Electromechanical Engineering 4 Light Angle · 2018. 1. 8. · Electromechanical Engineering 4 Team...

Electromechanical Engineering 4 Team AM8 SSV Part 2 Report

Electromechanical Engineering 4

Light Angle

Team AM8

SSV Part 2 Report

Coach: Mr. Hu Yunhao

Team menber:

Mingqian Zhu

Kaiyu Yang

Yu Chen

Meghashyam Adurti

Shaik Jasmine Sultana


1

Contents

1 Impact test with PCB 200C20 load cell .

2 Sankey Diagrams.

3. Strength Analysis and Calculations.

4. Technical Drawing.

5. Collisions Situations.

6. Others.


2

1 Impact test with PCB 200C20 load cell .

For the impact test on SSV ,we used a PCB 200C20 Load cell attached to a mass

corresponding to the mass of the ball with which the car collides during the race

conditions .The PCB quartz ,piezoelectric force sensors are measurement devices that are

capable of measuring the forces in a high frequency dynamic force and strain events such as

those which are encountered during

compression ,tension, Impulse ,reaction and in

our case Impact forces, With help of this

experiment we can simulate the real race

condition and get to know the impact forces

during the car colliding with the steel ball with

its max power ,also we can estimate the height

of the projectile after collision. The results of

the test help in optimization of the car

built .The outputs we amplified with a amplification factor of 100 .We performed a series of

impact tests on our SSV from various heights and recorded the impacts .The impact plots are

given below

Height of pendulum from ground = 17 cm

Peak value =0.02826 V


3


Peak value =0.101 V


Peak value =0.5093 V


4


Peak value =1.3 V

Impulse Calculations:

The calculations are carried out considering that the time taken by SSV to reach 10 m =

5.4sec. Since the car shows elastic collision .From the impulse equation.

Acceleration of SSV =1.30604 m/s2.

Force on SSV at 5.4 sec =1.567N

FX∆T= M(V1-(-V2))

∆T = M[V1+V2]/F

Velocity of the ball at 5.4 sec.

INTIAL

Height

Height of

pendulum ΔH cm Voltage (V)

sensitivity

(mV/kN) Amp Force (N) Energy

16 17 1 0.028 56.2 100 4.9822064 1.20785

16 20 4 0.101 56.2 100 17.97153 1.421

16 24 8 0.3 56.2 100 53.380783 1.7052

16 33 17 0.51 56.2 100 90.747331 2.34465

16 40 24 1.3 56.2 100 231.31673 2.842


5

∆T = 0.73[0+5.411]/1.56= 2.52sec.

Impulse on SSV = Issv= (mv-mu)/∆T =1.2(0-5.41129)=-6.4935 N.s

Impulse on Ball =I+ball+=0.73x5.41129=3.95N.s

∆I(Change in Impulsessv-ball)= 6.4935-3.95= 2.5435 N.s

Based on the calculation, the SSV was able to survive the impact without any

deformation .However during the test the car was not comprising of a fully functional gear

box and motor to drive the vehicle ,since the absorbed impacts were significantly small as the

car was free to move backward , the impact test minor changes in our design ,addition of

guide wheels to make the motion if the car linear to facilitate a good collision with the ball

and reach a good acceleration on track.

2 Sankey Diagrams.

Sankey Diagrams are used to analyze the energy flow through SSV by taking into

consideration of all the possible causes of loss and gain of power by every component of the

vehicle construction which includes the solar panel, DC motor, gear and transmission losses,

rolling resistance and drag losses.

Case 1 : When the SSV is running on top speed .

Solar panel :-

Light intensity = 800 w/m2

Area of the solar cell = 39mmx78mmx16=48672x10-6

m2= 0.048672m

2.

Psolar-> 800 x 0.048672 => 38.9W.

Wwheel = V/R => 45rad/sec(angular velocity).

Wmotor= 405 rad/sec

Eback emf => Kex Wmotor= 0.0085 x 405 = 3.443V

From Ia= Isc- Is(eu/MnUr

-1 ) = 0.88 A

Ra= 3.32 Ω ; Ua= RaIa= 6.3646 x 0.88 = 5.600W

Heat loss = P solar- Psolar panel = 38.9-5.6 = 33.3 W

Efficiency of the solar panel η = 5.6/38.9*100=14.5 %.


6

Power loss due to internal resistance of the Motor :-

Ia2Ra = 0.88

2 x 3.32 = 2.571 W

Pmotor = Psolar panel- Pinternal resistance = 5.6-2.57 = 3.03 W

Efficiency of motor => Pmotor / Psolar panel = 3.03/5.6=0.54 => 54 %

ηmotor = 54%

Power loss due to gear transmission :-

We assume the efficiency of the gears to be 90%

Power loss due to gear transmission between motor and wheel

=>0.1x Pmotor = 0.1*3.03 = 0.303 W

Pwheel => Pmotor x 90% = 0.9x3.03= 2.727W

Drag Losses:-

Fw= ½ x Cwx A x ρ x V3 = 0.5 x 0.5 0.02 m

2 x 1.293 x (1.8)

3 = 0.037 N

Pair= Fwx V = 0.037 x 1.8 = 0.0678 W

Rolling friction:-

Fw= Crrmg = 0.012 x1.2 x9.8 = 0.14112 N

Prolling = Fwx V => 0.14112 N x 1.8 = 0.254 W

Pgain = 2.727-0.254-0.0678 = 2.40525 W

Sankey Diagram For Full speed at Infinity:-

Solar Power

38.9 W

Heat Loss and reflection

33.3 W

Solar Panel power

5.6 W (14.3 %)

Internal resistance loss

2.571W

Motor Power

3.03 W(54%)

Gear loss 0.303 W

Pwheel

(90%)

2.727W

Air resistance 0.0678 W

Gain 2.4052W

Rolling friction (0.24 W)


7

Case 2 : SSV at Half Speed :

Solar panel :-

Light intensity = 800 w/m2

Area of the solar cell = 39mmx78mmx16=48672x10-6

m2= 0.048672m

2.

Psolar-> 800 x 0.048672 => 38.9W.

Wwheel = V/R => 22.5rad/sec(angular velocity).

Wmotor= 202.5 rad/sec

Eback emf => Kex Wmotor= 0.0085 x 405 = 1.7212V

From Ia= Isc- Is(eu/MnUr

-1 ) = 0.88 A

Ra= 3.32 Ω ; Ua= RaIa= 3.32 x 0.88 +1.721= 4.6428W

Heat loss = P solar- Psolar panel = 38.9-4.08 = 34.8.3 W

Efficiency of the solar panel η = 4.8/34.8x100=10.5 %.

Power loss due to internal resistance of the Motor :-

Ia2Ra = 0.88

2 x 3.32 = 2.571 W

Pmotor = Psolar panel- Pinternal resistance = 4.085 -2.571 = 1.514 W

Efficiency of motor => Pmotor / Psolar panel = 3.03/5.6=0.54 => 37 %

ηmotor = 37%

Power loss due to gear transmission :-

We assume the efficiency of the gears to be 90%

Power loss due to gear transmission between motor and wheel

=>0.1x Pmotor = 0.514 W

Pwheel => Pmotor x 90% = 0.9x3.03= 1.3626W

Air Drag Losses:-

Fw= ½ x Cwx A x ρ x V3 = 0.5 x 0.5 0.02 m

2 x 1.293 x (0.9)

3 = 0.00471N

Pair= Fwx V = 0.037 x 0.9 = 0.004241W

Rolling friction:-

Fw= Crrmg = 0.012 x1.2 x9.8 = 0.14112 N

Prolling = Fwx V => 0.14112 N x 1.8 = 0.1270W


8

Pgain = 1.36-0.00471-0.127 = 1.2280 W

Sankey Diagram 2:

3 Strength Analysis and Calculations.

This sheet is gotten from the impact test.

As we analysis that the force on Area B( ) is Normal force and the force on Area A( ) is

Shear force.

INTIAL

Height

Height of

pendulum ΔH cm Voltage (V)

sensitivity

(mV/kN) Amp Force (N) Energy

16 17 1 0.028 56.2 100 4.9822064 1.20785

16 20 4 0.101 56.2 100 17.97153 1.421

16 24 8 0.3 56.2 100 53.380783 1.7052

16 33 17 0.51 56.2 100 90.747331 2.34465

16 40 24 1.3 56.2 100 231.31673 2.842


9

The shear force and normal force of our solar car is

,

The normal force on B can be calculated by the force what we have got from the table. What

we should do now is to find the left element in the formula. Fist we try to find the Velocity of

the car.

From first question (1) ∆T= 0.73[0+5.411]/1.56= 2.52sec. . For our

solar car Now, all we

need to calculate the stress are known, we can easily get the stress value:

H/cm Δh/cm U/V F/N v/ 1 sm τ/MPa σ/MPa

1 17 1 0.028 4.98 2.01 0.00022 0.0249

2 20 4 0.101 17.97 2.36 0.00026 0.08985

3 24 8 0.3 53.38 2.85 0.000317 0.2669

4 33 17 0.51 90.75 3.90 0.000434 0.4575

5 44 24 1.3 231.32 4.74 0.000527 1.1566

According to the reference, we find the compressive strength is 10MPa.

(http://www.makeitfrom.com/material-data/?for=Medium-Density-Fiberboard-MDF)

And the maximum shear stress: 80MPa (We find it from text book.)

10MPa >> 1.156 and 80MPa >> 0.000527. So our car can stand the force when the height is

44cm at least.

Strength calculation and analysis before impact.

(1) For the gear

With I=0.90A

Torque constant=8.66mNm/A

Max efficiency=84%

http://www.makeitfrom.com/material-data/?for=Medium-Density-Fiberboard-MDF


10

Gear ratio i=9

The torque made by motor T=8.55*84%*0.90*9*10-3

N·m =0.0581742 N·m

We used 4 gears for the transmission system, and we can get that T=T1=T2.

According to the radius of the biggest gear 0.03m, we can get the force translated from

motor to the shaft.

F=T/r=0.0581742 N·m/0.03m=1.93914N

(2)The mechanical load on the shaft

W=mcar*g=1.3*9.81=12.753N

NC+ND=W=12.753N

We assume NA=NB.

0165.0106.0055.00

00

BDAC

DCBAy

NNNM

NNNNF

NA=6.3765N , NB =6.3765N ,NC =6.1359N , ND=6.6171N

The force distribution on the shaft:


11

The maximum bending moment:

I

RMMAX

·max ,where I= 4

4

1R , R=0.002m

Mmax=0.3762N/m

I

RMMAX

·max =59.9MPa

The torque:

T=TA+TB

0// BGAG


12

0112.0108.0

P

B

P

AIG

TIG

T

TA=0.02961 , TB=0.02856

The maximum torsion stress

PI

RT max

max

Where Ip=4

2

1R , R=0.2cm, T=0.02961N

max 2.357MPa

Shear force

0220.0108.00

00

ba

bax

FFM

FFFF

Fa=0.9872N, Fb=0.95194N


13

The maximum shear stress:

tI

QVMAX

max

Where RtRIRQ 2,4

1,

3

2 43

2

max

4

3

max

3

4

24

13

2

R

V

RR

RV

MAX

0.104798MPa


14

4 Technical Drawing.


15

5 Collisions Situations.

(a)

After the collision the speed of B and C is:

(b)

When , the potential energy of the spring reaches its maximum.

(c) The maximum potential energy of the spring is

(d) The object A can reverse the direction, and the maximum velocity of this direction is:


16

6 Others.

Material Cost

Time Cost

Electromechanical Engineering 4 Light Angle · 2018. 1. 8. · Electromechanical Engineering 4 Team...

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