# Homework Help: Interpreting negative pressure in pipe fluid flow

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1. Sep 19, 2016

### CivilSigma

1. The problem statement, all variables and given/known data
(This is more of a discussion question)

My question is, if we have a pipe between two reservoir A and reservoir B (Height of A > height of B) , then if flow in the pipe is to happen from A to B, what sign would the pressure sign be at any point in the pipe?

2. Relevant equations
Energy Equation:

$$\frac{P_1}{\gamma}+h_1+\frac{v_1^2}{2g} = \frac{P_2}{\gamma}+h_2+\frac{v_2^2}{2g} + h_f +h_L$$

3. The attempt at a solution
I think it makes sense that the pressure at any point in the pipe would be negative - because if we analyse the absolute pressures, we have atmospheric pressure pushing down on reservoir A and if the pressure along the pipe is negative, then $$P_{absolute, pipe} < P_{atm}$$ which implies flow from reservoir A to B.

Can someone please provide me with some discussion / their point of view?

Thank you.

2. Sep 19, 2016

### billy_joule

The lowest possible absolute pressure is zero, a perfect vacuum. Where no matter is present.
Draw a diagram and apply Pascals law.
All points in the water have higher static pressure than atmospheric.

No it doesn't.

3. Sep 20, 2016

### Staff: Mentor

You really haven't provided enough information, sakonpure6. Is the pressure at the entrance to the pipe and at the exit of the pipe atmospheric? Or, where is the outlet from tank A and the inlet from tank B situated?

If the pressure at the entrance of the pipe and exit of the pipe is atmospheric, then, at steady state, the pressure along the length of the pipe is atmospheric.

4. Sep 20, 2016

### CivilSigma

Hi Chester,

Here is the problem in its entirety and my solution:
http://imgur.com/ojWjvFB

Calculating the discharge,
first apply the energy balance between points 1 and 3 to get:

$$h_1-h_2 = h_f + h_L$$

From the question we know :
$$h_{f \, 1 \to 3} = h_{f \, 1 \to 2} + h_{f \, 2 \to 3}=1.5 m + 2.4 m = 3.9 m$$
Also,
$$h_L = \sum (K) \frac{v^2}{2g}$$
Where the coefficient K is the dimensionless factor. For sudden contraction at entrance, K=0.5 and for sudden expansion at exit K=1.
$$5m = \frac{(K_{entrance} + K_{exit} ) \cdot v^2}{2g} + 3.9m$$
$$v = \sqrt{ \frac{(5-3.9)\cdot 2 \cdot 9.81}{ (0.5+1)}}$$
$$v=3.79m/s$$

As a result,
$$Q=VA=\pi (0.05m)^2/4 \cdot 3.79 m/s = 0.00744 m^3/s = 7.44 L/s$$

Finding the Pressure at Point #2
Apply the energy equation from point 1 to 2 (Datum at point 1) to get:

$$h_1-h_2 = \frac{P_2}{\gamma_{oil}}+v^2/2g+h_f+h_L$$

Similar to above calculations for hf and hL,

$$-2m = \frac{P_2}{\gamma_w}+\frac{(3.79 m/s)^2}{2 \cdot 9.81 m/s^2} + 1.5m + 0.5 \frac{ (3.79 m/s)^2}{2\cdot 9.81 m/s^2}$$
$$P_2 = -4.598 m \cdot 820 kg/m^3 \cdot 9.81 m/s^2 \approx -37 kPa$$

So, what do you think about obtaining negative pressure at point 2 ? Did i make a mistake in my calculations?

Thank you.

5. Sep 20, 2016

### billy_joule

I haven't checked your math but you should expect a negative gauge pressure at the top of any siphon.
It is still a positive absolute pressure.

6. Sep 20, 2016

### CivilSigma

Why would that be the case? I don't really see why it would physically be negative. Could you please explain?

7. Sep 20, 2016

### billy_joule

It's only negative relative to the siphon entrance, it's still positive relative to a vacuum.

Pressure increases with increasing depth (ΔP = ρgΔh), the top of the siphon is higher than the entrance so the pressure is lower.