV 1.4 = const.
The torque fluctuations are smoothed out by:(i) multi-cylinder arrangement
(ii) the use of a flywheel
so that the torque output fro

21 79 296)(cfw_011 + = 435865 kJ Clearly the guess T = 2200 K was not correct, because the
result is not equal to zero. Repeating with enthalpy values for T = 2400 K, the imbalance in the
enthalpy equation becomes + 35473 kJ. Linear interpolation between

298 = 281751 kJ/kmol
The non-flow process is defined by: Q = W + (U 2 U1). But Q = 0 and W = 0 at constant volume.
Now U1 = U R1 and U2 = U P2 , so: 0 = 0 + (U P2 UR1 ) = ( UP2 UP0 ) + (UP0 UR0 ) + (UR0 UR1 )
= ( UP2 UP0 ) + U0 + (UR0 UR1 ) = ( ) ( ) + +

( ) ( ) [ ] 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

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