Smart Grid and Renewable Energy, 2012, 3, 96-103
http://dx.doi.org/10.4236/sgre.2012.32014 Published Online May 2012 (http://www.SciRP.org/journal/sgre)
1
The Scientific Revolution in Power Plants
Mohamed Hyder Seid-Ahmed Elfaki
Abu Halima (Block 1, No. 303), Khartoum, Sudan
Received October 21st, 2011; revised March 26th, 2012; accepted April 3rd, 2012
ABSTRACT
It has been made a proposal of new ideal cycle for power plants which is working by a turbine gas, for both closed and
open systems. It has been designed a special device for adding heat at constant volume. The aim of special device is to
decrease the amount of added heat for the new cycle. We have made a comparison between the simple gas turbine cycle
& the new cycle. The results has been shown, that the efficiency of new cycle is greater than the simple cycle of gas
turbine.
Keywords: I,C (Internal Combustion); E.C (External Combustion); S.C (Simple Gas Turbine Cycle); N.C (New Cycle)
1. Introduction
The research is concerning with a branch of mechanical
engineering which is called thermodynamic. It is always
has a relation with producing energy either for electricity
generation, cars, planes and trains by using the thermal
energy. The increasing of the thermal efficiency has been
considered as important parameter for mechanical engi-
neers who they want to specialize in the thermodynamic
science and power plants. They usually want to reach the
maximum efficiency as possible because they know the
effects of increasing of the thermal efficiency all over the
world in economical, environmental and development
sides.
2. The New Ideal Cycle for Internal
Combustion Power Plants
2.1. The Parts of New Ideal Cycle
The Figure 1 represents the (P-V) diagram of the new
ideal cycle for internal combustion power plants and Fig-
ure 2 represents the components of the new ideal cycle.
We see that the new ideal cycle has only 4 processes:
 Process (1-2): adding heat at constant volume by us-
ing special device.
 Process (2-3): adding heat at constant pressure by
using the injection of fuel.
 Process (3-4): isentropic expansion by using turbine.
 Process (4-5): rejecting heat at constant pressure by
using the special device.
2.2. Analysis of New Ideal Cycle
We know that the pressure at state 2 in Figure 1 is:
2
21
1
T
PP
T



(1)
Because 1
VV
2

so, the inlet temperature of turbine
is:
1
2
34
4
K
K
P
TT
P



 (2)
By substituting Equation (1) in Equation (2) we ob-
tain:
1
12
34
41
K
K
PT
TT
PT




(3)
The thermal equilibrium between process (1-2) & (4-5)
is given by:
P
1
2 3
4
5 V
Figure 1. The (P-V) diagram of the I.C cycle.
T
2
To Chimney
C
5
Special
Device
4
1
3
Figure 2. Components of the I.C cycle.

The Scientific Revolution in Power Plants 97
12 45
QQ




(4)




122 1454 5
air exh
VP
mTTmCT
C

 

T

(5)
We assume that (), then:
12 45
mm







21 45
air exh
VP
TT C TT
C  (6)
We solve Equation (6) for , so we obtain:
4
T

421
air
exh
V
P
C
TTT
C

5
T (7)
The specific work of special device is:


.2
air
SD
WT
R

1
T (8)
The specific work of turbine is given by:


34
exh
P
t
WT
C
T (9)
We substitute Equation (3) in Equation (8), so we
have:
1
12
4
41
1
exh
K
K
tP
PT
WCT PT













(10)
The specific heat added in the cycle is given by:


32
exh
aP
QC TT (11)
Also, we substitute Equation (3) in Equation (10), and
then we have:
1
12
4
41
exh
K
K
ap
PT
QC TT
PT










2
 (12)
The thermal efficiency is given by:
.tSD
th
A
WW
Q


 (13)
We substitute Equation (9), (10) & (12) in equation
(13), and then we will have:

1
12
42
41
1
12
42
41
1
exh
exh
K
K
Pa
th K
K
P
PT
CTRT T
PT
PT
CT T
PT






 














1ir
(14)
Notice:
 In this new ideal cycle, it must be 24
TT, and when
we assume that the effectiveness of exchanger is
100%, TT.
24
 The temperature at state 5 must be the minimum tem-
perature of gas exhaust5473TK

, to avoid the con-
densation of corrosive gases in the pipe.
Example (1-1):
 Data:
1
T= 300K, = 1150K,
3
T
p
r
ex
P
C
= 3, = 1 bar = 1.1
bar, = 0.718 KJ/Kg, = 1.11 KJ/Kg, =
1.005 KJ/Kg,
1
P4 P
C
air
V
Chair
P
air
K
= 1.4,
g
as
K
= 1.3
 Find:
th

for new cycle & S.G.T.C, Carnot

 Solution:
1) New cycle:
21
3300900 K
p
TrT

 


3
45
11.31
0.3
4
1150 912.3
31.1
K
K
p
T
TK
rP

 
 
 

 

54 21
0.718
912.3900 300
1.11
524.2Safe_Temp.
air
exh
V
P
C
TT TT
C
K
 




.21
0.287900300172.2 KJKg
SD air
WRTT




34
1.111150912.3263.8 KJKg
exh
tp
WC TT




32
1.111150900277.5 KJKg
exh
ap
QC TT
.263.8 172.233%
277.3
tSD
th
A
WW
Q


 
300
1 173.91%
1150
L
carnot
H
T
T
  
2) S.C (Open System):

11.4 1
1.4
23 1150 3410.6
p
TTr K







 




32 21
32
1.111150 912.31.005410.6 300
1.111150 410.6
18.6%
exh air
exh
PP tComp
th
AP
CTTCTT WW
QCTT

  
 
  


It seems that the efficiency of new cycle is greater than
the simple cycle of gas turbine.
The Table 1 indicates the difference value of
p
r &
3 to obtain other temperature, specific work of special
device & turbine, amount of added heat, work ratio, effi-
ciency of cycle and Carnot efficiency between the new
cycle (N.C) & S.C (Open System).
T
In the aircraft practice where the life expectancy of the
engine is shorter, the maximum temperature used are
usually higher than those used in industrial and marine
gas turbine units, more expensive alloys and blade cool-
ing allow maximum temperature of above 1600 K [1].

The Scientific Revolution in Power Plants
98
Table 1. Comparison between the new cycle & S.G.T.C
(Open System).
S.C N.C S. C N.C S. C N.C
T1 300 300 300 300 300 300
T2 411 900 429 1050 437 1119
T3 1200 1200 1400 1400 1500 1500
T4 942 942 1060 1060 1120 1120
T5 N.A 554 N.A 575 N.A 590
P1 1 1 1 1 1 1
P4 1.05 1.05 1.05 1.05 1.05 1.05
rp 3 3 3.5 3.5 3.73 3.73
Wcomp 111 N.A 130 N.A 138 N.A
WS.D N.A 172 N.A 215 N.A 253
Wt 287 287 377 377 423 423
Qa 876 333 1078 389 1180 423
Wnet 175 114 247 162 285 187
work
ratio 61.2 39.9 65.6 42.9 67.4 44.3
ηth 20.0 34.3 22.9 41.6 24.1 44.3
ηCarnot 75 75 78.6 78.6 80 80
N.A = Not Available.
3. The New Ideal Cycle for External
Combustion Power Plants
3.1. The Parts of New Ideal Cycle
The Figure 3 represents the (P-V) diagram of the new
ideal cycle for external combustion power plants and Fig-
ure 4 represents the components of the new ideal cycle.
We see that the new ideal cycle has only 5 processes:
 Process (1-2): adding heat at constant volume by us-
ing the special device.
 Process (2-3): adding heat at constant pressure by
using exchanger.
 Process (3-4): isentropic expansion by using turbine.
 Process (4-5): rejecting heat at constant pressure by
using the special device.
 Process (5-1): rejecting heat at constant pressure by
using exchanger which the working substance can be
water, air etc.
3.2. Analysis of New Ideal Cycle
We know that the pressure at state 2 in Figure 3 is:
2
21
1
T
PP
T



(15)
Because 1 and so, the inlet tempera-
ture of turbine is:
2 4
VV1 PP
1
2
34
4
P
TT
P





 (16)
3
14 5
V
2
P
Figure 3. The (P-V) diagram of for E.C cycle.
T
12 3
4
5
Special
Device
Exchange
r
Water
Figure 4. The components of the for E.C cycle.
By substituting Equation (15) in Equation (16) we ob-
tain:
1
2
34
1
T
TT
T





 (17)
The thermal equilibrium between process (1-2) & (4-5)
is given by:
12 45
QQ




(18)




1221454 5
Vp
mTTmCTT
C

 

(19)
We assume that (12 45 mm




), then:




21 45
Vp
TT CTT
C  (20)
We solve Equation (20) for , so we obtain:
5
T


1
54 21
TT TT 

 (21)
The specific work of special device is:


.2
air
SD
WT
R1
T

 (22)
The specific work of turbine is given by:


34
P
t
WTT
C
 (23)
We substitute Equation (17) in Equation (23), so we
have:
1
2
4
1
1
tp
T
WCT T


















(24)
The specific heat added in the cycle is given by:


32
exh
ap
QC TT (25)
Also, we substitute Equation (17) in Equation (25),
and then we have:

The Scientific Revolution in Power Plants 99
1
2
4
1
ap
T
QCT T
T












2



(26)
The thermal efficiency is given by:
.SD
th
A
WW
Q


 (27)
We substitute Equation (22), (24) & (26) in Equation
(26), and then we have:

1
2
42
1
1
2
42
1
1
air
air
P
th
P
T
CTRT T
T
T
CT T
T










 









1
(28)
Example (2-1):
 Data:
1
T= 300K, = 1250K,
3
T
p
r= 3, = 1 bar =
0.718 KJ/Kg, = 1.005KJ/Kg,
1
P
air
air
V
C
ai
P
Cr
K
= 1.4,
 Find:
th

for new cycle & S.G.T.C, Carnot

 Solution:
1) New cycle:
21
3300900 K
p
TrT 


3
4
1.4 1
1
0.4
1250 913.2
3
p
T
TK
r



 
 


5

54 21
air
exh
V
P
C
TT TT
C


51
0.718
912.3900300482.8 K
1.005
T




.21
0.287900300172.2 KJKg
SD air
WRTT



34
1.0051250913.2338.5 KJKg
air
tp
WC TT
 



32
1.0051250900351.7 KJKg
air
ap
QC TT
.338.5 172.247.3%
351.7
tSD
th
A
WW
Q


 
300
11 76
1250
L
carnot
H
T
T
 %
2) S.C (Closed System):

11.4 1
1.4
11
11 26.9%
3
th
p
r








The Table 2 indicates the difference value of
p
r &
3 to obtain other temperature, specific work of special
device & turbine, amount of added heat, work ratio, effi-
ciency of cycle and Carnot efficiency between the new
cycle & S.C (Closed System).
T
4. The Special Device for Adding Heat at
Constant Volume
4.1. Components of the Special Device
As we see in Figure 5, the special device constitutes
from 12 parts as follow:
1) Inlet pipe of air.
2) Outlet pipe of air.
3) Cam, which it is fixed.
4) Piston.
5) Exchanger-between air and exhaust gas.
6) Rotary part.
7) Exhaust gas outlet tube.
Table 2. Comparison between the new cycle & S.C (Closed
System).
S.C N.C S. C N.C S. C N.C
T1 300 300 300 300 300 300
T2 390 750 411 900 429 1050
T3 1000 1000 1350 1350 1510 1510
T4 770 770 986 986 1056 1056
T5 N.A 448 N.A 558 N.A 520
P1 1 1 1 1 1 1
rp 2.5 2.5 3 3 3.5 3.5
Wcomp 90 N.A 111 N.A 130 N.A
WS.D N.A 129 N.A 172 N.A 215
Wt 231 231 366 366 457 457
Qa 613 251 944 452 1086 462
Wnet 141 102 690 193 327 241
work
ratio 61 44.2 69.6 52.9 71.6 52.9
ηth 23 40.7 26.9 42.7 30.1 52.2
ηCarnot 70 70 77.8 77.8 80.1 80.1
N.A = Not Available
7 9
12
6
3 2 11
5
10

4
8
1
Figure 5. Components of the special device.

The Scientific Revolution in Power Plants
100
8) Exhaust gas inlet tube.
9) Connection rod of piston.
10) Spring.
11) Fixed part.
12) Cylinder.
4.2. The Working Mechanism of the Special
Device
 When the piston starts from the initial position, the
vacuum of cylinder will be occupied by the air from
the air inlet pipe as shown in Figure 6.
 After that, the temperature of air inside the cylinder
will increase because there is an exchanger between a
cool air and hot exhaust gas, see Figure 7.
 When the air reaches near the air outlet pipe, the all
air inside the cylinder will be hot. Then the piston
will compel the hot air to move outside the cylinder to
the outlet pipe by using cam. See Figure 8.
 When the all air has been rejected from the cylinder,
the above steps will be repeated for another new cool
air.
4.3. The Equations of the Special Device
Before we start to find the equations, the special device
which it is shown in Figure 5 has been modified to an-
other kind, because the new kind is easy to study, but we
must remember that the idea of the new kind is the same
as we explained in previous subsection. The new kind is
shown in Figure 9. The cam and air pipes are in front of
and behind the page respectively.
4.3.1. The Magnitude of th e F orce to Re ject the Air
from Cylinder
As we know, the pressure inside the cylinder of air will
increase as a result of arising in temperature, and the
specific volume is constant then:
maxair
fPA (29)
Also, we know that the area of cylinder is:
2
π
4
d
A (30)
By substituting Equation (30) in Equation (29), and
then we have:
2
max
π
4
air
d
fP (31)
When the connection rod hit the cams as show in Fig-
ure 10, the analysis of forces will be if we neglect the
force due to spring as shown in Figure 11, where “F” is
the force momentum of rotation of rotary part. “R” is the
reaction from the cam. “air
f
” is the force due to the air
inside the cylinder.
Cold air
Hot air

Figure 6. Cold air enters to the cylinder.
Cold air
Hot air

Figure 7. Increasing the temperature of air.
Cold air

Hot ai
r
Figure 8. Rejecting the hot air from cylinder.
Fixed
Part
Piston
&
Cylinder
Rotary

Figure 9. The new kind of the special device.
Direction
of Motion

Hot
air
Figure 10. The process of rejecting the hot air.
R
F
ai
r
f
Figure 11. The free body diagram of forces on cam.
The analysis of the force:
In x-axis:

cosFR

 (32)

cos
F
R

 (33)

The Scientific Revolution in Power Plants 101
In y-axis:


sin
air
fR

 (34)
By substituting Equation (33) in Equation (34) we ob-
tain:


sin
air
fR

 (35)
Then, the minimum magnitude of the force to reject
the air from the cylinder is:

min tan
air
f
F

 (36)
4.3.2. The Power of the Special Device
We know that the torque of the circular motion can give
by:
I

  (37)
where

, I and

are torque, momentum of inertia
and Angular acceleration respectively.
The power of circular motion is given by:
P

 
 (38)
where and
P


are power and angular velocity re-
spectively.
Also, we can express the torque as:
min
F
r

 (39)
where “r” is the radius which it is reached from the
center of the circle to the center of the piston.
By substituting Equation (39) in Equation (38) we ob-
tain:
min
PF r


 (40)
Now, we want to talk about an important parameter of
Equation (38), it is the value of the angular velocity
“

”.
As we see from Figure 12, the cylinder will move
from it is initial position 1 to reach position 2 at angle of
“

” mafter a period of time “t”, then:
r
t

  (41)
“t” will dependent on the time required to reach the air
from initial to final temperature.
By substituting Equation (41) in Equation (40) we
have:
min
r
PFrt


 (42)
Recall that:
π
180
ro

  (43)
If, we substitute Equation (43) in Equation (42). Then,
we have:
min
π
180
o
PFr
t



 (44)
Also, we can substitute Equations (36) & (29) in Equa-
tion (42), and then we have:

max
π
180 tan
o
PA
Pt r



 (45)
The last equation helps us to determine the power of
special device.
Example (3-1):
 Data: max P = 2 bar, d = 0.15 m, t =3 sec, r =
1.6 m, 120
 , 45  .
 Find: A, P
.
 Solution:

2
2
2
π0.15
π0.01767 m
44
d
A  

max
5
π
180 tan
π2100.01767 1.6 120
180tan(45) 3
3.95 Kw
o
PA
Pr
t



 



The Tab l e 3 indicates a different value of max , P

&
to obtain the power of special device. The data for
this table was given in the previous example.
d
We see from this table that the power of special device
can be neglected, because it is too smaller than the work
of turbine.
1
2


Figure 12. The angle

of the cylinde r .
Table 3. The power of special device.
Pmax
(bar) d(m) θ˚ A(m2)

rm


t (s)P

(Kw)
3 0.1545 0.0171.6 120 3 5.92
4 0.1545 0.0171.6 120 3 7.90
4 0.2 45 0.0311.6 120 3 14.0
4 0.3 45 0.0701.6 120 3 31.5
4 0.3 60 0.0701.6 120 3 18.2
4 0.3 80 0.0701.6 120 3 5.57

The Scientific Revolution in Power Plants
102
4.4. Concrete Method for Manufacturing the
Special Device
4.4.1. Component of Special Device Separately
The Figures 13-27 illustrate the major parts of the spe-
cial device. Any one of these parts have a different func-
tion which is distinguished from other.
Figure 14. Fixed part No. 2.
Figure 15. Piston No. 3.
Figure 16. Connected column [2] for fixed pa rt & b as e (N o . 4) .
Figure 17. Tube No. 5.
Figure 18. Connected column [1] for fixed part & cams (No.
6).
Figure 19. Rotary ring (No. 7).
Figure 20. Base of special device (No. 8).
Figure 21. Cylinder with cams (No. 9).
Figure 22. Pipe of inlet & outlet air (No. 10).
Figure 23. The column (No. 11).
Figure 24. Spring (No. 12).
Figure 25. The gear (No. 13).

The Scientific Revolution in Power Plants
103
1
9
Ground
Figure 26. Ring (No. 14).
Figure 27. Exchanger (No. 15).
Tube
Figure 28. Rotary part with tubes.
1 2
1- Exchanger
2- Pist on
Figure 29. The projection of piston and exchanger.
15
12
16
13
1
8
3
Cold air
Hot gas
2
Figure 30. System of piston & exchanger.
4.4.2. Final Form of the Rotary Part with Tubes
The rotary parts have many tubes which they are con-
nected with each other as shows in Figure 28. These
tubes contain the hot gases which come from exhaust
ases of turbine. g
Power
13
6
2
10
4
8
10
Figure 31. Arrangement of the special device.
4.4.3. Final Form of the Piston & Exchanger
The shape of piston & exchanger is very simple; we see in
Figure 29 the projection of the piston & exchanger. Fig-
ure 30 indicates how we can join the piston with rotary
part to do the function of adding heat at constant volume.
4.4.4. Arrangement the Components of the Special
Device
Figure 31 shows the method to arrange the components
of the special device with each other. The power of spe-
cial device comes from the power supply by using a gear.
5. The Conclusions
 The efficiency of new cycle is greater than the simple
cycle of gas turbine.
 The work ratio of new cycle is lower than the simple
cycle of gas turbine.
 The capacity of power plants will be smaller by using
the special device, rather than using two or three
compressor to increase the efficiency.
 We can make the life more comfortable and eco-
nomical for persons and factories by decreasing the
energy cost.
 We can also decrease the dangerous effects of exhaust
gases to the environment by making the fuel con-
sumption more economical.
REFERENCES
[1] T. D. Eastopand A. McConkey, “Applied Thermodynamic,”
Longman, Lomdon, 270p.