# Water flow rate generated by a pressure steam tank

I have a 5 $$m^3$$ capacity tank that generates steam with a heat exchanger. This tank is under 15 bar pressure. The water in the bottom part of the tank does not completely vaporize because there is always some fresh water filling it.

Under this bottom part, there is a purge valve connected to a 50 mm diameter pipe.

I would like to know how to estimate the water flow rate going though this purge pipe when the valve is open on a very short period of time (few seconds).

I tried to calculate the water velocity in the pipe with Bernoulli's equation but the numerical application gives me a really high value, which makes me think I made a mistake somewhere, or that this formula cannot be used for this installation:

$$V_{tank} = 0\,m/s\\ h_{tank} = 0,5\,m\\ P_{tank} = 1.500.000\,Pa\\ h_{pipe} = 0,5\,m\\ P_{pipe} = 0\,Pa\\ \rho = 866,89\,kg/m3$$

$$\frac12 \rho V_{tank}^2 + \rho g h_{tank} + P_{tank} = \frac12 \rho V_{pipe}^2 + \rho g h_{pipe} + P_{pipe}$$

$$V_{pipe}=\sqrt{\frac{2P_{tank}}\rho} = 58,3\,m/s$$

And, I would also like to know how to find the water flow rate over time while estimating the pressure drop?

Any help would be much appreciated

• Are you sure the pressure is 15 MPa? that's about 150 atm? For a 50 mm diameter pipe I would expect the speed to be high. Sep 26 at 18:16
• The pressure drop is 15 atmospheres plus height of the water. Using bernouli means there is no pressure drop through the valve and pipe. if you want the head loss through the pipe you need to use head loss calcs (or equivalent feet for the vale and use tables). Sep 26 at 18:40
• I'm sorry, it's 1.5 MPa, not 15. I did not make the mistake when calculating though. The pipe in itself is pretty short so I think that we can pretty much neglect the head loss. I actually have the head loss value of the valve, but I just wanted to expose here the "simplified" version of my reasoning to see if it was correct Sep 26 at 19:44

If I am not seeing it wrong you have a tank that has a head of

$$H_{vapor}= 15bars*10.2m/bar=153m$$

$$Q=C_d A\sqrt2gh$$

where

• Q = flow (cubic metres per second)

• $$C_{d} = coefficient\ of\ discharge$$

• A = area of orifice (square metres)

• g = acceleration from gravity (9.81 m/s^2)

• h = head acting on the centreline (m)

• $$A = \pi/4(D^2)$$

Assuming the discharge opening doesn't have sharp edges, we pic Cd=0.8

$$Q =\frac{0.8\pi}{4} (D^2)\sqrt{2gh}$$

After one second we have expanded the volume of 5m^3 by Qm^3. if we don't want to solve the dv/dt differential equation and ignore the friction of the 50mm pipe,

$$P_1V_1=P_2V_2= P_2(V_1+Q)$$

After one second if we assume the expanded volume V2 is 5m^3 + Q which is a reasonable estimate,

$$P_2=\frac{15bars*5m^3}{(5+Q)m^3}= 15bars \frac{5}{5+Q}m^3$$

• Thank you sir for your help Sep 28 at 16:34