Fluid Mechanics Help: Find Balloon Radius & Pipe Flow Rate

[eutopia]

A spherical helium-filled balloon is used to lift a load of 5905 N (in addition to the weight of the helium). Suppose the density of the air is 1.24 kg/m3. Find the minumum necessary radius for the balloon.


A 24.8 liter container is initially evacuated, and then 5.7 grams of water is placed in it. After a time, all the water evaporates, and the temperature is 115.8o C. Find the final pressure.


A water tower, vented to the atmosphere, is drained by a pipe that extends to the ground. The top surface of the water in the tank is 15.7 m above the ground. If the valve in the drain pipe is closed, what is the absolute pressure in the drain pipe at ground level?
Answer: 2.55E5 Pa
If the valve is now open and the cross-sectional area of the drain pipe is 2.24E-2 m2, find the volume flow rate in the drain pipe. Assume that the water speed at the top of the water in the tank is negligible.

 

[p53ud0 dr34m5]

show some work?

 

[eutopia]

I don't really know how to approach the first and second problem... like how to set up the equations or which one to use.

For the last one, volume flow rate = area x velocity, so I tried to find the velocity with Bernoulli's equation, but I didn't know how the variables are set up with it.

P1 + (1/2)pv1^2 + pgy1 = P2 + (1/2)pv2^2 + pgy2

 

[eutopia]

if P1 is the atmosphere pressure, then P2 is the pressure at the bottom of the pipe. since the velocity is neglible at the top, and at the bottom y2 becomes 0

P1 + pgy1 = P2 + (1/2)pv2^2... but then, if P2 already = P1 + pgy1.. then there would be no velocity... so I'm kind of confused.

 

[Pyrrhus]

Well You know where you would get the radius right? From the Volume of the Sphere, but first you need to set up the equation for the [load+balloon] system.

$$\vec{B} + \vec{W}_{balloon+load} = 0$$

Now express in terms you can work out.

 

[xanthym]

A water tower, vented to the atmosphere, is drained by a pipe that extends to the ground. The top surface of the water in the tank is 15.7 m above the ground. If the valve in the drain pipe is closed, what is the absolute pressure in the drain pipe at ground level?
Answer: 2.55E5 Pa
If the valve is now open and the cross-sectional area of the drain pipe is 2.24E-2 m2, find the volume flow rate in the drain pipe. Assume that the water speed at the top of the water in the tank is negligible.

I don't really know how to approach the first and second problem... like how to set up the equations or which one to use.

For the last one, volume flow rate = area x velocity, so I tried to find the velocity with Bernoulli's equation, but I didn't know how the variables are set up with it.

P1 + (1/2)pv1^2 + pgy1 = P2 + (1/2)pv2^2 + pgy2

-------------

if P1 is the atmosphere pressure, then P2 is the pressure at the bottom of the pipe. since the velocity is neglible at the top, and at the bottom y2 becomes 0

P1 + pgy1 = P2 + (1/2)pv2^2... but then, if P2 already = P1 + pgy1.. then there would be no velocity... so I'm kind of confused.

When the valve is opened, the situation becomes {P2 = P1}. Recalculate your results using this fact together with the other given values.


~~

 

[eutopia]

Well You know where you would get the radius right? From the Volume of the Sphere, but first you need to set up the equation for the [load+balloon] system.

$$\vec{B} + \vec{W}_{balloon+load} = 0$$

Now express in terms you can work out.

... I know the weight of the ballon, but what would the weight of the helium be? and what would B represent here?

When the valve is opened, the situation becomes {P2 = P1}. Recalculate your results using this fact together with the other given values.

Thanks! It makes sense now that I think about it.

 

[xanthym]

A 24.8 liter container is initially evacuated, and then 5.7 grams of water is placed in it. After a time, all the water evaporates, and the temperature is 115.8o C. Find the final pressure.

{Volume of Container} = V = (24.8 liter)
{Moles of Water in Container} = n = (5.7 grams)/(18 grams/mole) = (0.3167 moles)
{Temperature} = T = (115.8 degC) = (388.9 degK)
{Ideal Gas Constant} = R = (0.08206 Liter*atm/(mol*degK))

Use the Ideal Gas Law to calculate Pressure "P" in units of "atmospheres":
P*V = n*R*T


~~

 

[eutopia]

{Volume of Container} = V = (24.8 liter)
{Moles of Water in Container} = n = (5.7 grams)/(18 grams/mole) = (0.3167 moles)
{Temperature} = T = (115.8 degC) = (388.9 degK)
{Ideal Gas Constant} = R = (0.08206 Liter*atm/(mol*degK))

Use the Ideal Gas Law to calculate Pressure "P" in units of "atmospheres":
P*V = n*R*T


~~

I tried working with the ideal gas law, but forgot that to get the number of moles, i would take 5.7 and divide that by (1 for hydrogen + 1 hydrogen + 16 for oxygen) ^^ the 5.7 kind of just stared at me

 

[eutopia]

$$\vec{B} + \vec{W}_{balloon+load} = 0$$

I thought:
density = mass / volume

somehow find the mass of the helium ballon, and then I would get my volume.

But the mass is what got me stuck. Would the weight of the helium equal the weight of the load?... how else would you find the weight of the helium... and what does B stand for in the above equation?

 

[eutopia]

weight of helium + weight of balloon = weight of displaced air

density of helium = .179
density of air = 1.24

 

[xanthym]

weight of helium + weight of balloon = weight of displaced air

density of helium = .179
density of air = 1.24

Now solve the equation for spherical balloon radius "R":
{Weight of Helium} + {Weight of Balloon + Load} = {Weight of Displaced Air} ::: <--- Buoyant Force
::: ⇒ {ρ_helium*V_balloon*g} + {5905 N} = {ρ_air*V_balloon*g}
::: ⇒ {ρ_helium*(4*π*R³/3)*g} + {5905 N} = {ρ_air*(4*π*R³/3)*g}


~~

 

[Pyrrhus]

Eutopia, i meant what Xanthym said. B stands for buoyant force, remember we consider air a fluid.

 

[eutopia]

$$F = pVg$$

I guess that was the key formula I was missing

 

[eutopia]

but then again

$$pV = m$$!