A Methodical approach to calculate efficiency of rotary engine and ...

A Methodical approach to calculate efficiency of rotary engine and comparing with traditional reciprocating engine as well as with other technologies

Design and calculation developed by Fahim Mahmood B.Sc. in Mechanical Engineering from Bangladesh University of Engineering and Technology (BUET) Serving in PETROBANGLA under Ministry of Power, Energy and Mineral resources in Bangladesh

Contents 1) Section: 1 Introduction To The Efficiency Calculation……………….….1 2) Section: 2 Rotary Engine Efficiency Calculation…………….…………..3 3) Section: 3 Reciprocating Engine Efficiency Calculation………………..30 4) Section: 4 Comparison between different technologies…………………46 5) References……………………………………………………………….49

Design, estimate and calculation developed by Fahim Mahmood in favor of Patent US9528433 and its subsequent Cip publication US20160290221, on the fourteenth day of June, 2017

Section: 1 INTRODUCTION TO THE Efficiency Calculation The current technology revolving around the using of maximum force developed from gasoline ignited which is always acting along the direction of rotation of power shaft, consequently creates maximum torque and relevantly the efficiency is also elevated a lot from that of the existing technology of reciprocating engine. However to distinguish the degree of performance between two technologies, efficiency calculation is thought to be the best mode for comparison. The efficiency calculation of internal combustion engine based on Otto cycle is some what complicated. The traditional method which indicates the thermal efficiency based on compression ratio is entirely dedicated to compute the efficiency within the combustion chamber only and it is absolutely mute to define anticipated efficiency at power output shaft of an engine. Consequently it should not be mix up with the mechanical efficiency as

echanical efficiency < thermal efficiency. i.e. any internal

combustion engine in the world based on Otto cycle (Fig: 1 and Fig: 14) is not going to overcome the barrier of thermal efficiency. This equation of thermal efficiency [7] is  specific heat ratio and air-fuel mixture and

ℎ

=

−

−

, wherein k is the

is the volumetric compression ratio. However the specific heat ratio,

≈ . for

≈ . for air only. For EFI and GDI (gasoline direct injection) type engine

is used in all calculation. However in special case of GDI engines

≈ . can be used to calculate

parameters in compression cycle while for calculation in expansion cycle the value of

≈ .

is taken up . .

This mechanical efficiency is smaller than thermal efficiency because the thermal efficiency from the

combustion chamber is fractionated within the linking mechanism after the combustion chamber stage in order to achieve power at power output shaft and this loss of power within the linking mechanism is the sole purpose/field of discussion in this efficiency calculation wherein the traditional reciprocating engine exhibits 18% fuel efficiency and current invention exhibits 40% fuel efficiency @ 8.36 compression ratio for the each. We all know “Gas pressure exerts its pressure perpendicular to the exposed area which is soon after available as a form of force on the linking mechanism.” This thumb rule holds that the efficiency of any internal combustion engine is fully dependent on how efficiently the linking mechanism 1


can convert the thermal efficiency to mechanical efficiency and subsequently deliver it to power output shaft. So the sole purpose of any combustion engine is to keep the fractionation of force applied to the linking mechanism be as low as possible. This fractionation/vectorization of force can be defined as to split a force into a sine component and a cosine component if said force is applied on an inclined plane. The traditional reciprocating engine along with the Wankel engine having the linking mechanism which is continuously prone to higher degree of force fractionation in compare to that of the current invention and subsequently having much less efficiency than that of this inventive technology have. On the whole, the traditional reciprocating mechanism has successive internal parts wherein it is observed that the applied force from combustion passes through several inclined planes (wherein planes are making inclination angle along the direction of force) at the couplings if the observation is done carefully within the linking system against the situation where the current technology is utilizing the gas pressure from combustion through the minimum transitional coupling. In reciprocating engine, the linking mechanism is the coupling among piston, connecting rod and crankshaft while on the other hand the current technology has the linking mechanism among bar, main wheel and power shaft. So for the dissimilarities in the linking mechanism between said two systems, the efficiency also varies accordingly although both the systems operate on Otto cycle.

METHOD TO COMPUTE THE Efficiency The method to calculate the efficiency is to find out the average loss in the linking mechanism by the method of integral average and subsequently using it in later calculation. Additionally the average pressure is also calculated between consecutive stages and we all know that projected force is the product of average pressure over exposed surface area which is then later multiplied with loss factor and with distance travelled by that force in order to get the final product of work done or work to be done wherein the force is the net product arising from gas pressure exposed to difference in exposed surface area of the bars in the case for current invention and this force is finally available for delivery at power shaft in form of torque.

2


Section: 2

Rotary Engine Efficiency Calculation

This rotary engine[6]and[4] is legally bonded to patent US9528433[9] and to the cip (continuation in part) publication US20160290221, which is capable to run according to the operating principle/condition described below. In this system, each bar (prime mover and follower) having length= r unit which is also the radius of the cylindrical chamber within the casing. This rotary engine operates on classical Otto cycle according to the fig: 1. The main wheel having a radius of ŕ unit. The guider hinged wheels are arranged at an angle of ‘𝜶 ’ with respect to the main wheel center. This angle ‘𝜶 ’ is constant for all calculations in this literature. All variable angles counted in this literature are measured from positive X-axis in all attached figures unless particularly specified. The angle between the two bars at the center of cylindrical cavity or chamber is θ which is counted around origin O(0,0) in fig: 2. This angle θ counted from positive X-axis varies when the bar propagates by rotation, especially when the rotation of prime mover making the delimited combustion chamber rotatably travels from BDV to TDV and consequently the

3


angle between bars gradually increases from θ1 to π and doing vice versa when prime mover travelling from TDV to BDV and then returning to angle

𝜃

counted from positive X-axis. The center of main wheel

is shifted at a distance “d” from the center of central cylindrical chamber and along positive X-axis. These are the basic parameters for this engine design wherein all the angles are unitized by Radians and all temperature values are in absolute temperature Kelvin unless otherwise specially specified in this literature. The terms TDV stands for Top Dead Volume or maximum volume, BDV stands for Bottom Dead Volume or minimum volume and TP stands for Transition Phases between BDV and TDV. Apparently there is one TDV, one BDV and two TP’s within an engine module wherein the demarcated combustion chamber rotatably transformed between BDV and TDV through the TP’s and clearance distance between the casing and the main wheel is K (see fig:2). The angle between two junction points in main wheel (where guider hinged wheels are coupled with main wheel) is

at the center of main wheel.

O(0,0) is the rotating origin of the bars as well as the center of cylindrical chamber and modified origin M(d,0) is the rotating origin and the center of main wheel wherein M(d,0) is shifted by a distance of d from origin O(0,0) along positive X-axis. Bars rotate around origin O, making angle θ with positive Xaxis and main wheel rotates around modified origin M(d,0), making angle

with positive X-axis.

Conditions: At BDV, θ1
ŕ > ,

=

⁄ and

> ⁄

́ which

further ensures that compression ratio of this engine is dependent on the value of ́ and . Any engine design related to this rotary engine must have the angle,

=

⁄ in order to operate this rotary engine

by its core operating principle. The equation between the angles is derived as follows. At BDV main wheel creates creates angle

𝛼

with positive X-axis and at stage 3 (described later in this discussion) prime mover with positive X-axis and subsequently main wheel creates

4

with positive X-axis in same


+

stage. So,

𝛼

= ,∴

=

by putting the values of

𝜋

and

. Considering the other hand, .

+ ́+

= , which means

>

́

Equations: It is clear from fig: 2 that, = ́ cos

𝛼

.……………………………..……..(1)

And for any positive value of θ, = − ́ cos

𝛼

(cos 𝜃 + √cos 𝜃 + tan

𝛼

)

…..……. (2)

Wherein, “l” is a part of length of each bar which is exposed to combustible mixture of gas in current setup, i.e. “l” is within the combustion chamber. 𝑇 𝑉 ℎ ℎ

𝑇𝑃 ℎ

𝑉

=

−

=

𝜃 −

==

́ ́

−𝜃

𝜃 = tan−

tan

………………………………(3)

− sin −

𝛼

+ sin ́

…………………….……..(4) −

……..…..………………(5)

…..…………….……………………(6)

Wherein h is the distance between the parallel planes of central cylindrical chamber and this distance can also be the width of the bars. For constant volume combustion,

And,

ln

𝑇 𝑇

=

+

ln

+ . .

=

=

. ………………………………….(7)

………………………………………...(8)

Wherein “f ” represents fuel-mixture ratio =

.

for stoichiometric combustion and Q is the Heating value

(Lower Heating Value) of gasoline= 42700 kJ/kg. 5


For isentropic compression or expansion, the required equations for ideal Otto cycle are given below: ⁄

=[ ⁄ ]

⁄

=[ ⁄ ]

⁄

−

−

=[ ⁄ ]

ℎ

…………………………………….………...(9)

𝑖 ,

……………………………………………...(10) .……………………………….……………..(11)

= .

𝑖

≈ .

𝑖

Derivation of equation from 1 to 6:

𝑖

.

From fig:2 above it is clear that 𝛼

= ́ cos .As the main wheel rotates on its axis, so the radius of curvature for main wheel will be

́.

The exposed length (wherein the gas pressure is exerted on) to gas for the case of any bar will be the total length of the bar minus the length engulfed by the main wheel. Applying this geometrical phenomenon in −

equn of circle for main wheel we get, +

=> =>

=> c ∴

+

𝜃

=

−

−

−

+

+

= ́

+ − ́

+

[bar having a straight line equation where y=mx and m is slope]

− ́

=

= ,

∴

cos 𝜃 + cos 𝜃 √cos 𝜃 −

[m=tanθ and (1+m2)=1/cos2θ] ±√

= ́

−

=

cos 𝜃 + cos 𝜃 √cos 𝜃 − +

=>

=

cos 𝜃 + cos 𝜃 √cos 𝜃 − + sec

=> = − √

+

ℎ

,

́

− . co

co

𝜃

𝜃

.

− ́

(only + sign is accounted because θ is counted from + x axis)

=>

,

= ́

́

cos

= − = −√

(putting the value of y)

…………………………(12) +

6

[from fig:2, = √

+

]


= − √c

= − √ +

𝜃

= −c

= − cos 𝜃 − √cos 𝜃 − + sec 𝛼

=> = − ́

(Putting value of x)

(cos 𝜃 + √cos 𝜃 − + sec

𝛼

=> = − ́

𝜃


Wherein, angle between bars ranges from θ1 to π 𝛼

and = ́


𝛼

𝛼

𝛼

)

)

(Derived)

)...

On the basic construction criteria of this rotary engine, equation 3 can be derived as follows: In fig: 3, common area (B) engulfed by the main wheel and one half of casing. So, ́

B= =

− ∙ ∙ ́ sin ́

𝛼

− ∙ ́ sin

∙ ́ cos =

́

𝛼

− sin

Cross sectional area (A) is TDV/h from top view. A=

T V

On this basis equation 4 can be also derived as follows:

=

−

́

− sin

From fig: 3 common area D can be calculated as D= ∙

∙ ́ sin

= ∙

́ cos

𝛼

= ∙ ́ sin

𝛼

∙ ́ sin

𝛼

From fig:4 through fig:7 below, Cross sectional area (E) of BDV from top view, E=

ℎ



V

= ℎ

V

=

𝜃 −

𝜃 −

−

−

7


=

𝜃 − ∙ ∙ ́ sin

=

𝜃 −

=

𝜃 −

́

sin

́

+

−

+ sin

́

− sin

− sin

On the same basis, equation 5 can also be derived as follows: Comparing fig: 8 with fig: 2, TP/h = F can be calculated as So, F=

∴

ℎ

𝑇𝑃 ℎ

𝑇𝑃 ℎ

𝑖 ,

=

= {

=

− {

−

́ − } ́ −

−𝜃

=

−

−

́

́ +

− sin ́

=

}− {

−𝜃

Establishing equation between 1 and 1 SD= 

tan

𝛼

tan

 tan



As

𝜃

𝜃

=

𝛼

=

tan

= tan

= tan−

 𝜃 = tan−

𝜃

𝜃 − −

́

́

−

+ sin

}

[from fig: 2 above]

tan

𝜃

tan

𝛼

𝛼

tan

𝛼

is constant, consequently 𝜃 is also constant.

Alternative method to find volume of combustion chamber Prime mover makes 𝜃 and follower makes 𝜃́ with positive X-axis. From fig: 3, we can find that Volume created due to prime mover position, =

𝜃−

=

𝜃−

. sin 𝜃 −

́ sin 𝜃 cos 𝜃 + √cos 𝜃 +

Volume created due to follower position,

𝜃́ −

=

8

. ́ sin 𝜃́ −

− ́

́

́

−


Now, sin 𝜃 = ́ sin

𝜃́ −

=

́ sin 𝜃́ cos 𝜃́ + √cos 𝜃́ +

and ́ sin 𝜃́ = ́ sin ́ wherein

−

So 𝜃́ can be evaluated for a value of 𝜃 where ́ = =

,

=

𝜃

−

́

𝜃

=

−

(cos 𝜃 + √cos 𝜃 + ) −

(𝜃 − 𝜃́ ) −

́

− and

wherein and ́ is distance from junction point

for prime mover and follower bar respectively.

∴ volume created in a stage, =

= ́.

́ cos 𝜃́ + √cos 𝜃́ +

́ (cos 𝜃 + √cos 𝜃 + ) by putting the values of

of main wheel to origin

−

𝜃

+

́

𝜃́

cos 𝜃́ + √cos 𝜃́ +

−

́ {sin 𝜃 (cos 𝜃 + √cos 𝜃 + ) − sin 𝜃́ cos 𝜃́ + √cos 𝜃́ +

́

−

}−

𝜋 ́

+

𝜋

……

Now by taking integral average of equn (13) we get average volume change between adjacent/subsequent stages which is formulated below wherein 𝜃 and 𝜃 represents first stage angle and second stage angle

respectively created by prime mover with positive X-axis and 𝜃́ and 𝜃́ represents first stage angle and

second stage angle respectively created by follower simultaneously with positive X-axis ̅̅̅̅̅̅ =

=

𝜃 −𝜃

𝜃 −𝜃

×

𝜃 ∫ (𝜃́ −𝜃́ ) 𝜃

√cos 𝜃́ + ×

[ (𝜃́ −𝜃́ )

}− ́

𝜋

} 𝜃́ . 𝜃

{(𝜃́ − 𝜃́ )(𝜃

+√cos 𝜃 + ) 𝜃 +

Different stages

́

𝜃 ∫𝜃́ { (𝜃 − 𝜃́ ) −

́

𝜃 −𝜃

́ {sin 𝜃 (cos 𝜃 + √cos 𝜃 + ) − sin 𝜃́ cos 𝜃́ +

−𝜃 )− 𝜃 −𝜃

́

𝜃́

− 𝜃́

}−

́ (𝜃́ −𝜃́ )

𝜃

∫𝜃 sin 𝜃 (cos 𝜃 +

𝜃 𝜋 ∫𝜃́ sin 𝜃 (cos 𝜃 + √cos 𝜃 + ) 𝜃́ ] − ́ … … … … … … … … …

The power output cycle or expansion cycle happens in four significant stages and are typically the core design principle of this engine in order to operate in proper working condition. The stage 1 is the stage when the prime mover making angle

𝜃

with positive X axis or 𝜃 =

𝜃

or when the combustion chamber

is in stage of BDV and the corresponding follower bar angular position is in 𝜃́ = 9

−𝜃

. In stage 2 the prime


mover’s angular position is at 𝜃 =

𝜋

mover’s angular location is at 𝜃 =

while the follower angular position is in 𝜃́ = . In stage 3 the prime

𝜃 and the follower angular position is in 𝜃́ = . Similarly in stage 4

the prime mover’s angular position is at 𝜃 =

𝜋

𝜋

and the follower angular position is in 𝜃́ = . At stage 4

the expansion cycle is likely to finish. It is also observable that the force arising from gas pressure always act at a middle position of exposed portion of bar into the combustion chamber and this force will act along the dotted circle ‘C’ in fig: 2. So the curvilinear distance S12 is the distance when the prime mover travels angular from

𝜃

𝜋

to from +ve X-axis. Similarly S23 is the distance when the prime mover travels

𝜋

angular from to . In the similar way S34 is the distance when the prime mover travels angular from

to

𝜋

. So the value of

S12, S23 and S34 will be as follows in light of fig: 9 below: Radius of the dotted circle,

As,

=

+ − ́

+

+ ́

+

− − ́

=

Because dotted circle passes at half way mark between the circular casing and the main wheel. ∴

=

+ ́

and consequently dotted circle C is shifted ́

chamber, .i.e. ́ =

.

, ̂ = − = − { − ́ cos ́

∴ ̂ = + cos

𝛼

unit from center of central cylindrical

𝛼



𝛼

)

𝛼

)}

…………(15) [in fig:2]

Wherein ̂ is the radial distance measured from the center of cylindrical chamber to any point of

dotted circle C periphery by an angle 𝜃 with positive X-axis and at this distance from casing center

the force arising from the gas pressure acting perpendicular on the exposed area of bar (prime mover or follower) and the resultant force is working along the dotted circle C. 10


Consequently, ̂ sin 𝜃 = When, 𝜃 =

𝜃

ℎ

,

=

sin

……………………………(16)

and when 𝜃 =

= cos −

So from fig: below,

ℎ

́

,

=

=

−

−

So curved distance travelled by resultant force between different adjacent stages, =

∙

,

=

∙

and

Operation of this Engine

=

∙

on the other hand,

=

,

=

and

=

.

The operation of this engine is pretty straight forward. The after-combustion gas has a perpendicular impact on free exposed end of bars and obviously the bars will rotate along the dominant force projected on those bars. The gas will create perpendicular impact on bar-end because gas has a natural tendency to create perpendicular impact on an exposed surface and this perpendicular impact always working along the direction to the tangent of the curvature formed by the power shaft rotation. So force applied by the gas will be dominant to rotate the bars if and only if, there is a difference (

́) between

exposed surface area of the bars. It is the main theory around which this efficiency calculation stands. This engine also comprises torque enhancer pocket which is also able to convert heat energy to work even if, when there is no energy conversion from the bars only. So torque enhancer pocket itself increases work in expansion cycle, reduce compression work and subsequently increases overall engine efficiency even more from engine efficiency calculated below. A basic knowhow of working principle can be achieved by the attached references at the end of this manuscript.

Typical (standard) engine with units = . unit (preferably in meter)

́ = . unit (m) =

𝜋

rad (constant for this type of engine design)

Main wheel shifting length, And 𝜃 = .

rad

= . unit (m)

[from (1)] [from (6)] 11


Normal atmospheric condition =

° =

=

. KPa= 101.3 kN/m2

From pv diagram of Fig:1 =

=

= .

ℎ m3

=

T T

= .

.

ℎ m3

ℎ m3

−

=

. ℎ . ℎ

=>T2 = 298

. −

= 563.39 K

[Temperature after full compression] 

=

𝑇 𝑇

= 101.3 x

−

 . .

.

=

𝑇 𝑇

x

−

= 1600.26 kN/m2

[pressure after full compression] So, T2 = 563.39 K = 1600.26 kN/m2 In combustion stage Volume remains constant in this stage and thus rendering T

=T

and, T3 = T2 + .Q. Wherein, fuel-air ratio =

.

v

, So fuel-mixture ratio will be,  = 12

.


Q = Lower Heating Value (LHV) of gasoline = 42700 kJ/kg R = 0.287 kJ/kg.K =

𝑅 −

=

.

. −

 T3 = 563.39 + As, ∴

T

=T x

=

.

.

T

And

=T

= . .

kJ/kg.K

x 42700 x

= 3405.34 K

.

x 1600.26 = 9672.57 kN/m2 =

The four stages in Expansion cycle (in view of Fig: 1) At Stage 1: = P3 = 9672.57 kN/m2 = 3405.34 K

At Stage 2:

= =

At Stage 3:

𝜃

𝜃 𝜃́ = − in equ

+

= 4.06h unit3 [when 𝜃 = 𝜋

=

At Stage 4:

[when 𝜃 =

+

= 11.71h m3 [when 𝜃 =

𝜃́ =

in equ

𝜃 and 𝜃́ = in equ

]

]

]

= TDV = 17.38h m3 [the unit is assumed to be meter] [when 𝜃 =

𝜋

𝜋 and 𝜃́ = in equ

]

Fig: 4 representing the stage 1 when the combustion chamber is in BDV or the minimum volume.

13


Fig: 5 representing the stage when the combustion chamber is traveling from BDV to the stage 2. Fig: 6 representing the stage when the combustion chamber is traveling from stage 2 to stage 3. Fig: 7 representing the stage when the combustion chamber is traveling from stage 3 to the stage 4 or TDV. Fig: 8 representing the placement of TP or transition phase in the cyclic process of combustion chamber. Average volume change calculation between adjacent stages From stage 1 – 2 By putting values of 𝜃́ = 𝜃 =

𝜋

, 𝜃́ = , 𝜃 =

𝜃

and

in equn (14) and evaluating definite integral we = .

get,

−𝜃

ℎ

From stage 2– 3

𝜃 Now putting values of 𝜃́ = , 𝜃́ = ,

𝜃 =

𝜋

and 𝜃 =

in equn (14) and evaluating definite

integral we get,

From stage 3– 4

= .

Now putting values of 𝜃́ = we get,

=

.

ℎ

ℎ 𝜃

𝜋

, 𝜃́ = , 𝜃 =

and 𝜃 =

14

𝜋

in equn (14) and evaluating definite integral


Average pressure calculation between adjacent stages From stage 1 – 2 =

For isentropic operation, wherein

is corresponding pressure at average volume

∴

,

=

.

×

=

.

=

.

×

.

.

kN/m

.

ℎ ℎ

From stage 2 – 3 and doing similar step of previous step =

=

×(

.

kN/m

) =

.

. ×( .

ℎ ) ℎ

.

From stage 3 – 4 and doing similar step of previous step =

=

≈

×

.

.

×

.

kN/m

ℎ . ℎ

.

Calculation for average exposed area to gas pressure between adjacent stages From stage 1 – 2 Taking the average integral of exposed length in light of equn (2), over the range from, 𝜃 =

̅̅̅̅̅̅ =

−

∫ { − ́

𝜃

𝜃=

𝜋

then for prime mover,

cos 𝜃 + √cos 𝜃 + tan 15

} 𝜃


=𝜋

−

𝜃

𝜋⁄

∫𝜃 ⁄ { − ́

Applying the value of 𝜃 ,

̅̅̅̅̅̅ = .

𝛼

𝛼


)} 𝜃

, , ́ and solving definite integral we get, m

follower bar travels from 𝜃 = −

𝜃

𝜃=

́ So for follower bar, ̅̅̅̅̅̅ =

− −

𝜃

́ ∴ ̅̅̅̅̅̅ = .

for when prime mover travels from 𝜃 =

𝜃

cos 𝜃 + √cos 𝜃 + tan

∫−𝜃 { − ́

𝜃=

𝜋

} 𝜃

m

Cumulative average exposed length difference or the net effective average length of exposed area to ́ = ̅̅̅̅̅̅̅̅ − ́ gas=̅̅̅̅̅̅ − ̅̅̅̅̅̅

=

From stage 2– 3

.

− .

In the similar way over the range from, 𝜃 = ̅̅̅̅̅̅ =

−

=

𝜋−

𝜋

𝛼

𝜋 𝜋

Applying the value of

𝜃=


m

then,


∫ { − ́

∫ { − ́

𝜋

m = .

} 𝜃

𝛼

)} 𝜃

, , ́ and solving definite integral we get, ̅̅̅̅̅̅ = .

follower bar travels from 𝜃 =

́ So for follower bar, ̅̅̅̅̅̅ =𝜃

to 𝜃 =

𝜃

𝜃

, when prime mover travels from 𝜃 = cos 𝜃 + √cos 𝜃 + tan

∫ { − ́

−

m

̅̅̅̅̅̅ ́ = .

𝜋

𝜃=

} 𝜃

m

Cumulative average exposed length difference or the net effective average length of exposed area to gas= ̅̅̅̅̅̅ − ̅̅̅̅̅̅ ́ = ̅̅̅̅̅̅̅̅ − ́

From stage 3– 4

=

.

− .

In the similar way over the range from, 𝜃 =

m= . 𝜃= 16

m

𝜋

then,


̅̅̅̅̅̅ = =

− 𝜋


∫ { − ́

−𝜋

𝜋

𝛼

∫𝜋 { − ́



} 𝜃 𝛼

)} 𝜃

, , ́ and solving definite integral we get,


𝜃

́ =𝜋 So for follower bar, ̅̅̅̅̅̅

𝜃=

−

́ ∴ ̅̅̅̅̅̅ = .

𝜃

̅̅̅̅̅̅ = . 𝜋

𝜋

m

for when prime mover travels from 𝜃 = 𝛼

∫𝜃 { − ́


𝛼

)} 𝜃

𝜃=

𝜋

m


=

.

− .

m= .

Travelling distance calculation

m

In view of fig: 8 and fig: 9 when the prime mover experiences an angular travel from 𝜃 = to 𝜃 =

𝜃

, i.e. during a travel from stage 0 to

stage 1, the force arising from pressure travels a curved distance

in fig: 8 wherein the dotted

́ ́, circle C with its center M

is shifted ́ from

origin O(0,0). So from the equation (15) and (16), = .

= .

m wherein, ̂ = . and

= .

Similarly from stage 1 to stage 2, From stage 2 to stage 3,

= .

m and

= .

=

m

17

.

m [see fig:8 and fig:9]


=

From stage 3 to stage 4,

=

m =

From stage 4(stage 1 in compression cycle) to stage 5 (stage 2 in compression cycle), From stage 5(stage 2 in compression cycle) to stage 0(stage 3 in compression cycle), where

=

= .

= .

m m

is the distance when the prime mover travels from stage 0 to stage 4 in compression cycle.

Gross expansion work done, 𝑾𝒆

Thus pressure projected on exposed area (ℎ × ) and then multiplied with = ̅̅̅̅̅̅ × ̅̅̅̅̅̅̅̅̅̅  −  =

=

=

.

× .

. ℎ+ .

×ℎ×

×ℎ× . .

ℎ+

ℎ kN-m or kJ

+ ̅̅̅̅̅̅ × ̅̅̅̅̅̅̅̅̅̅  −  + .

ℎ

.

× .

×ℎ×

×ℎ× +

to get the term work done.

+ ̅̅̅̅̅̅ × ̅̅̅̅̅̅̅̅̅̅  −  .

× .

×ℎ×

×ℎ ×

If an observation can be done in above equation, we can see that maximum work is done on stage 2

to 3. The average pressure is active through traveling distance within two conjugative stages such that total work done due to fluctuating pressure exposed to the bar end within that travelling distance is equivalent to work done by the average pressure active all through that travelling distance.

The four stages in Compression cycle The calculation in compression cycle is little complicated though. Some critical analysis are also needed in this concern such as compressive force exerted on middle of exposed on follower bar end is firstly needed to go through inclination factor in order to transmit through main wheel and then secondly the required force is needed to transmit through main wheel to arrive at prime mover by another inclination factor. This system is graphically presented in fig: 10 below where angle of this inclination of both bars with the tangent of main wheel at junction point= 𝜑=

− 𝜃 {when prime mover creates this angle} and

𝜑́ = ́ − 𝜃́ {when follower bar creates this angle}

18


Wherein

& 𝜃 are relating to prime mover and ́ & 𝜃́ are relating to follower while these angles are

evaluated for an instantaneous position of prime mover and follower from the reference axis such as positive X-axis.. Multiplication factor Multiplication factor arising from the situation where an impact from follower bar is received in prime mover bar which happened due to an angle of inclination between bar and the tangent of main wheel at coupling point or junction point and it’s phenomenon is also figured in fig: 10 below. So logically power transmitted through this junction point will be the end product of cosine of that angle at junction. This multiplication factor is significant at compression stage as the prime mover enforces the follower bar to compress the gas or the work needed to be done in combustion chamber. However this

loss/multiplication factor is not accountable in expansion stage as the gas itself having a perpendicular

19


impact on prime mover i.e. work is done on the prime mover with a perpendicular impact of gas whether or not the follower bar is encountering loss at junction/fulcrum; in fact it causes no effect to prime mover rotation at expansion. On the whole it can be stated that if the gas with a perpendicular effect on the bar reaches directly to prime mover bar then there will be no multiplication factor and subsequently if the gas pressure having a perpendicular effect on bar (other than the prime mover) and produces power that arrive at prime mover or power shaft, then a loss factor to be accounted in the calculation and that is the reason behind why this engine having only one combustion chamber per engine module in order to extract maximum power per unit of fuel consumed. This force multiplying factor is formed at junction between prime mover and main wheel as well as also at a junction between follower bar and the main wheel At Stage 1: = P1 = 101.3 kN/m2 =

=𝑖 𝑖 𝑖

= 298 K

=

At stage 1, =

𝜃=

∴𝜑=

−

𝜋

= ⁄

=−

.

𝜋

And for follower bar, ́ =

∴ 𝜑́ =

−

𝜋

=

𝜋

𝛼

,

ℎ m3 𝑖

, 𝜃́ = ⁄ ,

At Stage 2:

At stage 2, =

𝛼

=

−

for prime mover figured in fig:

𝒃,

20


𝜃=

−

and

𝜃

,

𝜃

, ∴𝜑= =

At Stage 3:

𝛼

−

, ́ = +

=

𝜃

−

𝜃́ = ,

= 11.71h m3

𝜋

=− .

∴ 𝜑́ =

−

=

In similar manner, at stage 3, =

𝜃=

, ∴ for prime mover,

𝜑=

and for follower bar, 𝜃́ =

𝜋

,

=

∴ 𝜑́ =

𝜋

+

−

=

́ =

𝜋

−

𝜋

=

=−

𝜋

= 4.06h m3

At Stage 4: = BDV = .

ℎ m3

At stage 4 for prime mover,

∴ 𝜑= ́ =

= 𝛼

∴ 𝜑́ = (

−

−

+

𝜃

𝜃=

𝜃

,

and for follower bar,

𝛼

−

+

−

𝜃́ =

+

𝜃

−

𝜃

,

) = −(

−

𝜃

)

Calculation for average exposed area to pressure between adjacent stages From stage 1 –2 Taking the integral average of exposed length in light of equn (2), over the range from, 𝜃 =

𝜋

𝜃=

−

𝜃

then for prime mover,

21


̅̅̅̅̅̅ = =

− 𝜋−

∫ { − ́ 𝜃

−

𝜋−

𝜋 × ∫𝜋


𝜃

Applying the value of 𝜃 ,

𝛼

{ − ́

} 𝜃


, , ́ and solving definite integral we get,



𝜋−

𝜋 𝜋

𝜋

̅̅̅̅̅̅ = .

𝜃=

∫𝜋 { − ́

𝛼

)} 𝜃

m

for when prime mover travels from 𝜃 = 𝛼


̅̅̅̅̅̅ ́ = .

𝛼

𝜋

𝜃=

)} 𝜃

−

𝜃

m


From stage 2– 3

=

.

− .

m =− .

In the similar way over the range from, 𝜃 = ̅̅̅̅̅̅ =

=

−

𝜃

𝜋

𝛼

∫ 𝜋−𝜃 { − ́


𝜃


∫ { − ́

𝜋− 𝜋+

−

m

𝜃=

then, } 𝜃


𝛼

)} 𝜃




𝜃=

𝜋

− 𝜋

𝜋

𝜋

m

for when prime mover travels from 𝜃 =

∫𝜋 { − ́

𝛼

́ ∴ ̅̅̅̅̅̅ = .


𝛼

−

)} 𝜃

𝜃

𝜃=

m

Cumulative average exposed length difference or the net effective average length of exposed area to gas= ́ ̅̅̅̅̅̅ − ̅̅̅̅̅̅ = ̅̅̅̅̅̅̅̅ − ́

=

.

− .

m=− .

22

m


From stage 3– 4 In the similar way over the range from, 𝜃 = ̅̅̅̅̅̅ =

=

−

∫ { − ́

𝜋+

𝜃

− 𝜋

𝜋+

∫𝜋


𝜃

𝜋

follower bar travels from 𝜃 = to 𝜃 =

+

𝜃

𝛼

{ − ́

𝜃

+

then,

} 𝜃


𝛼

)} 𝜃


Applying the value of 𝜃 , 𝜃=

𝜃=

𝜃=

.


𝜋−

𝜃

−

𝜋

−

𝜋−

∫𝜋

𝜃

𝜃

m

for a situation arises when prime mover travels from


{ − ́

̅̅̅̅̅̅ ́ = .

} 𝜃

m

Difference in average exposed length or the net effective average length of exposed area to gas= ̅̅̅̅̅̅ − ̅̅̅̅̅̅ ́ = ̅̅̅̅̅̅̅̅ − ́

=

.

− .

m=− .

m

𝜃 =

−

Average volume change calculation between adjacent stages From stage 1 – 2 𝜋

By putting values of 𝜃́ = , 𝜃́ = , 𝜃 = integral we get,

From stage 2– 3

=

.

ℎ

Now putting values of 𝜃́ = , 𝜃́ = = .

definite integral we get, From stage 3– 4

Now putting values of 𝜃́ = definite integral we get,

𝜋

𝜋

, 𝜃́ = = .

𝜋

ℎ

ℎ

−

,𝜃 =

𝜃

−

,𝜃 =

23

𝜃

𝜃

in equn (14) and evaluating definite

𝜃 =

𝜃 =

in equn (14) and evaluating

+

𝜃

in equn (14) and evaluating


Average pressure calculation between adjacent stages =( =

=

Now

= , consequently =

=

Again

. ×

. ×

.

.

, consequently =

. ×

.

.

=

ℎ ℎ

.

ℎ ℎ

.

=

×

)

×(

. .

ℎ ℎ

=

) =

.

×

=

.

kN/m2

kN/m2

.

For direct fuel injection (mode2) and electronic fuel injection,

kN/m2 ,

Average multiplying factor calculation between adjacent stages

≈ .

Prime mover is driven in expansion cycle while follower is driven by prime mover in compression cycle. As prime mover drives follower in compression stage, prime mover is doing more work than the work required to be done by the free end of follower bar. So for both prime mover and follower bar torque remains the same, i.e. 𝑖

Wherein,

𝑖

=

×

𝑖

×

𝑖

=

𝑖

=

=

×

. .

×

is force required at prime mover to make required force at follower bar end which

is passing along dotted circle C in fig: 10a below and mover at any stage (say at stage 1).

So multiplication factor,

. .

=

𝐼. .

+

́ ×𝐼.́ .

= 𝐼.

. .

.

is the force multiplication factor for prime

+

́ ́ . ×𝐼.

[

𝑖

𝑖 :

]

If method of integral average between successive stages (say between stage 1 &2) is applied then, ̅̅̅̅̅̅ . .

= 𝐼.̅̅̅̅̅̅̅̅̅̅ + .

̅̅̅̅̅̅̅ ́ ̅̅̅̅̅̅̅̅̅̅ ́ . ̅×𝐼.

…………………………… 24


Wherein, t is the length of prime mover from center O(0,0) to junction point which can be found from equn (19) above and ̅̅̅̅̅̅̅̅̅̅ . . and ̅̅̅̅̅̅̅̅̅̅ .́ . are the average inclination factor for prime mover and the

follower bar respectively. The integral average is applied to ̅, ̅̅̅̅̅̅̅̅̅̅ . . and ̅̅̅̅̅̅̅̅̅̅ .́ . for a given angle 𝜃 change between successive stages [such as between stage 1&2] . The inclination factor, . .

= cos 𝜑 = cos

example]

− 𝜃 [for demonstration purpose stage 1 is shown as an

From stage 1 – 2 For prime mover, angle 𝜃 changes from

𝜋

−

to

𝜃

and

changes from

to

−

𝛼

while prime mover travels from stage 1 – 2 in compression cycle while 𝜃 is measured around origin O (0,0) and

is measured around modified origin M(d,0) from positive X-axis. 𝜋

Similarly for follower, angle 𝜃 changes from to

and

changes from

travels from stage 1 – 2 in compression cycle. (see fig: 10a through 10b) So from equn (19), ̅ between stages is ̅=

𝜋−

𝜃

−

̅̅̅̅ . .

=

̅̅̅̅ .́ .

=

𝜋

𝜋−

́

× × ∫𝜋 𝜋

𝜋− −

𝜋×

𝜃

𝜋−

(cos 𝜃 + √cos 𝜃 + ) 𝜃 = . .

7

𝜋−

𝜋 × ∫𝜋

−

𝜋

𝜋−

∫𝜋

.

7

m

cos

−𝜃

Solving the definite integral in view of equn (17) we get, ̅̅̅̅ . . 𝜋−

𝜋

×

𝜋−

𝜋

𝜋

𝜋

× ∫𝜋 ∫ 𝜋 cos

−𝜃

𝜃.

Solving the definite integral in view of equn (17) we get, ̅̅̅̅ .́ .

=

.

= +

̅̅̅̅̅̅̅̅̅̅ . . × . 25

.

+

= .

= .

So from equn (18) we get,

̅̅̅̅̅̅ . .

𝜃.

̅̅̅̅̅̅ ́

̅ × ̅̅̅̅̅̅̅̅̅̅ .́ .

× .

= .

to

while follower


From stage 2 – 3 For prime mover, angle 𝜃 changes from

−

𝜃

to

and

−

changes from

while prime mover travels from stage 2 – 3 in compression cycle. Similarly for follower, angle 𝜃 changes from

𝜋

to

and

changes from

to

𝜋

𝛼

to

while follower

travels from stage 2 – 3 in compression cycle. (see fig: 10b through 10c) So from equn (19), ̅ between stages is ̅=

̅̅̅̅ . .

𝜋− 𝜋+

=

𝜋

́

× × ∫ 𝜃 (cos 𝜃 + √cos 𝜃 + ) 𝜃 = . 𝜋−

𝜃

𝜋− 𝜋+

×

𝜃

𝜋−

𝜋

𝜋

𝜋

× ∫ 𝜋−𝜃 ∫ 𝜋 cos

m

−𝜃

. 𝜃

Solving the definite integral in view of equn (17) we get, ̅̅̅̅ . . ̅̅̅̅ .́ .

=

𝜋

−𝜋

×

𝜋

−𝜋

𝜋

𝜋

× ∫𝜋 ∫𝜋 cos

−𝜃

= .

. 𝜃


= .


̅̅̅̅̅̅ . . =

From stage 3 – 4

.

= +

For prime mover, angle 𝜃 changes from

̅̅̅̅̅̅̅̅̅̅ . . × .

.

+

= .

× . +

to

̅̅̅̅̅̅ ́ ̅ × ̅̅̅̅̅̅̅̅̅̅ .́ .

𝜃

and

changes from

+

to

while prime mover travels from stage 3 – 4 in compression cycle. Similarly for follower, angle 𝜃 changes from

𝜋

to

−

𝜃

and

changes from

while follower travels from stage 3 – 4 in compression cycle. (see fig: 10c through 10d) So from equn (19), ̅ between stages is ̅=

́

𝜋+

× ×∫𝜋 𝜋+ − 𝜋 𝜃

𝜃

(cos 𝜃 + √cos 𝜃 + ) 𝜃 = . 26

m

𝜋

to

𝛼

−

𝛼


̅̅̅̅ . .

=

̅̅̅̅ .́ .

=

𝜋− 𝜋+

𝜃

×

𝜋

𝜋−

𝜋+

×∫𝜋

𝜃

7𝜋

∫ 𝜋 cos

. 𝜃

−𝜃

Solving the definite integral in view of equn (17) we get, ̅̅̅̅ . . 𝜋−

𝜃

−

𝜋×

𝜋−

𝛼

−

𝜋−

𝜋 × ∫𝜋

𝜃

𝜋−

∫𝜋

𝛼

cos

= .

−𝜃

. 𝜃


= .


̅̅̅̅̅̅ . . =

Gross compression work done = ̅̅̅̅̅̅ × ̅̅̅̅̅̅̅̅̅̅  −  =

≈− Efficiency,  =

We know,

𝑖

+ .

≈−

× − .

.

ℎ−

.

.

×ℎ× .

× −. .

× .

.

× .

.

ℎ

+

̅̅̅̅̅̅ ́

̅ × ̅̅̅̅̅̅̅̅̅̅ .́ . = .

× .

+ ̅̅̅̅̅̅ × ̅̅̅̅̅̅̅̅̅̅  − 

× ̅̅̅̅̅̅ . .

×ℎ× .

ℎ−

𝑖

̅̅̅̅̅̅̅̅̅̅ . .

× ̅̅̅̅̅̅ . .

×ℎ×

ℎ kN-m or kJ

= .

+

×ℎ×

̅̅̅̅̅̅ × ̅̅̅̅̅̅̅̅̅̅  −  .

=

+

× .

+

.

× − .

𝑖

/

Mass of fuel at stoichiometric ratio, =

=

.

×

𝑖

ℎ×{

LHVof gasoline =

×

. .

− 𝑖

}× .

/

×

𝑖 .

kg = .

∴ Energy from fuel or Heat input = .

ℎ kg

ℎ×

= 27

.

ℎ kJ

×ℎ× ×ℎ× .

× ̅̅̅̅̅̅ . . × .

+


=

.

ℎ+ −

= .

.

ℎ

.

=

ℎ

.

%

So findings can be made very logically that this rotary engine technology is well ahead efficient than the existing technology not only in the sector of fuel efficiency but in the sector of torque also as the inventive technology is dealing with perpendicular direct propulsion of prime mover by the gas itself. One of the major reasons of high efficiency of this technology, can be primarily recognized that in expansion cycle, the expansion force along dotted circle ‘C’ on prime mover, travels total Length +

+

=

.

m against compression force total travelling length

𝑖

=

=

+

+

= . m along dotted circle ‘C’, i.e. the expansion force have more angular duration and flexibility

than compared to the compression force, consequently a duration of compression cycle is smaller than a

duration of expansion cycle. Moreover this technology is involving in using single combustion chamber per single engine module in compared to using multiple combustion chambers per single engine module and it is evident from above figures and calculation that prime mover already receives direct impact from gas and it is not an issue for the engine whether or not follower receives loss in expansion cycle as like the principle described in fig: 10 as the power is not transmitted from follower to prime mover in expansion cycle. The follower bar is used for only sealing purpose or for providing support to combustion chamber and it does not contribute in power generation in expansion cycle of this engine. Subsequently the engine is not incurring the loss factor in expansion cycle due to any inclination of follower with the tangent of main wheel at junction point. However, if the 2nd chamber (non-combustion chamber) were used as to be another combustion chamber then the gas will need to propel the follower bar first to drive the prime mover via main wheel in one of the combustion chambers and relevantly loss factor will also need to be accounted in calculation of expansion cycle, i.e. work from expansion is needed to be counted with multiplication factor in logical form if follower bar contributes to power generation of this engine. Moreover gas pressure projected on follower bar to drive prime mover, will definitely resulting the dotted circular curved path (circle C in fig: 2 to 9) of force to be shrunk from dotted circle “C” to periphery of 28


the main wheel, i.e. effective radius of torque for power shaft will be decreased and subsequently work and efficiency per unit of fuel ignited will be decreased drastically too. So using of multiple combustion chambers per engine module is thus discouraged in designing an efficient rotary engine wherein now-adays fuel efficiency and torque output is the primary issue in the power generation sector. In order to evaluate the torque, an integral average of ̂ can be found by equn (15) between successive stages which can be further multiplied with the projected average force applied at prime mover bar-end between

successive stages. In particular, the original design comprises torque enhancer pocket which increases expansion work and reduces compression work; subsequently increases efficiency further from this calculation although it was not shown here. For better performance, this engine can also be coupled at a suitable combination with an electric motor in order to design an excellent hybrid power system which ultimately retaining overall fair fuel efficiency in almost all condition of traffic. Another great feature of this rotary engine is that it can be used with a cycle of cogeneration process [5] (figured below in fig: 11) wherein the waste heat can be recycled through one or more heat exchanger(s) to produce a super-heated steam and this steam can drive a turbine in next step. This type of process concurrently opens up a wide range of choices for the designers where a miniature turbine rotary output can be connected to drive an

A/C compressor or an alternator or the both in suitable combination in order to generate additional energy from waste heat which is also known as waste heat management. 29


6Section:

3 Reciprocating Engine Efficiency Calculation

Each piston having a bore diameter = 2d unit which is also the diameter of the cylindrical casing or chamber. The crankshaft having a radius of d unit and length of connecting rod = l unit The connecting rod makes an angle ‘α’ with the plane of linear movement of the piston and the crankshaft making an angle 𝜃 with the same plane wherein 𝜃

.

ranges between − tan−

The angle between the crankshaft and connecting rod is ( −

+ tan−

− 𝜃)= . These above angles α and

𝜃 varies when the piston travels from BDC to TDC and vise versa. The volume of the combustion

chamber at BDC is TDV and the volume of combustion chamber at TDC is BDV. These are the basic parameters for this engine design wherein all the angles are unitized by Radians and all temperature values are in absolute temperature Kelvin unless otherwise specially specified in this literature. The term TDV stands for Top Dead Volume or maximum volume, BDV stands for Bottom Dead Volume or minimum volume or clearance volume. This reciprocating engine is specially four stroke engine that follows the principle of classical Otto cycle which is figured in fig: 13 and the pv diagram in fig:13 is frequently used later in this section to calculate pressure at specific volume of combustion chamber.

Conditions: At TDC, At BDC,

=

=

𝜃=

𝜃=

+

,

ℎ 30

and

= , , ,…….∞


In ideal condition for perfect operation, l>

= tan−

and

and

is the maximum angle

formed between connecting rod and piston linear movement and also when power transmission rate is maximum in an expansion cycle or compression cycle. Distance between piston coupling and crankshaft coupling that changes due to ups and down motion of piston, ́ = +

− cos −

cos 𝜃 =

− cos

+

− cos 𝜃 ….………………………(1)

Changes in value of this length resulting the piston linear displacement. (Fig:13)

Loss factor calculation: This reciprocation engine incurring a significant loss between the piston and crankshaft when piston moves or crankshaft rotates by the influence of gas pressure or by forward momentum respectively. This loss is factorized below as loss factor and is figured in Fig: 13. . . = cos

. . = cos

× sin{ − × sin

+𝜃 }

+ 𝜃 ….….….(2)

Three different stages:

For compression cycle, at Stage 1 in view of Fig: 15 below, the piston is in BDC. At Stage 2, when the piston is in a position of maximum power transmission or when, at the stage 3, when the piston position is in TDC.

=

and

Similarly, for expansion cycle at Stage 1 in view of Fig: 14 below, the piston is in TDC and it is approaching toward stage 3 via stage 2 wherein at Stage 2, the piston is in a position of maximum power

31


=

transmission or when 𝜃=

𝜋

−

.

. For both compression and expansion cycle at stage 2,

=

and

Loss factor at different stages

At Stage1 . .

=

. .

= cos

. .

=

=

as

At Stage2

𝜃= . =

as

At Stage3

=

as

𝜃=

𝜃=

+

𝜋

−

.

.

Average loss factor calculation between adjacent stages: The generic formula for average loss factor in view of equn (2), ̅̅̅̅̅ . =𝜃 =𝜃

−𝜃

− −𝜃

×𝛼

𝛼

×𝛼

𝜃

× ∫𝛼 ∫𝜃 cos

−𝛼

𝛼

× ∫𝛼 cos

−𝛼

× sin

× {cos

+𝜃

+𝜃

𝜃.

+𝜃 }

− cos

……..(3)

wherein θ and α are calculated from vertical axis AB in respective compression and expansion cycle in fig:

and fig:

.

Average loss factor between stage 1and 2 in expansion cycle 𝜋

For stage 1 to 2 in equn (3), 𝜃 = , 𝜃 = − ∴ ̅̅̅̅̅ . .

=𝜃

− −𝜃

−

=𝛼

=𝛼

𝑎 𝑎

×𝛼

×

×

𝜋

−𝛼

−𝛼 𝑎

𝜋

𝛼

× ∫𝛼 cos

−𝛼 𝑎

×∫

𝛼 𝑎

×∫

𝛼 𝑎

× {cos

{cos

{cos

,

+𝜃

× cos

× sin

=

and − cos 𝜋

+ −

−

Average loss factor between stage 2 and 3 in expansion cycle 𝜋

For stage 2 to 3 in view of equn (3) projected on fig:14, 𝜃 = − 32

=

.

+𝜃 }

− cos

+ cos

}

}

..………………….

,𝜃 = ,

=

and

= .


∴ ̅̅̅̅̅ . .

− −𝜃

=𝜃

=

𝜋

×𝛼

×𝛼

+𝛼 𝑎

=𝛼

𝑎

×

𝜋

𝛼

× {cos

× ∫𝛼 cos

−𝛼

𝑎

× ∫𝛼

× ∫𝛼

+𝛼 𝑎

𝑎

× cos

{cos

𝑎

+𝜃

{cos

× sin

+

+𝜃 }

− cos

− cos

−

× cos }

−cos

𝜋

+ −

}

…………………...

Calculation of average length travelled by piston between adjacent stages: (in expansion cycle) Average length travelled by piston between stage 1and 2 by taking integral average of (1) in this section ̅̅̅̅̅̅ ́ = =

𝛼 𝑎

×

𝜋

𝛼 𝑎

×

𝜋

=𝛼

×

𝜋

×

𝜋

𝑎

=𝛼

𝑎

𝜋

−𝛼 𝑎

{

−𝛼 𝑎

×∫

∫

−𝛼 𝑎

×∫

{

− cos

−𝛼 𝑎

×∫

{

− cos

−𝛼 𝑎

×[

− cos

𝜋 𝜋

−

+ (

−

+ (

𝜋

− sin

− cos 𝜃 } 𝜃.

+

−

+

𝜋

𝜋

−

− sin

−

− cos

×

{

𝜋

𝜋

−

)} )}

−

− cos

}]

……………………………………(6)

In the similar way for stage 2 to 3 ̅̅̅̅̅̅ ́ =

− 𝛼 𝑎 −

=𝛼

=𝛼

𝑎 𝑎

× ×

𝜋

×

𝜋

𝜋

+𝛼 𝑎

×∫

∫ 𝜋−𝛼

+𝛼 𝑎

×∫

{

𝜋

+𝛼 𝑎

×{

𝑎

{

− cos − sin

− cos 𝜋

+

𝜋

+

− cos 𝜃 }. 𝜃

+

+ (

𝜋

+

×

+

𝜋

+ sin +

𝜋

−

+ cos

)} }

……………………………………..(7)

Calculation of average length travelled by piston between adjacent stages: (in compression cycle)

33


In compression cycle piston travels from BDC to TDC. Again at BDC is stage 1, TDC is stage 3 and stage 2 is the intermediate stage between BDC and TDC wherein stage 2 proposition is the same as that for proposition of stage 2 in expansion cycle. See fig: 15 below. As stage 1 in compression cycle is the stage 3 in expansion cycle, so average length travelled by piston from TDC will be between stage 1and 2 by taking integral average of equn (1) in this section ̅̅̅̅̅̅ ́ =

−𝛼

=

−

=

−𝛼

=

−𝛼

𝛼 𝑎 𝑎 𝑎

𝑎

×

×

𝜋

×

𝜋

×

𝜋

𝜋

+𝛼 𝑎

+𝛼 𝑎 +𝛼 𝑎

+𝛼 𝑎

×∫

×∫

∫

{

𝜋 𝜋

− cos

[

×[

+𝛼 𝑎

− cos

[

×∫

𝜋

− cos

+

+

+ {

+

+ {

𝜋

− sin

+

+

𝜋 𝜋

− cos 𝜃 } 𝜃. − sin

+

− cos

+

}] }]

𝜋

×

……………………………………………………………………(8)

𝜋

+

{ +

− cos

}]

In the similar way for stage 2 to 3 ̅̅̅̅̅̅ ́ =

−

=

−𝛼

=

−𝛼

− 𝑎 𝑎

− 𝛼 𝑎

×

𝜋

×

𝜋

×

𝜋

−𝛼 𝑎 −𝛼 𝑎

−𝛼 𝑎

×∫ ×{

𝜋

×∫

∫ 𝜋+𝛼

{

− cos − sin

……………………………………..(9)

𝑎

{

𝜋

− cos −

𝜋

−

+ + (

𝜋

+

×

− cos 𝜃 }. 𝜃 −

𝜋

+ sin −

𝜋

+

+ cos

)} }

Practical engine concept with standard ratios Practically a reciprocating engine is generally a ‘square’ engine to enhance efficiency where this square’ engine refers to a stroke to bore ratio be 1:1. Moreover the rod ratio is 1.75:1 for standard engine design wherein this rod ratio is the ratio between the connecting rod and the stroke length and it ranges generally 1.4 to 1.8. 34


Calculation of affected parameters of the engine = tan−

= .

, wherein = tan−

= tan−

.

, ∴ = .

= .

Compression ratio is chosen as 8.36 (for ease of comparison with rotary engine)

=> => ∴

= .

∴

𝑉+𝜋

.

×

= . 𝑉

×

= .

=

(let us assume the unit is meter)

Normal atmospheric condition =

=

° =

. KPa= 101.3 kN/m2

From pv diagram = T T

= .

=

= .

−

=

As,

m3 m3 =>T2 = 298

. −

. .

= 563.48 K =

𝑇 𝑇

−



=

x

= 101.3 x So, T2 = 563.48 K and

𝑇 𝑇

−

.

. .

= 1601.32 kN/m2

= 1601.32 kN/m2

In combustion stage Volume remains constant at this stage.  T = T 35


and, T3 = T2 + .Q. wherein, fuel-air ratio =

v

, So fuel-mixture ratio,  =

.

.

Q = Lower Heating Value (LHV) of gasoline = 42700 kJ/kg R= 0.287 kJ/kg.K 𝑅 −

=

.

=

 T3 = 563.48 +

T T

=

x

.

kJ/kg.K

x 42700 x

.

T =T =

= .

. −

.

= 3405.43 K

.

x 1601.32

= 9677.68 kN/m2 = =

The three stages in Expansion cycle At Stage 1 in view of fig: 14 below: = P3 = 9677.68 kN/m2 = 3405.43 K =

At Stage 2: ={ .

Now,

= .

= .

− cos .

+

m3

=[ ∴

]

=

= 9677.68 x

[

.

] .

− cos

𝜋

− .

.

= 1514.53 kN/m2 36

}×

+


Now, =>

T T

−

= =

−

x .

= 3405.43 x

.

.

= 2219.71 K

Average pressure from stage 1 to 2 As we all know pressure is solely dependent on volume change in a combustion chamber along with temperature change and for isentropic process as per fig: 12, the average pressure is dependent on the average volume. However this average volume change is dependent only to average displacement of the piston when it travels from stage 1 to 2 and this volume change is independent to the BDV or clearance volume as piston displacement does not affect in the amount of volume of BDV, i.e. BDV does not transform with piston displacement shown in fig: 15 and BDV is the addition or exception to average displacement of piston. So it can be surely stated that average volume of combustion chamber within stage 1 to 2 is directly proportional to the average distance travelled by the piston and the corresponding pressure to average volume is the average pressure between stage 1 and stage 2. So from equn (5), ̅̅̅̅̅̅ ́ = .

́ = ̅̅̅̅̅̅ = ̅̅̅̅̅̅ ×

We know,

+

̅̅̅̅̅̅̅

≈ .

= [̅̅̅̅̅̅̅]

37

m3


∴ ̅̅̅̅̅̅ =

[̅̅̅̅̅̅̅]

=

Average loss factor from stage 1 to 2

.

[

.

.

]

.

=

kN/m2

Average loss factor is a term which represents the resultant work output factor after the loss from piston to crankshaft via connecting rod. To calculate this factor from stage 1 to 2, the equation (3) becomes equn (4) and after solving the definite integral and putting value of ̅̅̅̅̅ . .

= .

=

.

, we get

% i.e. nearly 36% of the total work from piston at stage 1 to 2 is lost

due to the linkage between piston to connecting rod and connecting rod to crankshaft. This loss factor is the main ingredients for traditional reciprocating engine of being less efficient than the current innovation. Although traditional efficiency calculation based on compression ratio only may show a reciprocating engine efficiency of 60% but it does not accounting the loss in the linking among piston, connecting rod and crankshaft because the traditional calculation based on compression ratio is just only measuring the efficiency within the combustion chamber or measuring the efficiency before said linking mechanism and that’s why the traditional theoretical efficiency deviates much from the practical scenarios. So it is extremely important for any new internal combustion engine to account the loss in linking mechanism to compare the efficiency with that of the traditional one. Now from fig: 15 above it is clear that average pressure, ̅̅̅̅̅̅ travels a distance of and 2.

∴ At Stage 3: ={ . = .

− cos

m3= TDV

+

− cos

=

×𝜃

=

}×

+

= .

38

×{ −

}

between stage 1


Again, So, ∴

=

=[

]

[

]

= 9677.68 x

.

.

.

= 612.22 kN/m2 Now, =>

T T

−

= =

−

x .

= 3405.43 x

.

.

= 1801 K

Average pressure from stage 2 to 3 The average displacement of the piston from stage 2 to 3, is proportional to an average displacement volume of the piston in the combustion chamber and this corresponding pressure to average displacement volume is the average pressure between stage 2 to 3. In reciprocating engine the BDV or clearance volume does not transform, i.e. BDV or clearance volume is independent to the constraint required by the stroke length, as BDV is not the derivative of 𝜃 (crankshaft angle with sliding plane of piston). So the

only transformation happened in the volume, precisely due to variation in the stroke length and that’s why average stroke length is calculated in order to get the average stroke volume, as x sectional area,

is

uniform throughout the whole stroke length and that average volume will also indicate the corresponding pressure to be the average pressure between desired adjacent stages. ́ = . So from equn (6), average stroke length=̅̅̅̅̅̅ ̅̅̅̅̅̅̅

∴ ̅̅̅̅̅̅ =

́ = ̅̅̅̅̅̅ = ̅̅̅̅̅̅ ×

= [̅̅̅̅̅̅̅]

[̅̅̅̅̅̅̅] =

.

[

m

+

.

.

≈ .

]

.

39

=

m3

.

kN/m2


Average loss factor from stage 2 to 3 Average loss factor is a term which represents the resultant work output factor after the loss from piston to crankshaft via connecting rod. To calculate this factor from stage 2 to 3, the equation (3) becomes equn (5) and after solving the integral while putting value of ̅̅̅̅̅ . .

= .

=

.

, we get,

% i.e. 40% of the total work from piston at stage 2 to 3 is lost due to the

linkage among the piston, the connecting rod and the crankshaft.

From fig: 14 it is clear that the average pressure, ̅̅̅̅̅̅ travels at a distance of ∴

=

×𝜃

𝜋

=

×{ +

Gross expansion work done = ̅̅̅̅̅̅ ×

× ̅̅̅̅̅ . .

=

+

=

.

=

× .

× .

kJ

}= .

× .

× .

+ ̅̅̅̅̅̅ × +

.

×

× ̅̅̅̅̅ . .

× .

×

× .

If an observation can be done in above equation, we can see that maximum work is done between stage 1 to 2.

The three stages in Compression cycle Referring to fig:15 below at Stage 1: = P3 = 101.3 kN/m2 = 298 K

At Stage 2:

=

= .

The value of volume,

m3

for compression cycle at stage 2 when the piston moves on its travelling

from TDV to BDV, is also same to the value of the

of the expansion cycle. Only difference between

40


the volumes

of expansion and

compression cycle is that the

of expansion

cycle increased from BDV while

of

compression cycle is decreased from the TDV. ,

and

= .

∴

=

m3 =[

]

[

. .

= 101.3 x

]

.

= 250.62 kN/m2 Now,

T T

=>

−

= =

−

x

= 298 x

.

. .

= 367.28 K Average pressure from stage 1 to 2 The average displacement of the piston from stage 1 to 2, yields an average displacement volume of the piston in the combustion chamber and the corresponding pressure to average displacement volume is the average pressure between stage 1 to 2 and the average stroke length at stage 1 to 2 in compression cycle is same as the average stroke length at stage 2 to 3 in expansion cycle. So from equ n(8), ̅̅̅̅̅̅ ́ = .

and average volume in combustion chamber is ̅̅̅̅̅̅.

́ ∴ ̅̅̅̅̅̅ = ̅̅̅̅̅̅ ×

+

≈ .

m3

And,

̅̅̅̅̅̅̅

∴ ̅̅̅̅̅̅ =

41

= [̅̅̅̅̅̅̅] [̅̅̅̅̅̅̅]


=

. ×[

=


.

. .

]

kN/m2

.

𝜋

For stage 1 to 2 in equn (3) in compression stage and further from fig:16, 𝜃 = , 𝜃 = + and

∴ ̅̅̅̅̅ . .

=

.

=𝜃

=

=

− −𝜃

𝜋

−

×𝛼

+𝛼 𝑎

𝜋

+𝛼 𝑎

−𝛼

×𝛼

×𝛼

𝛼

× ∫𝛼 cos ×∫

𝑎

𝛼 𝑎

×∫

𝑎

𝛼 𝑎

× {cos

{cos

{cos

+𝜃

× cos

× sin

− cos 𝜋

+ +

+

+𝜃 }

− cos }

+ cos

,

=

}

..………………….

Average loss factor is a term in compression cycle which is the resistance of linking mechanism and the crankshaft needs to overcome the resistance to compress the air-fuel mixture in the cylinder, i.e. the crankshaft needs to do more work than the work done by the piston alone to compress the air-fuel mixture. In order to calculate this factor from stage 1 to 2, after solving the integral by applying value of in the equation ̅̅̅̅̅ . .

= .

, we get, =

It is observable that ̅̅̅̅̅ . .

.

%

of compression cycle is not the same value of ̅̅̅̅̅ . .

of expansion

cycle. The reason behind that, in expansion cycle piston drives the crankshaft by linking mechanism but in compression cycle, crankshaft drives the piston by linking mechanism. From fig:14 and fig:15, it is clear that the average pressure, ̅̅̅̅̅̅ travels at a distance of

which is same as the distance travelled by average pressure in expansion cycle from stage 2 to 3.

At Stage 3:

∴

=

× −𝜃

={ . =

= − cos

= .

× {−

𝜋

+

− cos

m3

42

+

}=− . }×

+

m


=[

]

=

[

] = 101.3 x

.

Average pressure from stage 2 to 3

.

.

= 1601.29 kN/m2

The average displacement of the piston from stage 2 to 3 yields an average displacement volume of the piston in the combustion chamber and the corresponding pressure to average displacement volume is the average pressure between stages 2 to 3. So from equn (9), ̅̅̅̅̅̅ ́ = .

́ = ̅̅̅̅̅̅ = ̅̅̅̅̅̅ × ̅̅̅̅̅̅̅

= [̅̅̅̅̅̅̅]

∴ ̅̅̅̅̅̅ =


=

+

. × [

.

.

[̅̅̅̅̅̅̅] ]

.

≈ .

=

.

𝜋

For stage 2 to 3 in view of equn (3) projected on fig:16, 𝜃 = + ∴ ̅̅̅̅̅ . .

=𝜃

=

=

− −𝜃

𝜋

×𝛼

−𝛼 𝑎

𝛼 𝑎

×

−𝛼

×𝛼

𝜋

𝛼

× ∫𝛼 cos × {cos

𝑎

−𝛼 𝑎

× ∫𝛼

× ∫𝛼

𝑎 𝑎

{cos

{cos

+𝜃

× cos

× sin

− cos

+

+

m3

kN/m2

,𝜃 = ,

+𝜃 }

− cos

× cos

−cos

}

= 𝜋

+ +

= .

and

}

…………………...

In the similar manner for compression work done by the crankshaft at stage 1 to 2, to find out the compression work needs to be done by crankshaft in stage 2 to 3, the average loss factor is therefore calculated below by applying the value ̅̅̅̅̅ . .

= .

=

and solving the definite integral in equn (11), we get,

. %

So the crankshaft requires less energy in stage 1 to 2 than that of stage 2 to 3 to compress mixture in combustion chamber during compression cycle since the projected gas pressure on piston is required to be divided by the loss factor calculated above. 43


From fig:14 and fig:15, it is clear that the average pressure, ̅̅̅̅̅̅ travels at a distance of

and this

distance travelled during compression cycle in stage 2 to 3 is same as the distance travelled in expansion cycle from stage 1 to stage 2. ∴

=

× −𝜃

×

× 𝐿. ̅̅̅̅ .

Gross compression work to be done = ̅̅̅̅̅̅ ×

=

.

=−

.

=−

×

.

× − .

−

kJ

.

=

× {−

+ ̅̅̅̅̅̅ ×

×.

kJ

+

.

×

×

𝜋

−

}=− . × 𝐿. ̅̅̅̅ .

× − .

×

m

.

If an observation is done in above equation, we can see that an event of maximum work to be done by the crankshaft occurs between stage 2 and 3. It is also noticeable on above equations that the loss factor is multiplied to the pressure exerted in expansion cycle but the exerted pressure is divided by the loss factor in compression cycle. The reason behind that, in expansion cycle, work is done to the crankshaft after the loss in linking mechanism but however, in compression cycle, the work is required to be done by the crankshaft to compress the mixture in the cylinder or to make the required changes in volume of combustion chamber, i.e. in expansion cycle, the work from piston passes through loss factor to arrive at crankshaft and in compression cycle, work needs to be done by crankshaft in sufficient amount that, with the help of forward momentum of flywheel, by overcoming the loss factor crankshaft can thus be able to make required changes in volume of combustion chamber. 𝑖

Now, Efficiency,  =

We know, ∴

=

= .

𝑖

×

= . 𝑖

×

×{

/

. .

𝑖 ℎ𝑖

𝑖

}× .

×

− 𝑖

+

𝑖 .

𝑖 .

= .

kg 44

𝑖


,

∴

=

= .

. =

𝑖

+ − . .

𝑖

=

= . =

.

.

×

/ kJ

%

The above calculations are based on ideal Otto cycle and let us assume that the bearing of engine is frictionless. However in reality the pv diagram (Fig: 12) does not follow the ideal Otto cycle 100% and in reality the above calculated efficiency is further decreased to about 15% for a chosen compression ratio of 8.36. The reason behind the decreased efficiency is that among the above calculated efficiency, a fraction of power is also lost to drive camshaft, to drive gear train and also for the uncontrollable heat transfer or loss via cylinder walls wherein the transmission loss is not only applicable to this type of engine, rather any mechanism that uses gear reduction unit, torque converter etc will experience a slight decrease in net efficiency. Although the efficiency of this type of engine can be increased if the compression ratio is increased according to traditional calculation for efficiency but this compression ratio however cannot be increased limitless because compression ratio more than 13 is likely to develop a tremendous elevated temperature in combustion chamber besides auto ignition and subsequently increases the overall costing of the vehicle or power generation system. Moreover excessive temperature in combustion chamber is harmful because this high temperature resulting significant amount of NOx emission and the environment is vulnerable to these pollutants. That’s the major reason behind among various other reasons; the engine based on Otto cycle does have a compression ratio ranging from 8 to 13.

45


Section: 4 Comparison between different technologies Pros and cons of EV (Electric Vehicle) Pros  EV is extremely efficient in converting battery charge to wheel  EV has almost noiseless and vibration less operation  EV produces almost zero emission for local environment  EV has less complicated design and subsequently has lower cost to manufacture and lower maintenance.  EV can control the variation of speed on road very efficiently  EV does not require time to warm up the engine Cons  EV requires lots of time to charge battery  EV makes similar pollution to that of I/C engine globally  EV has net efficiency almost 35% if the calculations are based from well to wheel.  EV has comparatively less torque and less power density than that of I/C engine.  EV has a great chance of getting idle in long and heavy duty drive if the EV is not hybrid  EV may not be suitable for racing purpose  EV is not a standalone feature/system and it has to depend on the electricity supply. Therefore any malfunction in supply due to natural reason/insufficient production EV remains idle as electricity is not physically available on earth. **Note: Although someone can argue that EV can be efficient if it uses renewable energy. But it is true that it blocks other electrical equipment(s) to use that renewable energy or enforces those electrical items to use energy from fuel. Furthermore any renewable source on earth is not a stable means of power generation.

46


Pros and cons of Fuel Cell (can be used in Electric Vehicle) Pros  FC (Fuel Cell) is efficient in converting fuel to electricity (almost 55%)  FC has almost noiseless operation if it is installed into EV  FC can charge battery and drive an EV  FC installed in EV can control the variation of speed on road very efficiently  EV with FC can make a good hybrid vehicle Cons  EV with FC having a net efficiency not more than 43% if the calculations are based from well to wheel.  EV with FC has comparatively lower torque and power density than that of I/C engine.  EV with FC requires a lot of time to warm up the engine  Not be suitable for racing purpose  FC produces significant emission for local and global environment  FC has complicated design and subsequently may require moderately higher cost to manufacture and maintenance  FC may require frequent cleaning of carbon deposit from FC chamber.  FC may not be able to adopt the method of cogeneration.

Pros and cons of this Rotary Engine Pros  This Rotary Engine is very efficient in converting thermal efficiency into mechanical efficiency which has almost 50% efficiency @compression ratio 12, in view of well to wheel calculation  This Rotary Engine may exhibit extreme torque and power density  This type of engine is ideal for heavy duty drive 47


 This rotary engine can adopt the method of cogeneration to elevate the overall fuel efficiency up to 68% (exchanging heat from exhaust gas by heat exchanger to produce super-heated steam and then by driving turbine to generate electricity in the power generating system)  Engine warm up time or wait time is not very long  This type of engine with an electric motor can be an excellent hybrid combination  This rotary engine may be suitable for racing purpose  This rotary engine mechanism is not incurring reciprocating motion and subsequently there is no need for change of direction of momentum in the system. Cons  The rotary engine require gasoline and air mixture to operate 

The rotary engine requires maintenance although not frequent

 The rotary engine may produce significant emission for local and global environment, however using of hydrogen fuel may eliminate a majority of that emission and will make it a green engine.  The rotary engine has complicated design and subsequently may require costing to manufacture and for maintenance  The compression ratio of the rotary engine should not exceed 13, otherwise it will create harmful emission of NOx

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References [1] Well-to-wheel analysis of direct and indirect use of natural gas in passenger vehicles http://www.sciencedirect.com/science/article/pii/S0360544214008573

[2] Wells to wheels: electric car efficiency https://matter2energy.wordpress.com/2013/02/22/wells-to-wheels-electric-car-efficiency/ [3] The derivations are made in view of basic mechanical engineering properties of thermodynamics and in view of trigonometry as well. Some definite integrals are evaluated by scientific calculators.640640640640640640 [4] The rotary engine described in this literature can present working principle in the url below https://www.youtube.com/watch?v=JiAa5qFfa4M&t This rotary engine presented above is legally covered by the patent US9528433 and by a cip publication US20160290221. [5] Cogeneration https://en.wikipedia.org/wiki/Cogeneration [6] Comparison of efficiency between traditional reciprocating and this rotary engine can be found in brief in the URL below https://www.youtube.com/watch?v=pCMrbP8kTVc&t [7] https://en.wikipedia.org/wiki/Thermal_efficiency [8] The patent literature is available in the url below https://www.google.com/patents/US9528433 [9] Pressure-volume diagram of Otto cycle in the url below https://en.wikipedia.org/wiki/Otto_cycle

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