Question 1.6.6: Orifice Flow with Constrained Exponent Consider the data of ...
Orifice Flow with Constrained Exponent
Consider the data of Example 1.6.5. Determine the best-fit value of the coefficient b in the square-root function
f = b h^{1/2}| Height h (cm) | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
| Time t (s) | 7 | 7.5 | 8 | 8.5 | 9 | 9.5 | 11 | 12 | 14 | 19 | 26 |
First obtain the flow rate data in ml/s by dividing the 250 ml volume by the time to fill:
f = \frac{250}{t}
Referring to Example 1.5.3, whose model is y = b x^{m}, we see here that y = f , h = x, m = 0.5, and a = b. From Equation (1) of that example,
b = \frac{\sum^{n}_{i=1} {x^{m}_{i} y_{i}}}{\sum^{n}_{i=1} {x^{2m}_{i}}} (1)
a = \frac{\sum^{n}_{i=1} {h^{0.5}_{i} y_{i}}}{\sum^{n}_{i=1} {h_{i}}} (1)
The MATLAB program to carry out these calculations is shown next.
h = (1:11);
time = [26, 19, 14, 12, 11, 9.5, 9, 8.5, 8, 7.5, 7];
flow = 250./time;
a = sum(sqrt(h).*flow)/sum(h)
f = a*sqrt(h);
J = sum((f - flow).^2)
S = sum((flow - mean(flow)).^2)
r2 = 1 - J/S
The result is a =10.4604 and the flow model is f =10.4604 \sqrt{h}. The quality-of-fit values are J = 5.5495, S = 698.2203, and r² = 0.9921, which indicates a very good fit.