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

[sci0x]

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

	=	58376.72 Pa
	=	0
	=	4 m
	=	1.01 x 10^5 Pa
	=	X
	=	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

 

[BvU]

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 ?

 

[Chestermiller]

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

	=	58376.72 Pa
	=	0
	=	4 m
	=	1.01 x 10^5 Pa
	=	X
	=	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?

 

[Lnewqban]

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?

 

[sci0x]

Thanks for your replies

	=	121000
	=	0
	=	4 m
	=	1.01 x 10^5 Pa
	=	X
	=	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

 

[Chestermiller]

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).

 

[sci0x]

Thanks for this