transformer

1. TECHNOLOGICAL UNIVERSITY(MYITKYINA)
DEPARTMENT OF ELECTRICAL POWER ENGINEERING
DESIGN CALCULATION OF
DISTRIBUTION TRANSFORMER (315KVA)
BE Mini Thesis
Title Defense
1
Presented by;
Mg Sai Seng Pha
Supervised by;
Daw
Date – 28.8.2022

2. Outlines of Presentation
Introduction
Aim and Objective
Implementation Program
Outlines of thesis
Design Calculation of Distribution Transformer (315V)
Conclusion
Future Plan
2

3. CHAPTER 1
INTRODUCTION
• A crankshaft related to crank is a mechanical part able to perform a
conversion between reciprocating motion and rotational motion.
• In a reciprocating engine, it translates reciprocating motion of the
piston into rotational motion; whereas in a reciprocating compressor,
it converts the rotational motion into reciprocating motion.
• In order to do the conversion between two motions, the crankshaft
has "crank throws" or "crankpins", additional bearing surfaces whose
axis is offset from that of the crank, to which the "big ends" of the
connecting rods from each cylinder attach.
• It is typically connected to a flywheel to reduce the pulsation
characteristic of the four-stroke cycle, and sometimes a torsional or
vibrational damper at the opposite end, to reduce the torsional
vibrations often caused along the length of the crankshaft by the
cylinders farthest from the output end acting on the torsional elasticity
of the metal. 3

4. History of crankshaft
• During the background research, emphasis was given on identifying the morphology of
the crankshaft, the manufacturing techniques that exist now and the thermal treatments in
different stages of the production. The kinematic-kinetic analysis and the balancing
method are also included in the background research.
• Crankshaft is one of the critical components for the effective and precise working of the
internal combustion engine.
• Study is done on the crankshaft and found that there are tremendous stress induced in
crankshaft.
• The Industrial revolution which started in 1760 and finished in 1840 brought into the light
new manufacturing techniques and opened the road for revolutionary inventions, which
are used even today.
• One of the most innovative inventions is the gasoline and diesel engine which made a
huge contribution in the process of the world.
4

5. Project goals
• This project goal is to identify if the crankshaft can be designed in
individual parts in a way that fulfills all the technical requirements.
• Firstly to analyze the single cylinder crankshaft from an engineering
point of view, discussion about single cylinder crankshaft and
secondly, to understand the functionality of the geometry and design
consideration and the third, calculated on single cylinder crankshaft.
• The main goal is to understand exactly on crankshaft and how it
works, how to maintain and how to calculate the stresses on the
crankshaft, vibration modes and dynamics force analysis.
5

6. Diagram of single cylinder crankshaft
6

7. Chapter 2
Consideration of calculation
• Compare von-mises stress in first and second stage
• Calculation total torque on third stage
• Compare with stress and analysed
7

8. Diagram of angles consideration on crankshaft
8

9. Failure analysis of crankshaft(Chapter 3)
Pressure Calculations:
Density of petrol (C8H18) ;
ρ = 750 kg / m3
= 750 × 10-9 kg / mm3
Operating Temperature ; T = 20℃
= 20 + 273K
= 293.15 K
Mass of displacement; m = ρ × V
m= (750 × 10−9
) × (124 ×103)
m= 0.093 kg
Molecular mass of petrol ;
M = 114.228 × 10-3 kg / mole
Gas constant for petrol ;
R = 72.7868 × 103 J / kg / mol.K
We know that PV = mRT
P × 124 × 103 = 0.093 × 72.7868 × 103 × 293.15
P =
0.093×72.7868×1000×293.15
124×1000
P =16.003MPa
9

10. Design Calculations
Gas Force ( Fp) ;
Fp = P × A
Fp=16.003× (
π
4
×D2)
=16.003×(
π
4
×53.52)
=16.003×(
π
4
× 2862.25)
=16.003×2248.006
=35974.84N
=35.97484 × 103 N
10

11. Moment on pin;
Mmax =
Fp
2
×
lc
2
=
35974.84
2
×
56
2
Mmax= 503.65 × 103 Nmm
Section Module of crankpin ;
Z =
π
32
×(dc)3
= 0.0982 × 283
= 2155.69 mm3
Torque obtained at maximum power of Suzuki Access 125 Engine ;
P =
2πNT
60
6.5× 735=
2π×7000×T
60
(1ps=0.735KW)
T=
6.5×735×60
2π×7000
=6.517 Nm
T =6.517× 103 Nmm
11

12. Von mises Stress ;
𝜎𝑣𝑜𝑛=
Meq
Z
So, Equivalent bending moment
Meq = [(kb × Mmax)2+
3
4
(kt × T)2]
1
2
Where,
kb= Combined shock & fatigue for bending = 1
kt= Combined shock & fatigue for torsional = 1
= [(1 × 503.65 × 103
)2
+
3
4
(1 × 8.8672 × 103
)2
]
1
2
=[(2536.63 × 108
) +
3
4
(0.5897 × 108
) ]
1
2
=[2537.22 × 108]
1
2
= 50.37082× 104
Meq= 503.708× 103 Nmm 12

13. Now,
σvon =
Meq
Z
=
503.708×1000
2155.69
σvon =233.644MPa
Equivalent twisting moment ;
Teq = (Mmax
2
+ T2
)
1
2
=[ 503.65 × 103 2
+ (6.517 × 103
)2
]
1
2
=[ 253663.323 × 106
+ (42.471 × 106
)]
1
2
=[2.53705794 × 1011
]
1
2
=[253705.794 × 106]
1
2
=503.692× 103Nmm
13

14. Teq =
π × dc
3
× τ
16
503.692×103=
π×21952×τ
16
τ =
503.692 × 103 × 16
π × 21952
=116.858 N mm2
Calculation of vibration mode analysis of crankshaft
Force on the piston:
Bore diameter (D) =53.5mm,
Fp= Area of the bore ×Max. Combustion pressure
= (π/4) ×D2×Pmax
=π
4 ×53.52×
2.5
=5.62KN
We know that;
sin ∅ =
sin θ
(l
r)
=
sin 40
4 14

15. Which implies ∅ = 0.161°
We know that thrust in the connecting rod
FQ=
Fp
cos ∅
From this we have,
Thrust on the connecting rod;
FQ =
5.62
cos 0.161
=5.62KN
Thrust on the crankshaft can be split into tangential component and the radial component.
Tangential force on the crank shaft,
FT = FQ sin(θ+ ∅)
=5.62× sin(35+8.24)=3.62KN
Radial force on the crank shaft,
FR = FQ cos (θ+ Ø)
=5.62× cos(40+0.161)=4.295KN 15

16. Reactions at bearings (1 & 2) due to tangential force is given by,
HT1=HT2 =
FT×b1
b
(since b1=b2=b/2)
=
3.62×53.5
2
53.5
=1.81KN
Similarly, Reactions at bearings (1 & 2) due to radial force is given by,
HR1 = HR2 =
FR×b1
b
(since b1=b2=b/2)
=
4.295×53.5
2
53.5
=2.15KN
16

17. Design of crankpin
Let dc = Diameter of crankpin in mm.
We know that the bending moment at the centre of the crankpin,
Mc = HR1 x b = 2.15 × 53.5 = 115.025 KN-mm
Twisting moment on the crankpin,
Tc=HT1×
L
2
= 1.81×
55.2
2
=170.68KN
From this we have the equivalent twisting moment
Te = Mc
2
+ Tc
2
= 115.0252 + 170.682 =205.821KN-mm
We know that equivalent twisting moment (Te);
Te =
π
16
× (dc)3× τ
τ =
Te × 16
π × (dc)3
17

18. 116.858 =
205.821×103×16
π×(dc)3
Shear stress value is limited to τ = 116.858 N/mm2
so dc = 20.77 mm
Since this value of crankpin diameter (dc= 20.77 mm) is less than the when the crank
is at top dead centre already calculated value of crankpin diameter, therefore, we
shall take, dc=21 mm.
Diameter of the crank pin =21 mm
Design of crank pin against fatigue loading
According to distortion energy theory
The von mises stress induced in the crank-pin is,
Mev = (kb × Mc)2+ 3
4 (kt × Tc)2
= (2 × 115.025)2+ 3
4 (1.5 × 170.68)2
=319.5KN-mm
Here, Kb = combined shock and fatigue factor for bending (Take Kb =2)
Kt = combined shock and fatigue factor for torsion (Take Kt =1.5)
Mev =
π
32
× (dc)3× σv
319.5× 103=
π
32
× 213 × σv
σv= 351.41 N/mm2 18

19. Calculation of dynamic force analysis of crankshaft
n= l
r =150/37.5 = 4
Inertia forces due to reciprocating masses:
Divide the mass of the rod into two dynamically equivalent parts
Mass of the crank pin, ma= (m× b)/l= 2×55.2/150 =0.736kg
Where, m is mass of the rod
Mass at the gudgeon pin, mb= 2−0.736= 1.264 kg
Total mass of the reciprocating parts, m=2.5+1.264=3.764 kg
Inertia force due to reciprocating masses, Fb= mrω2 cos θ+cos 2θ
n
=3.764×0.0375×(2π ×1800/60)2 ×
cos 40+cos 80
4
= 4059.5N
Force on the piston, Fp=2×106 ×
𝜋
4
× 53.5 2=5620N
Net force,F=Fp−
Fb
=5620−4059.5
=1560.5N
Torque due to this force,T=F [sin θ +
sin 2θ
2 n2−sin2θ
]
T=1560.5 sin 40 +
sin 80
2 42−sin240
T=1197.7Nm
19

20. Torque to consider the correction couple;
αc = −ω2 sin θ
n2−1
n2−sin2θ
3
2
=−
2π×1800
60
2
× sin 40
42−1
42−sin240 3 2
=− 5567 rad/s2
L = b+d = 55.2+53.5=108.7 mm
Tc=mb𝛼𝑐(l−L)
cos θ
n2−sin2θ
=3.764 × 55.2 × (−5567) × (150 −108.7)×
cos 40
42−sin240
=− 9269.047KNm
20

21. where m is mass of the rod,
Torque due to mass at A;
Ta=magr cos θ
=0.736×9.81×37.5×cos 40
=207.411Nm
Total torque on the crankshaft;
Net torque on the crank shaft =T – Tc + Ta
=1197.7−(−9269.047)+207.411
=10674.16Nm
21

22. Chapter 4
Discussion and conclusion
• In this calculations, three parts of calculations provided.
• The first part is calculated about failure analysis of crankshaft, the second
part is calculated about vibration mode analysis of crankshaft and the third
is about dynamics force analysis of crankshaft.
• In the first part, the crankshaft failure occurs due to decreased in the fatigue
strength.
• The second part is calculated about force on the piston and resulted reaction
at bearings due to tangential force and radial force.
• The third part is calculated about inertia forces due to reciprocating masses.
• Torque to consider the correction couple is involved in this part of
calculation. Finally, total torque on the crankshaft is calculated.
22