Can someone help me figure out how to solve the following thought problem which involves magnetic induction, electrolysis of water, and buoyancy?

A system containing a spool of wire is dropped into a well of seawater of unknown depth (to be determined in the problem). The well is in a magnetic field, and a current is induced in the spool of wire as it falls. The induced current in turn drives an electrolytic reaction which generates hydrogen and oxygen gases which are collected in a balloon attached to the spool.

As more gases are generated, they displace more and more water increasing the buoyancy of the combined spool + balloons system.

Assume that the system is at Standard Temperature (25 C) and Pressure (1 Atmosphere) at the surface of the well and temperature of the sea water in the well is also 25 C.


Using these equations:

Buoyancy acceleration:

g(ρV-m) where: a = acceleration, V= volume of displaced water,
a= -----------, m= mass of system, ρ = density of H20 (0.999505 at 20° C)
m+ ρV g is the gravitation acceleration constant (9.8m/sec2)

Faraday's Law:

-N d\(\Phi\) ε= emf in volts,
ε= ------ N = number of turns of wire,
dt \(\Phi\) = electromotive force in volts,

ideal gas law

pV =nRT p = pressure in pascals, V = volume of gas in cubic meters,
n = amount of gas in moles,
R = gas constant (8.314472 J·K−1·mol−1),
T = absolute temperature in degrees Kelvin

Nernst Equation (for electrolysis):

E(cell) = E(0) - RT ln Q where E(cell) is the cell potential (emf)
--- E(0) is the standard cell potential at the temperature
zF z is the number of moles of electrons transferred
F = Faraday constant (= 9.648 533 99(24)×104 C mol−1)

Also Assume:

Density of sea water = 1025 kg/m³
pressure increases by 1 atm with each 10 m of depth in sea water.
1 atm = 01.325 kPa

Electrolysis of 1 mole of H20 requires 282.1 kJ of energy yielding 1 mole H2 and 1/2 moles of O2 gas.

mass of 1 mole of H2 = 1.0794g.
1 mole of O2 gas weighs = 16g.

The system:
1 cylindrical spool of 16 gauge magnet wire (with 1000 windings). The inner diameter of the spool is 2.5 cm and the height of the spool is 5 cm. The spool is weighted at the bottom and a deflated rubber balloon is stretched over the entire spool. The 2 ends of the wire are bare and turned up inside the balloon and will act as electrodes. The whole assembly of spool, wire and balloon occupies 100 cubic cm of space and weighs 1 kg.

The spool is released at water level and begins falling down the middle of a well that is 10 cm in diameter, with height to be determined according to the problem defined below.

On opposite sides of the well are 2 bar magnets. Each magnet is 1.25 cm wide, and .3125 cm thick, and stretches the height and depth of the well. Each is magnetized through its thickness, with one magnet having its N pole on the inside of the well and the other magnet having its S pole on the inside of the well. The surface field strength of each magnet separately is 1000 gauss.

As the spool falls through the magnetic field, it generates an electrolysis current that generates H2 and O2 gases that inflate the balloon.


Since the spool-balllon system occupies currently displaces 0.1L of water, 1.015 L of gas will have to be produced to displace an additional 1L of sea water (at a total weight of 1.025kg) and reach zero buoyancy.

The speed of descent is determined by buoyancy. The rate of acceleration at each point is g *(V - m) / (m + V) where V = the total volume of the system (including the .1 L for the spool component), m is total mass of the system ( including the 0.999505 kg spool) and g is the gravitational constant of 9.8 m/sec2.

Note that volume, V, occupied by the gas produced varies with Pressure of the surrounding water at its current depth.

Obviously, the rate of descent is going to be decelerating as more gas is produced, and so is the number of field lines the spool falls through each second, so the rate of production of gas will also decline. Also since pressure is increasing with depth, the volume of a fixed amount of gas will be less at greater depths with increased pressure.

The questions to be answered are:

1) How deep does the well have to be for the system to fall all the way to its point of neutral buoyancy without hitting the bottom of the well first?

2) How long will it take from release until spool-balloon system reaches neutral buoyancy?

3) What is the shape of the curve of distance traveled (depth) vs. time plot, and what is the equation governing it?

4) What is the shape of the volume of gas produced vs. time plot, and what is the equation governing it?

5) What is the impact of doubling the surface strength of the magnets to 2000 gauss, or halving it to 500 gauss? Of doubling the number of turns of wire to 2000 turns, of halving it to 500 turns?

 

Hmm.. I got 7.

Just kidding. I fell asleep half-way through the post.

This is a very complex problem. I would almost be easier to build the model rather then do the math.

What have you gotten so far?

 

It's a good problem, as it includes multiple things going on -- this is the kind of problem you'll run into in the real world.

As in all problems, it's important to organize your thoughts and develop your strategy on how you'll solve the problem. Here, you're dealing with a motion problem -- you're being asked for a distance and a time. If you're beyond a freshman physics class, then you know the principles you need: figure out what the system is, then figure out all the forces acting on that system. Then you can write down the equation of motion as \(\ddot x = \frac F m\) where x is the displacement. I've written it that way to emphasize it will be a differential equation that you'll have to solve. Note that \(\dot x\) (the velocity) will be in one of the components of F because the electrolysis current depends on the velocity. Since there is a gas being generated, you'll have to also ask if the system's mass is a constant (i.e., does the water for electrolysis come from outside the system as it's needed (like an "inverse" rocket) or is it carried along with the spool?).

Once you get the EOM (equation of motion), you can then try to solve it exactly or integrate it numerically on the computer.

Make sure you understand the physics of each of the component parts of the problem -- if you don't, you won't write down the correct EOM -- and no amount of correct solving will give you the correct answer if you start with the wrong equation.

 

retched: I get your point. I'll report on separate parts of the problem that I've figured out in separate responses. This response focuses on the amount of gas that has to be produced and the energy required to produce it as a function of depth.

someonesdad: Thanks! Those are good points.

The way I thought about it, the water is not part of the mass of the spool-balloon system. I just thought of it as water that is in the empty center of spool. Technically it needn't be part of the system, since it doesn't displace any water -- it just is water -- and it is open to the rest of the water of the well, so as gas develops, it could be displaced by the developing gas.

But I can see how it might be simpler to make it a closed system, with constant weight of 1kg as you suggest. There just needs to be enough water inside the closed system to convert to more than 2x the amount of gas needed to achieve buoyancy (1 L at the pressure of the maximum depth will be generated on the way down, and an equal amount in moles will be created on the way up -- though it will occupy more than 2L at surface pressure).

Since the closed system weighs 1 kg, it must create 1L of gas (at depth d) to achieve neutral buoyancy, and will generate twice as much by the time it reaches the surface again (however it will then occupy more space because it will be at lower pressure as it rises).

From the Ideal Gas Law we know that at S.L.C [25oC (298K), 101.3kPa (1 atm)], 1 mole of an ideal gas has a volume of 24.47L

Therefore at surface pressures we would need 0.04087 moles of gas to equal 1 L of gas.

Also 1 L of gas requires 2/3 mole of H2O to be dissociated. So we will need to dissociate 0.02725 mole of H2O.

We also know that each mole of H20 converted requires 282.1 kJ of energy, so we need 7.686 kJ of energy to create enough gas for buoyancy at the surface 0.7686 kJ for each additional meter of depth.

Therefore, energy required, E(d), in kilo-joules required to achieve neutral Buoyancy as a function of depth d in meters is:

E(d) = 7.686 (1 + .1 d) kJ

Now we need to figure out the function of energy generated by magnetic induction as a function of distance. But I'll leave that for a future response.

 

What is the amount of power produced per inch if the speed of decent was constant?

 

I think this question involves PV work since it deals with pressure, volume, and energy. Work = -PV (where V is the displaced volume). Also, one mole of a gas at STP is 22.42 litres, not 24.47L.

 

zero_coke:
You are correct that 1 mole gas at Standard Temperature & Pressure (STP) = 22.42 L.
But this problem is set at Standard Laboratory Conditions (SLC) not STP.
STP and SLC conditions are defined as:
STP: temperature = 0C (273K) and pressure =101.325 kPa
SLC: temperature = 25C (298K) and pressure = 101.325 kPa

At the colder temperatures of STP, 1 mole of ideal gas occupies 22.42 L.
At the warmer temperatures of SLC, 1 mole of ideal gas occupies 24.47L.

Obviously, using STP or SLC should affect the final numerical results slightly, but the underlying governing equations and solutions should be the same regardless of whether you use STP or SLC (or any other arbitrary temperature and pressure just by applying the ideal gas law).

But thanks for pointing that out, since it could have been an easy mistake to make.

 

retched:

Momentarily I'll be posting my current solution for the 2nd part of the problem, the governing equation for Energy Produced as a function of time and depth.

That isn't a solution for Power, which would have to integrate Energy Produced over time, and I haven't figured out what the depth vs. time function is.

Figuring out the depth vs. time function will require employing the buoyancy acceleration equation, along with cumulative energy produced, the number of moles of gas dissociated using that energy, and the volume occupied by that many moles of gas at the current depth. That still remains to be done.

By the way, the problem is given in SI units (e.g. meters) not imperial units (inches). But obviously conversions between the two could be made.

I would not expect the descent to be constant, since it is governed by the gravitational acceleration constant and by ratio of the weight of the volume of gas produced to the weight of the water displaced.

i would expect that Power would be governed by Faraday's Law. If my calculations in the part I am about to post are correct, the Energy produced, Em(d,t) is:

Em(d,t) = 54.8947 (d/t)^2 joules

Since Power = Energy * Time,

P(d,t) = 54.8947 d^2/t joules/second = 54.8947 d^2/t watts

If the rate of descent were constant, then d/t is a constant too (we can call it velocity, v).

P(d,t) = 54.8947*d*v watts

However, I may have miscalculated the magnetic flux strength (see note in my next posting), in which case this could be in error.

 

Summary of Previous step (step 1): Solve for Energy Consumed:
The volume of gas produced is a function of the Cumulative Energy Consumed to dissociate that many water molecules, independent of the time it takes to generate that energy. In addition to the number of molecules that have been dissociated, the volume is also only dependent on temperature (which is constant) and pressure (which varies with depth). Therefore we can define the Cumulative Energy Required to achieve neutral buoyancy, E(d), strictly as a function of depth, d, which we did in the previous step:

EQ 1: E(d) = 7.686 (1 + .1 d) kJ

Summary of Current step (step 2): Solve for Energy Produced:
Next we want to figure out the equation for the Cumulative Energy Produced through magnetic induction, Em. However, the total Energy produced is growing cumulatively over time. So Em will depend on time. In addition, the amount of electrical energy induced depends on the number of flux lines the wires cross, and the velocity each wire crosses those flux lines. Since velocity is also defined as distance / time, Em must be expressed as a function of both distance (or depth), d, and time, t): Em(d,t). Below I show how I derived the governing equation of the Em(d,t) function.

EQ 2: Em(d,t) = 54.8947 (d/t)^2 joules

In a future step, we will use the Buoyancy acceleration equation to determine how depth, d, varies over time, t. Then we can integrate E(d) and Em(d,t) and find the solution where the integral E(d) = the integral of Em(d,t).

Detail: Determining the governing equation for Em(d,t)

EQ 3: EMF = BLv

We are given that the magnetic field of each magnet at its surface is, B = 1000 gauss = .1 Tesla = .1 Newton / amp * meter (see note 1)
We are also given the Length of wire, L = 1000 turns * 2.5 cm diameter * pi = 7,854 cm = 78.54m
Also, by definition velocity, v, equals distance, d, over time, t: v = d/t
Therefore, EMF(d,t) = ( .1 N / amp * m ) * ( 78.54 m ) * v m/sec = 7.854v volts = 7.854 * d/t volts

The Energy produced through electromagnetic induction, Em, in joules is the EMF * the charge, q, in coulombs:

EQ 4: Em(d,t) = EMF(d,t) * q(d,t) coulomb volts = EMF * I(d,t) amp sec volts = 7.854 * d/t * I(d,t)

Now we need to figure out the coil resistance,

EQ 5: R = rL/A.

We are using a spool of 16 gauge copper wire wrapped 1000 times around a 2.5 cm diameter spool.
That means its length L = 1000 * 2.5 * pi = 7,854 cm = 78.54m in length.
16 gauge wire has a cross section, A = 1.3 sq. mm = 1.3x10-6 sq. meters
Material resistance of copper wire is 1.68×10−8 Ω·m

R = 1.68×10−8 Ω·m * 78.54m / 1.3x10-6 sq. meters = 1.1237 ohms

Now we can calculate, the current, I, using Ohm's law

EQ 6: I = V / R,

where V = voltage in volts, and R = resistance in ohms
I(d,t) = 7.854 * d/t volts / 1.1237 ohms = 6.9894 d/t amps

Substituting I(d,t) back into EQ 4 we get

EQ 7: Em(d,t) = 7.854 * d/t * 6.9894 * d/t = 54.8947 (d/t)^2 joules

In the next step of the problem to solve, we need to calculate the depth, d of the spool-balloon system at time, t.

Note 1: the problem reports two magnets on opposite sides of a well, 10 cm apart, each stretching from the surface to the bottom and with each magnet having a 1000 gauss field at their surface. I have used this 1000 gauss field number in these calculations, but I am not sure if that is appropriate, or if not, how to determine the field strength where the wire spool is falling. If that number is significantly different from 1000 gauss, the equations above could be solved incorrectly.

 

This is giving my brain quite the workout.

If the poles are swapped for each magnet on the well wall, the average should still be 1000 gauss. If they were not swapped, there would be no gas production.

Knowing the change in speed as a result of the amount of gas produced, acceleration, and magnetic forces on the spool, (being non-ferrous I would suspect the eddys have no or negligible effects on the acceleration of the spool)

I was trying to solve as there was NO water, but the gas was still produced.

Dont let me throw you off, your doing great.

If the well was empty, and you had a meter connected to the wire ends, what would be the amount of power produced, not dealing with the buoyancy.

After that could be determined, and figuring on linearity, the waters effect on the negative acceleration and could be calculated in. Regardless of the speed the spool moves, it is the distance that determines the gas developed. That should be linear.

...ouch...

 

I wonder how much the interaction between the coil and the magnet would slow it down. Maybe you have been asked to ignore that.

 

I just mentioned that.

The nonferrous copper would create eddy currents but would not be attracted to the magnets.

I had to wipe my molten brain matter off my shoulders after thinking through this.

 

I think we crossposted. I was thinking more that the coil has a current through it which is powering the electrolysis. Surely that will slow it down in real life.

 

I have a simulation in excel now. I'm not sure it is right though, as I am somewhat skeptical of the results it predicts.

I am using the following equations to calculate the magnetic drag from the induced current.

vmd = m · Fmd
Fmd = B^2 · L^2 · v / R

vmd is the velocity of the magnetic drag vector, which is opposite in sign to the velocity of coil system.

v is the velocity of the system (computed from V(t-1) + acceleration due to buoyancy plus hydrodynamic drag (also opposite in sign to direction of travel) + magnetic drag)

m is the weight of the system (1kg in the model)

Fmd is the force of the magnetic drag

B is the strength of the magnetic field on the wire (1000 gauss in the model)

L is the length of the wire coil (I calculated 78.54m by multiplying the 1000 turns by 2.5 cm diameter by pi).

R is the Resistance of the complete circuit (Resistance of the wire 1.01 ohms + Resistance experience by the current between the two electrodes 8.333 ohms based on the resistivity of seawater)

 

Here are the governing equations in my model:

DISTANCE, VELOCITY AND ACCELERATION
0) d(t) = d(t-1) + velocity * Δt; negative values of d are depth, positive are height
1) velocity(t) = velocity(t-1) + Δvelocity(t-1...t)
2) Δvelocity(t-1..t) = accelerationbuoyancy(t-1) + velocitydrag(t)
3) velocitydrag(t) = velocityfluid_drag(t) + velocitymagnetic_drag(t)
4) accelerationbouyancy(t) = g (ρfluid · VTotal - msystem ) / (ρfluid · VTotal + msystem ) ; where fluid = liquid for all dt <=0, air for all dt > 0

GAS PRODUCTION FROM ELECTROLYSIS
5) VTotal = Vgas + Vsystem Volume of gas produced
6) Vgas = n· Rgas· T / pressured
7) pressured = pliquid(d) in liquid where dt < =0 ; = pair(d), if air where dt > 0
8) pliquid(d) = p1atm + g · ρliquid · dt ; in liquid where dt < 0; p1atm = sealevel pressure
9) pair(d) = p1atm · e-m · g · dt/ 0.001 · k · T , in air where dt > 0
10) nt = nt-1 + Δnt; n is the number of moles of gas produced
11) Δnt = Ngas · Einduction

INDUCED EMF AND CURRENT THAT POWERS ELECTROLYSIS
12) Einduction = Iinduced * EMF / Δt
13) Iinduced = EMF / Rtotal
14) EMF = Bwire · Lwire · velocity(t)

HYDRODYNAMIC AND AERODYNAMIC FLUID DRAG
15) Ffluid_drag = bfluid_drag_coefficient · velocityt-1
16) bfluid_drag_coefficient = bsphere · bratio(shape/sphere)
bsphere = 6 · ∏ · ηfluid · rsphere ; where fluid = liquid for all dt <=0 , air for all dt > 0
17) rsphere = ( 3 · VTotal / 4∏ ) -3

MAGNETIC DRAG
18) velocitymagnetic_drag = msystem · Fmagnetic_drag
19) Fmagnetic_drag = Bwire2 · Lwire2 · velocity(t) / Rwire

 

Here are the constants and parameters used in creating a mathematical model

CONSTANTS AND MODEL PARAMETERS
Name Value Units Description

Constants 1.38E-23
g 9.81 m/s2 gravitaional acceleration constant for earth in meters/sec2
Rgas 0.008314472 m3·kPa/°K·mol ideal gas constant in Liter·kiloPascal/°K·mole
NA 6.02E+23 mol-1 Avogadro's number
kgas 1.38E-26 m3·kPa/°K Boltzmann gas constant = 1.3806503 × 10-23 m2 kg s-2 K-1
∏ 3.14159 unitless Pi
e 2.71828 unit euler's number (base of natural exponent)
A16 gauge 0.0000013 m cross sectional area of 16 guage wire in square meters
ρCu 8940 kg/m3 density of Copper at 25° C (298°K) in kilograms per cubic meter
rCu 1.68E-08 Ω·m resistivity of copper wire in Ω·m
ρH2O 997.044 kg/m3 density of H2O at 25° C (298°K) in kilograms per cubic meter
VH2O 0.000018 m3/mol volume of 1 mole of H2O in cubic meters per mole
mH2O 0.018 kg/mol mass of 1 mole of H2O in kilograms per mole
µH2O 1.00E-03 Pa s dynamic viscosity of H2O
νH20 1.00E-06 m2 s−1 kinematic viscosity of H20
rH2O 182000 Ω·m resistivity of H2O in ohm meters
VoltageH20 1.23 V The standard potential of the water electrolysis cell is 1.23 V at 25 °C.
ρseawater 1025 kg/m3 density of seawater at 25°C (298°K) in kilograms per cubic meter
Vseawater 0.000018 m3/mol volume of 1 mole of sea water in cubic meters per mole
mseawater 0.018 kg/mol mass of 1 mole of sea water in kilograms per mole
rseawater 0.21 Ω·m resistivity of sea water in ohm meters
Voltageseawater 1.229 V theoretical potential for seawater electrolysis is 1.229 V
µseawater 1.08E-03 Pa s dynamic viscosity of seawater
νseawater 1.05E−06 m2 s−1 kinematic viscosity of seawater
rdrinking water 2000 Ω·m resistivity of drinking water varies with impurities between 20 and 2000
ρair 1.184 kg/m3 density of air at 25°C (298°K) in kilograms per cubic meter
µair 4.51E-06 kg·m/s dynamic viscosity of air in kilogram meters per second
mamu_air 28.95 g/mol average molecular weight in unified atomic mass units, u, where u = 1g/mol

Parameters calculated values are in black, editable parameters in blue
the Laboratory Conditions
T 298 °K Temperature in Kelvin (Standard Laboratory Conditions = 298°K = 25°C)
patm 101.3 kPa pressure at surface (1 atmosphere) in kiloPascals
Δt 1 s sampling rate in seconds

the well:
dwell m depth of well
wwell 0.1 m width of well

the magnets:
Bsurface 0.1 T magnetic field strength at surface in Teslas (1 Gauss = 0.0001 Teslas)
Bwire 0.1 T magnetic field strength experienced by wire coil
wmagnet 0.0125 m width of each magnet
dmagnet 0.003125 m depth of each magnet
hmagnet 0 m height of each magnet

the wire: 16 gauge copper magnet wire
nturns 1000 number of turns (unit-less)
dcoil 0.025 m diameter of 1 coil in meters (2.5 cm = 0.025 m)
Lwire 78.54 m length of wire in meters
Awire 0.0000013 m cross sectional area of wire in square meters (16 gauge wire)
compositionwire Cu symbol chemical composition of wire (can be used to look up resistivity, etc.)
rwire 1.68E-08 Ω·m resistivity of wire in Ω·m
Rwire 1.01 Ω Resistance of wire in Ω
ρwire 8940 kg/m3 density of wire in kilograms (1 kg·m3 = 0.001 g·cm3 = 0.001 kg·L )
mwire 0.91 kg mass of wire in kilograms
dwire 0.00000041 m diameter of wire in meters
hcoil 0.00041 m height of coil in meters
Vcoil 0.0000325 m3 volume of coil (including interior) in cubic meters (1 Liter = 1000 m3)


the spool system
Vmisc 0.8999675 m3 volume of miscellaneous other parts of spool system (e.g. balloons)
mmisc 0.09 kg mass of miscellaneous other parts of spool system (e.g. balloons, liquid)
Rmisc 0.00 Ω resistence in ohms of any other component (e.g. electrodes)
Vsystem 0.90000 m3 total volume in cubic meters of spool system (Vcoil + Vmisc )
msystem 1.00 kg total mass in kilograms of spool system (mcoil + mmisc )
Rsystem 9.35 Ω total resistence in ohms of spool system (Rwire + Rliquid + Rmisc)

The electrolytic equation:
eelectrolyte 1 mol H2O moles of Liquid Electrolyte solution consumed by electrolysis
gasanode(+) 0.5 mol O2 gas produced at anode by electrolysis of e moles of electrolyte in moles
gascathode(-) 1 mol H2 gas produced at cathode by electrolysis of e moles of electrolyte in moles
gascombined 1.5 mol gas moles of combined output gasses
fratio 1.5 ratio of moles of output gas generated over input gas consumed (unit less)

the electrolyte:
Eliquid to gas 282.10 J/mol Energy in Joules/mol to required to consume 1 mole of liquid
Egas 188.07 J/mol Energy required to generate 1 mole of gas in Joules per mole
Ngas 0.005317263 mol/J number of moles of gas produced by one Joule of energy
ρliquid 997.044 kg/m3 density of electrolyte liquid at Temperature T above
Vliquid 0.000018 m3/mol Volume occupied by 1 mole of electrolyte liquid at Temperature T above
mliquid 0.018 kg/mol mass of 1 mole of electrolyte liquid
rliquid 0.208333333 Ω·m resistivity of electrolyte liquid
delectrodes 0.03 m distance between electrodes in meters
Rliquid 8.33333332 Ω Resistance of electrolyte liquid
Voltagepotential 1.229 volts voltage potential of electrolyte
µliquid 1.08E-03 Pa s dynamic viscosity of seawater
νliquid 1.05E−06 m2 s−1 kinematic viscosity of seawater