A IS-cup coffee pot (see Figure 5.5-6) was placed under a water faucet and filled to the IS-cup line. With the outlet valve open. the faucet’s flow rate was adjusted until the water level remained constant at 15 cups, and the time for one cup to flow out of the pot was measured. This experiment was repeated with the pot filled to the various levels shown in
the following table:
(a) Use the preceding data to verify Torricelli’s principle for the co
t? obtain a relation between the flow rate and the number of cups in the pol. manufacturer wants to make a 36-cup pot using the same outlet valve but i C OE:n that a cup will fill too quickly, causing spills. Extrapolate the relation develop and predict how long it will take to fill one cup when the pot contains 36 cup .
(a) Torricelli’s principle in equation form is f = rVI/2, where f is the flow the outlet valve in cups per second, V is the volume of liquid in the pot in cu a constant whose value is to be found. We see that this relation is a power fun the exponent is 0.5. Thus if we plot 10g10 (J) versus 10g 10 (V), we should obtain line. The values for f are obtained from the reciprocals of the given data for t, f = 1/ t cups per second.
The MATLAB script file follows. The resulting plots appear in Figure 5. volume data is entered in the array cups, and the time data is entered in meas-times
% Data for the problem.
cups = [6,9,12,15];
meas_times = [9,8,7,6];
meas_flow = l./meas_times;
% Fit a straight line to the transformed data.
p = polyfit(10g10(cups) ,log10(meas_flow) ,1);
coeffs = [p(l), 10Ap(2)];
m coeffs (1)
b = coeffs(2)
% Plot the data and the fitted line on a loglog pl~t ~8 r
% how well the line fits the data .
.x = [6:0.01:40);
y = b*x.Am;
log log(x,y,cups,meas_flow, ‘0’) ,grid,x label(‘Volume (cups)’
y label(‘Flow Rate (cups/sec)’),axis([5 15 0.1 0.3))
The computed values are In = 0.433 and b = 0.0499, and our derived relation i = 0.0499Vo.433
• Because the exponent is 0.433, not 0.5, our model does not agree
with Torricelli’s principle, but it is close. Note that the first plot in Figure 5.5-7 shows that the data points do not lie exactly on the fitted straight line. In this application it is difficult to measure the time to fill one cup with an accuracy greater than an integer
second, so this inaccuracy could have caused our result to disagree with that predicted by Torricelli. (b) Note that the fill time is II!,the reciprocal of the flow rate. The remainder of the MATLAB script uses the derived flow rate relation f = O.0499V°.433 to plot the
extrapolated fill-time curve 1/ f versus
1.Plot the fill time curve extrapolated to 36 cups.
subplot(2,1,2) plot(x,l./y,cups,meas_times, ‘o’),grid,x label(‘Volume(cups) ‘) y label(‘Fill Time per Cup (see) ‘),axis([5 36 0 10)) % % Compute the fill time for V = 36 cups. V = 36; ‘f_36 = b *VA m The predicted fill time for one cup is 4.2 sec. The manufacturer must now decide if this time is sufficient for the user to avoid overfilling. (In fact, the manufacturer’ did construct a 36-cup pot, and the fill time is approximately 4 see, which agrees with our