Resulting discharge coefficients with accepted values

EM 341 DISCHARGE OVER WEIRS# 7
OBIECTIVE: To obtain empirical equations for the flow of water over sharp edged weirs
rectangular and v-shaped, and to compare resulting discharge coefficients with
accepted values.
RELATIONSHIP AND THEORY
Empirical formula: Q kH”
Rectangular theoretical: Q= (2/3 2g b)H
actual: Q=(C, 2/3 /2g b)H32
30 Vee theoretical: 0= 8/15 2g tanH2
-C, 8/15 2g tan& actual:
H=height at surface upstream where velocity is negligible, above the crest of the weir.
DATA: Rectangular width, b = 30 mm (1/8 ft); Vee half angle = 15.1°
VALUES OF Co
Weir From The From The From This
Literature Vendor Experiment
Rectangular 68 0.59
30 Vee 64 0.62
REQUIREMENTS:
LAB: Starting with the largest flow rate possible, make runs at AH = 4 to 8 mm (0.02 to 0.04 ft)
intervals until the flow dribbles, recording H + Ho, volume and time.
Record initial and final temperatures. Check y with instructor.
33
34 I FLUID MECHANICS LAB
REPORT:
For each weir,
Plot log Q vs. log H. Force slope to the theoretical value.
Write the empirical equation.
Evaluate Co.
State which of the three Co’s you consider the most credible and why.
Flow over Weirs
H
Datum
Ho
Rectangular weir:
dO=V.dA, dA =b.dh (A)
Applying Bernoulli Equation between (1) & (2)
(B) *2g 2g
Assuming
P P Pam Z H, z, =H–h, v, =0(upstream)
From equation (B):
=0
P+ PumH +-P + (H-h) + 2g 2g du
=2gh
H-b
=H
EM 341 DISCHARGE OVER WEIRS#7I 35
From equation (A):
do=V.dA = 2gh.b. dh
Iriangular weir
H-h -(H-h tan
Total width of element: 2x=2(H-h) tan|
do=V.dd =2gh.2(H-h) tan
J0
8 2g.tanH
Coefficient of discharge: C, = H
Qtheo H-h
SLOPE = 3/2
sLOPE = 5/2
EM 341 DISCHARGE OVER WEIRS: 7137
Pipework Energy Losses
OBJECTIVE: To determine the loss coeficients of various fittings in pipe flow and to compare
these values with the accepted values.
THEORY: Head losses in various pipe fittings are known to be of the form:
h Kv/(2g)
where K is the loss coefficient of a particular fitting.
5.1 Dimensions of pipes and fittings
Diameter of smaller bore pipe D 22.5 mmA, =3.97×10 m
Diameter of larger bore pipe D =29.6 mmA, = 6.88×10 m?
Radius to centre line of elbow R =14 mm
Radius to centre line of bend R, = 45 mm
5.2 Diferential Piezometer readings and loss of total head
If the measured flow rate is Q (1/s), then the velocities vi and vn along the pipes of cross-section
areas Ai and Az (m?), are:
=10 IA4 (m/s) and V, =10*Ql4, (m/s)
The laboratory readings are recorded as follows in Table (note that the reading for the enlargement
is negative):
Differential Piezometer Reading (mm)
Volume Time (s) Q (/s) Mitre 1 Elbow Enlargement Contraction Bend 9
(liters) 2 3-4 5-6 7-8 10
38 1 1LUID MECILANICSLAB
Calculation of loss coefficients K
Loss of Total Head (mm)
(litres/s) (m/s) (m/s) itre| Elb. Volume Vi V2 Mitre Elbow Enlargement Contraction Bend
(1-2) 34 2g 5-6 7-8 9-10 2g
(mm) (mm)
EM 341 DISCHARGE OVER WEIRS7 1 39
Lar radus
bend Elbcw
Conirac!ion Enlarcemer
Air veu
-tude
Ortrol
vave
Mitre To measuring Ã…
tar Fro stupp’
Report:
Graph loss of total head, for a fitting, vs. V/2g to determineK.
First graph should represent data for Mitre, Elbow & Bend.
Second graph should represent data for Enlargment
Third graph should represent data for Contraction.
Draw a graph of K vs. Re (Re = VD/, assume v = 1xl0 m*/s) for Mitre.
QUESTIONS:
How would you scale graphs if you were to use log log graph sheets? How would you
determine K?
1.
There is, for example, a difference in the elevations of stations 3 and 4. Did you take into
consideration this difference? How?
N
09 JQ
8
ealTTT
|
EM 341 DISCHARGE OVER WEIRS 7 |41
.
1 Ka 0.35 acc
900 smooth 2 0.19 (15)
bend
4 0.16
5 0.21
8 0.28
10 0.32
Threads: 22? 2?? 22?
Fire
Fittings
Approval by permissiorn of the American Society of Hearing. But ??2
Uncertainty in Experimental Measurements
Now an example of uncertainty:
P E.I E =100+2 Volts I=10+0.2 Amp
Nominal value of power:
= 100×10 =1000 Watts
Worst possible case, P =98×9.8 =960.4 Watts.
Best possible case, P =102 x10.2 =1040.4 Watts
So, the uncertainty in calculations of Power is +4.04%, -3.46%.
Quite unlikely that power would be in error by the amounts because voltmeter variations
probably could not correspond with amperemeter variations.
This leads to new method of computing uncertainty.
Kline and McClintock developed the following formula, for 20 to 1 odds, for the resulting uncertainty in the variable R:
42 1 FLUID MECIHANICS ILAB
OR
were K =R(%, X2,*,x,) and x, x,,.., x, are independent variables and , W2***, W, are
the uncertainties in the values of , X***, Xn*
First some basic formulas:
It is well established that the scatter in methodical experimental measurements forms a Gaussian
distribution called “Normal Distribution”. In such a case for the collected data, it means 95.4% of
the data has a value a value of xt20. Here, Xm is the mean value and o represents standard
deviation:
2-x where x, = x amdn = totality of data points.
30 X-2a x- p 20 X+3o
Gaussian of Normal error distribution
Pr)=
p(xm) is called a measure of precision of data.
The larger value of a will give a larger spread.
EM 341 DISCHARGE OVER WEIRS:7 | 43
O = 0.683 2a =0.954 3a=0.997
Large number of experimental data follow Gaussian Distribution.
References:
Experimental Methods for Engineers by J. P. Holman, p-47
Uncertainty in results of Herschel Venturi
c ep volxtlx
volltime 4.x(h-h) 2
A2g(h-h) 428
C=C(vol =V, t, h, h,)
4 2g
-A)
3
–4)0–coh-h)
OCV 1
Oh, J2 42g Ay-D-cn-A)”
Uncertainties for odds twenty to one:
-| JJJ W
a-Ar’) {G=)-Ga-tr) w.
Assume
44 t0.74 sec w, = 0.74 sec
H
&


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