Bernoulli Equation — Discharging water from a partially-filled sealed tank

In summary, the conversation involved discussing a past exam question about an enclosed hot water tank with a 0.2 bar g pressure in the head space and a 50mm diameter outlet. The velocity in the tank was assumed to be negligible and the hydrostatic head was 4m. The conversation included calculating the absolute pressure, atmospheric pressure, and using the Bernoulli equation to solve for velocity. It was pointed out that there were errors in the calculations, including using the wrong values for pressure and diameter in the equation. The correct values were then used to solve for the velocity, which was found to be 10.926 m/s. It was also discussed how increasing the outlet diameter to 100mm would affect the initial discharge
  • #1
sci0x
83
5
Homework Statement
An enclosed hot water tank is filled to a height of 4 m with water at a temperature of 70°C.
The pressure in the head space is 0.2 bar g. The 50 mm diameter outlet from the bottom of
the tank is closed off by a butterfly valve. If this valve was suddenly opened to allow the
contents of the tank to discharge to the atmosphere, calculate stating all assumption, the
initial discharge velocity of water from the tank.

If the discharge outlet was increased to 100 mm what effect would that have on the initial
discharge velocity? Explain why and show your reasoning.

Data
Density of water at 70°C = 978 kg m-3
Gravitational acceleration g = 9.81 m s-2
1 bar = 10^5 Pa
Relevant Equations
Bernoulli Equation
Hi there, I'm doing a past exam paper Q and i'd like some help.

Assumptions are: The velocity in the tank is negligible and the hydrostatic head is 4m.

Pressure in the vessel:
Gauge pressure
1 bar g = 10^5 Pa
0.2 bar g = 20,000 Pa

Hydrostatic Pressure: (4)(9.81)(978) = 38,376.72 Pa

Absolute Pressure = 20,000 Pa + 38.376.72 Pa = 58,376.72 Pa

Atmospheric pressure = 1.01 x 10^5 Pa

P_{1}
=58376.72 Pa
v_{1}
=0
h_{1}
=4 m
P_{2}
=1.01 x 10^5 Pa
v_{2}
=X
h_{2}
=0.05 m
My Calcs:
Bernoulli Eq to solve for velocity:
P1+1/2ρ⋅V^21+ρ⋅g⋅h1= P2+1/2ρ⋅V^22+ρ⋅g⋅h2
58376.72 Pa + 1/2(978)(0^2) + (978)(9.81)(4) = 1.01 x 10^5 Pa + 1/2(978)(v^2) + (978)(9.81)(0.05^2)
96753.44 = 101023.9855 + 489V^2
-4270.545 = 489V^2 (The figure on the left is negative, so must be gone wrong)
8.7332 = v^2
2.955 m/s = v
 
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  • #2
sci0x said:
Absolute Pressure = 20,000 Pa + 38.376.72 Pa = 58,376.72 Pa
Hi,

Can you reconsider the absolute pressure, given that the pressure in the head is 0.2 Barg ?
 
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  • #3
sci0x said:
Homework Statement:: An enclosed hot water tank is filled to a height of 4 m with water at a temperature of 70°C.
The pressure in the head space is 0.2 bar g. The 50 mm diameter outlet from the bottom of
the tank is closed off by a butterfly valve. If this valve was suddenly opened to allow the
contents of the tank to discharge to the atmosphere, calculate stating all assumption, the
initial discharge velocity of water from the tank.

If the discharge outlet was increased to 100 mm what effect would that have on the initial
discharge velocity? Explain why and show your reasoning.

Data
Density of water at 70°C = 978 kg m-3
Gravitational acceleration g = 9.81 m s-2
1 bar = 10^5 Pa
Relevant Equations:: Bernoulli Equation

Hi there, I'm doing a past exam paper Q and i'd like some help.

Assumptions are: The velocity in the tank is negligible and the hydrostatic head is 4m.

Pressure in the vessel:
Gauge pressure
1 bar g = 10^5 Pa
0.2 bar g = 20,000 Pa

Hydrostatic Pressure: (4)(9.81)(978) = 38,376.72 Pa

Absolute Pressure = 20,000 Pa + 38.376.72 Pa = 58,376.72 Pa

Atmospheric pressure = 1.01 x 10^5 Pa

P_{1}
=58376.72 Pa
v_{1}
=0
h_{1}
=4 m
P_{2}
=1.01 x 10^5 Pa
v_{2}
=X
h_{2}
=0.05 m
My Calcs:
Bernoulli Eq to solve for velocity:
P1+1/2ρ⋅V^21+ρ⋅g⋅h1= P2+1/2ρ⋅V^22+ρ⋅g⋅h2
58376.72 Pa + 1/2(978)(0^2) + (978)(9.81)(4) = 1.01 x 10^5 Pa + 1/2(978)(v^2) + (978)(9.81)(0.05^2)
96753.44 = 101023.9855 + 489V^2
-4270.545 = 489V^2 (The figure on the left is negative, so must be gone wrong)
8.7332 = v^2
2.955 m/s = v
This application of the Bernoulli equation has two errors. For the two points of application you have selected, the correct values are:
$$P_1=101000+20000=121000\ Pa$$
$$h_2=0$$

Questions?
 
Last edited:
  • #4
You are including hydrostatic pressure twice in the left terms (for calculating absolute pressure and in the head term).
The instrument is measuring the differential pressure between the head space of the tank and the atmosphere.
You are using the diameter of the pipe as h2.

Recheck Bernoulli’s conditions inmediately upstream and downstream the valve.
What total pressure (upstream the valve) is pushing the water to flow through and what total pressure (downstream the valve) is resisting that flow?

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html#beq

Do you have the three forms of energy upstream the valve, when there is no movement?
What about downstream an instant after the valve is suddenly and fully opened?
 
Last edited:
  • #5
Thanks for your replies


P_{1}
=121000

v_{1}
=0

h_{1}
=4 m

P_{2}
=1.01 x 10^5 Pa

v_{2}
=X

h_{2}
=0

P1+1/2ρ⋅V^21+ρ⋅g⋅h1= P2+1/2ρ⋅V^22+ρ⋅g⋅h2
121000 + 0 + (978)(9.81)(4) = 1.01x10^5 +1/2(978)V^2 + 0
159376.72 = 101,000 + 489V^2
58376.72 = 489 V^2
119.379 = V^2
10.926 m/s = V (hope this is right)

If the discharge outlet was increased to 100 mm what effect would that have on the initial
discharge velocity? Explain why and show your reasoning.

Larger discharge outlet will have increased initial discharge velocity because a lot more pressure will be lost due to friction in the smaller outlet

I could be wrong here but:
50mm Diameter
Velocity = Flow/Cross Sec Area
10.926 = Flow /3.14(0.025)^2
0.02145 m^3/sec = Flow

100mm Diameter
Velocity = Flow/Cross Sec Area
10.926 = Flow/3.14(0.05)^2
0.0858 m^3/sec = Flow

There is 4 times less flow in the 50mm Diam outlet
 
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  • #6
The Bernoulli equation does not include viscous friction and assumes steady state flow, so talking about initial velocities in this context does not make sense. Within the framework of the Bernoulli equation, the exit velocities will be the same (although, as you showed, the volumetric flow rates will differ).
 
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  • #7
Thanks for this
 
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FAQ: Bernoulli Equation — Discharging water from a partially-filled sealed tank

What is the Bernoulli Equation?

The Bernoulli Equation is a fundamental principle in fluid dynamics that relates the pressure, velocity, and height of a fluid in a system. It states that as the velocity of a fluid increases, its pressure decreases and vice versa.

How does the Bernoulli Equation apply to discharging water from a partially-filled sealed tank?

In this scenario, the Bernoulli Equation can be used to calculate the velocity of the water as it exits the tank. As the water flows out of the tank, its velocity increases and the pressure decreases, allowing it to flow out of the tank.

What factors affect the discharge rate of water from a partially-filled sealed tank?

The discharge rate is affected by the size of the opening, the height of the water in the tank, and the properties of the fluid, such as its density and viscosity. These factors can be incorporated into the Bernoulli Equation to calculate the discharge rate.

Can the Bernoulli Equation be used for any type of fluid?

Yes, the Bernoulli Equation can be applied to any type of fluid, as long as it is incompressible and the flow is steady and inviscid (no friction). This includes liquids such as water and gases such as air.

Are there any limitations to using the Bernoulli Equation for discharging water from a partially-filled sealed tank?

Yes, the Bernoulli Equation assumes that the flow is ideal and there is no energy loss due to friction or turbulence. In reality, there will always be some energy loss, so the calculated discharge rate may not be completely accurate.

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