# Archived Fluid Dynamics final one

#### momogiri

Water flows steadily from an open tank as in the figure below. The elevation of point 1 is 10.0m, and the elevation of points 2 and 3 is 2.00m. The cross-sectional area at point 2 is 0.0480m^2; at point 3 it is 0.0160m^2. The area of the tank is very large compared with the cross-sectional area of the pipe.

Part A is "Assuming that Bernoulli's equation applies, compute the discharge rate in cubic meters per second."
Which I've solved using Bernoulli's principle

so Q_3 = 0.200m^3/s

Part B is "What is the gauge pressure at point 2?"
Now this one I'm stuck with, but here's what I've done:

A_2*v_2 = A_3*v_3
(0.048)v_2 = (0.2)(0.016)

v_2 = 0.0666667m/s

so using Bernoulli's principle..(and cancelling some stuff)

P_2 + 0.5*rho*v_2^2 = P_3 + 0.5*rho*v_3^2
which becomes
P_2 + (0.5)(1000)(0.0666667)^2 = 101300 + (0.5)(1000)(0.2)^2
thus P_2 = 101317.7778

And I figured since gauge pressure meant P_G = P - atm
I subtracted 101317.7778 by 101300
which got me 17.7778 as my gauge pressure...
So what am I doing wrong? I don't seem to be getting the right answer here...

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#### MexChemE

For part A, we apply Bernoulli and the continuity equation from point 1 to point 3
$$A_1 v_1 = A_3 v_3$$
$$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_3 + \frac{1}{2} \rho v_3^2 + \rho g h_3$$
Since the tank is open, $P_1 = P_3$, so we have
$$v_3^2 - v_1^2 = 2g (h_1-h_3)$$
$$v_3^2 \left(1 - \left( \frac{A_3}{A_1} \right)^2 \right) = 2g (h_1-h_3)$$
Now, since the area of the tank is very large compared to the area of the jet, $1 - \left( \frac{A_3}{A_1} \right)^2 \approx 1$, then
$$v_3 = \sqrt{2g (h_1-h_3)} = \sqrt{2 \left( 9.8 \ \frac{m}{s^2} \right) (10 \ m - 2 \ m)} = 12.52 \ \frac{m}{s}$$
So the flow rate is
$$Q = (0.016 \ m^2) \left(12.52 \ \frac{m}{s} \right) = 0.2 \ \frac{m^3}{s}$$

For part B, we apply Bernoulli and the continuity equation from point 2 to point 3
$$A_2 v_2 = A_3 v_3$$
$$P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 = P_3 + \frac{1}{2} \rho v_3^2 + \rho g h_3$$
Points 2 and 3 have the same elevation, so we have
$$P_2 + \frac{1}{2} \rho v_2^2 = P_3 + \frac{1}{2} \rho v_3^2$$
We can calculate $v_2$ with the continuity equation
$$v_2 = \left( \frac{0.016 \ m^2}{0.048 \ m^2} \right) \left( 12.52 \ \frac{m}{s} \right) = 4.17 \ \frac{m}{s}$$
Now we calculate the pressure at point 2
$$P_2 = P_3 + \frac{1}{2} \rho (v_3^2 - v_2^2) = 101325 \ Pa + \frac{1}{2} \left(1000 \ \frac{kg}{m^3} \right) \left( \left(12.52 \ \frac{m}{s} \right)^2 - \left(4.17 \ \frac{m}{s} \right)^2 \right) = 171006 \ Pa$$
Finally, we calculate the gauge pressure
$$p_2 = P_2 - p_{atm} = 171006 \ Pa - 101325 \ Pa = 70 \ kPa$$

"Fluid Dynamics final one"

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