The empirical coefficient of discharge?

Question

In the lab you examined two types of flow measurement structure a thin plate weir and a venturi meter. Below is a set of data colleced from an experiment carried out with a venturi meter. The table contains data on the difference in water level(H) in the piezometer tubes located at the entry to the venturi meter and its narrowed section and the measured volumetric flow rate (Q). The diameter of the venture meter was 60mm at its entry and its diameter was 25mm at its narrowest section.Given the data below estimate the empirical coefficient of discharge.

H(mm) - Q(I/s)

50 - 0.48

100 - 0.57

150 - 0.83

200 - 0.95

250 - 1.06

300 - 1.17

350 - 1.25

Answer 1

The theoretical formula for discharge is:

Q = c Av

Where:

Q = the rate of discharge

c = coefficient of discharge

A = the area of the discharge opening

v = the velocity of discharge

However the velocity of dicharge is related to the the pressure head by:

v =(2gH)^(1/2), hence the discharge can be exprssed in terms of h as:

Q = c A sqrt(2gH), therefore;

c = Q/(A sqrt(2gH))

This means that c is proportional to Q/sqrtH.

The next step is to check  which of the sets of data does not fall within normal variations by dividing the value of Q by the sqrt of H. Thus:

H  ___    Q  _____  Q/sqrtH

50 ___0.48 _____  0.0679

100__ 0.57 _____  0.057

150 __0.83  _____ 0.0678

200 __0.95  _____ 0.0672

250 __1.06  _____ 0.0670

300 __1.17  _____ 0.0675

350 __1.25  _____ 0.0668

The above shows that the set of data at H = 100mm is outside of the norm and may be discarded.

The mean value of Q/sqrtH may be obtained by linear regression. But since the tablated values except the discarded one, are almost the same then the average of this values may be used. Hence, Q/sqrtH = 0.0674

For H = 1000mm (1meter) Q = 0.0674sqrt1000 = 2.131 l/sec

=0.002131 cu.m /sec  

The area of the o*****e is:

A = pi((25/1000)^2)/4 in sq.m

Thus the coefficient of discharge is;

c = Q/(Asqrt( 2gH))

= 0.002131/(pi(0.025)^2)/4)sqrt(2x9.81x1))

c= 0.98

Answer 2

I get it to be around 0.97, and there's something wrong with the 2nd data point, it gives an answer of 0.81.  0.68l/s instead of 0.57 gives a more consistent discharge coefficient.

If you combine Au = constant with p+0.5*rho*u^2 = constant you can get an equation for flowrate q = Au in terms of pressure difference, pipe areas and discharge coefficient.  

q^2 = 2*g*h*C^2*(A1^2*A2^2)/(A1^2-A2^2)

g = 9.81, h = height of water, metres, C = discharge coeffiicent and A1 and A2 are areas at 60mm and 25mm.