TAM 335 Lab Report 4 PDF

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TAM 335 Lab Report Lab Title:

Elementary Laboratary Procedure

Student Name : Rong Jin

Lab Section: ABJ

Objective: To measure the resultant force applied by a curved rectangular duct to the air flowing within the duct, and to determine if a control-volume analysis based on the linearmomentum principle is valid. And to determine whether the tangential velocity of the air flow in the bend can be modeled by a simple idealization.

Sample Calculation: Choose flow rate #1, box pressure = 8.3mb, upstream pressure = 1.81mb

Known: Pouter3=2.04mb, Pouter4=3.33mb, Pinner3=-2.43mb, Pinner4=-5.69mb, douter3=0.05m, douter4=0.0893m, dinner3=0.05m, dinner4=0.0696m, θouter 3=180 ° , θouter 4 =202.5° , θinner 3=0 ° , θinner 4=22.5° , b=0.1m, r2=0.1m, r1= 0.05m, Pbox = 8.3mb, P0= 1.81mb

Use integration method:

P ¿ P P ¿ P ( θinner 1) +(¿ ¿ outer 2∗cos ( θ outer2 ) )∗∆ d inner /2 ¿¿ (¿ ¿ outer 2∗cos ( θ outer2 ) )∗¿ ∆ d outer /2−b ¿ (θouter 1 )+¿ ¿¿ F x =−b ∫ Pouter n x ds−b ∫ Pinner nx ds=−b ¿

o s ( 1 8 0 ) +3. 3 3 * c os ( 2 0 2 . 5 ) ) * ( 0 . 0 8 9 3 0 . 0 5 ) * 1 0 0 / 2-0 . 1 *( 2 . 4 3 * c os ( 0 ) =c 5 . 6 9 * c o s ( 2 2 . 5 ) ) * ( 0. 0 6 9 6 0 . 0 5 ) * 1 0 0 / 2 =1 . 0 0 5 4 +0. 7 53 3 =1 . 7 5 8 7N P ¿ P P ¿ P outer 2∗sin ¿ θ +(¿ ( θouter 2) )∗∆ d inner /2 ( inner1) ¿¿ (¿ ¿ outer 2∗sin ( θouter 2) )∗¿ ∆ d outer /2−b ¿ ( θ outer1 ) +¿ ¿¿ F y =−b ∫ Pouter n y ds−b∫ P inner n y ds =−b ¿ = s i n ( 1 8 0) +3 . 3 3 * s i n( 2 02 . 5 ) ) * ( 0. 0 8 9 30 . 0 5 ) * 1 00 / 2-0 . 1 * ( 2 . 43 * s i n ( 0 ) 5 . 6 9 * s i n ( 2 2 . 5 ) ) * ( 0 . 0 6 9 6 0 . 0 5 ) * 1 0 0 / 2 =0 . 2 50 4 +0 . 2 1 3 4 =0 . 4 63 8N

Si n c et h e r ea r et ot a l l yt e ns e t so fd a t a ,t h et o t a lFxa n dFye q u a l st ot h es u mma t i o no f a l lt h eFxa ndFyc a l c u l a t e di ne a c hs e t . ∑ F x =6.89 N

∑ F y =7.39 N

Us eCo nt r o lVo l u mea n a l y s i s : F x =2∗ ( Pbox−P 0 )∗ A F y = ( 2∗P box−P0 )∗ A

A=( r 2 r 1 ) * b=0 . 0 0 5 m2 F x =2∗ ( 8.3−1.81)∗A∗100=6.49 N F y = ( 2∗8.3 −1.81 )∗A∗100=7.39 N

Fr e e–v o r t e xMo d e l : Cp ( r )=1−

[ ( )]

2

r 2−r 1 ∗1 /r 2 r2 ln r1

Ta ki n gd a t ao fr =5 5mm

[ ( )]

2

Cp ( r )=1−

0.1 −0.05 2 ∗1/0.055 0.1 ln 0.05 =0 . 7 2

Da ta Collection & Analysis: 1. Table-1. Force found by intergration Flow rate #1

Flow rate #2

Flow rate #3

Fx (N)

6.89

4.43

1.62

Fy (N)

8.22

5.25

1.94

Flow rate #1

Flow rate #2

Flow rate #3

Fx (N)

6.49

4.21

1.57

Fy (N)

7.39

4.79

1.80

2. Table-2. Force found by volume analysis

3. Table-3. Coefficient of pressure Rad Dist (mm)

Cp (Exp)

Cp (Free Vortex)

Cp (Solid)

55

-0.835130971

-0.720139654

0.462222222

60

-0.6394453

-0.445395126

0.36

65

-0.380585516

-0.231579279

0.248888889

70

-0.232665639

-0.061922949

0.128888889

75

-0.115562404

0.07494712

0

80

0

0.186965242

-0.137777778

85

0.103235747

0.279803121

-0.284444444

90

0.218798151

0.357602166

-0.44

95

0.285053929

0.423443496

-0.604444444

4. Figure-1. Fy, Fx plot

Obj ect31

5. Figure-2. Coefficient of pressure vs Radius

Obj ect34

Conclusion: From figure 1, the (fx,fy) calculated by integration and volume analysis methods has linear relationship. The line generated by volume analysis method is slightly offset from the line generated by integration method. It is because when using volume analysis method, it assumes the flow in the bend is steady and two dimensional and ignores the secondary flow effect and separation effects.

For figure 2, after connecting all the experimental data points, it is clear that the line formed by experimental data has similar shape as the free vortex model. And it is totally different from the solid model. Therefore, the free vortex model is valid.

Questions: Q2. Tatm = 24 °C ; Patm =998mb= 99.8 kPa

Assuming ideal gas: PV=MRT R=8.314/M(air)=8.314/(28.97*10^-3)=287 kg/K Rho(air) = M/V =P/(R*T) = 99.8*10^3/(287*(24+273))= 1.17 kg/m^3 When flow rate #1 U01 =

√

2 ( Pbox−P 0 ) r h o ( air )

=

√

2∗ (8.3−1.81 ) ∗100 m =33.3 s 1.17

=

√

2∗ (5.37−1.16 )∗100 m =26.83 s 1.17

=

√

2∗ (2.02−0.45 )∗100 m =16.38 1.17 s

When flow rate #2 U02 =

√

2 ( Pbox−P 0 ) r h o ( air )

When flow rate #3 U03 =

√

2 ( Pbox−P 0 ) r h o ( air )

Q6. The free vortex model suggests that the velocity distribution in the curved portion of the duct is not steady and two dimensional, instead the distribution is more like a free vortex located at the center of the of the curvature of the bend. Therefore, the secondary flow effect and separation effects are accounted in the free vortex model. Since free vortex model is valid, the approximation used in control volume analysis is not legitimate.

Q7. As the flow rate decreases the models of air flow become more accurate.

Extra Q. Flow rate Q = A*U0 A = cross-section area of the duct = (r2-r2)*b = 50*100*10^-6 = 0.005 m2 Q1 = A*U01= 0.005*33.3 = 0.1665 m3/s Q2 = A* U02=0.005*26.83 = 0.13415 m3/s

Q3 = A* U03=0.005*16.38 = 0.0819 m3/s

F=

√ Fx 2 + Fy2

Force by integration: F1= F2= F3=

√ Fx 2 + Fy2 √ Fx 2 + Fy2 √ Fx 2 + Fy2

= = =

√ 6.892+ 8.222=10.73 N √ 4.432 + 5.252=6.87 N √ 1.622 +1.94 2=2.53 N

Force by volume control analysis: F4= F5= F6=

√ Fx 2 + Fy2 √ Fx 2 + Fy2 √ Fx 2 + Fy2

= = =

√ 6.492+ 7.392=9.84 N √ 4.212 + 4.792 =6.38 N √ 1.572+ 1.802=2.39 N

Relative errors: R1 =

F 1−F 4 10.73−9.84 =0.083 = 10.73 F1

R2 =

F 2−F 5 6.87 −6.38 = =0.0713 F2 6.87

R3 =

F 3−F 6 2.53 −2.39 =0.0553 = 2.53 F3

Figure-3. Force vs Flow Rate

Force vs Q 12

Force (N)

10 8 integration control volume analysis

6 4 2 0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

flow rate (m^3s)

Figure-4. Relative error vs Flow Rate

Relative Error vs Flow rate 0.09 0.08

Relative error

0.07 0.06 0.05 0.04 0.03 0.02 0.01 0

0

0.1

0.2

0.3

0.4

0.5

flow rate (m^3/s)

0.6

0.7

0.8

0.9...