Mass Transfer Driving Force....

For our purposed we need to look into two film resistance theory (...i think) also known as two-film theory. i am currently working through Fundamentals of Momentum Heat and Mass Transfer by Welty, Wicks, Wilson and Rorrer. it has a good chapter on one phase mass transfer and two phase mass transfer so from a liquid to say gas! the other book is literally written in a language i don't understand.

Hygroscopic dehumidification for Air Conditioning with Solar Absorption Refrigeration

If we could make the solar ammonia absorption ice maker work continuously as discussed we could also use the same system for air conditioning, but this would also require dehumidification of the air.

Dehumidification could be achieved by using a hygroscopic material to absorb moisture from the air. Then this wet material could be dried by in the sun to be recycled to pick up more moisture from the air in a continuous process and shown below:

The small amount of electrical power required to drive the fans and cycle the hygroscopic material could be provided by a car battery.

DF Research project meeting 21/7/10

Desert Fridge Research Project meeting minutes 21/7/10

Attendance (Skype): Muddassar Rashid (MR), Qasid Safir (QS).

Discussion:

· Heat transfer in 1D model of DF

o MR: Why does area term not appear in any equation? QS: Because we are using heat flux (i.e W/m2)

o MR: For heat transfer across composite wall you have not used overall resistance as in methods seen previously. QS: instead we have used overall heat transfer coefficient which is reciprocal of resistance.

· Mass transfer in 1D model of DF

o QS: unable to use ficks law in form presented here to determine mass of vapourising water as this requires knowledge of diffusion length term. Instead we could try using mass transfer driving force approach as used in other examples of sweat cooling analysis. However do not understand how this equation used for mass transfer driving force is derived. Exmplained in convective heat and mass transfer by W Kays but includes 14 pages of differential equation. Need to read. MR: agreed.

· Mass Transfer by Wesselingh and Krishna

o MR: This book explains that there are three factors which influence mass transfer driving force: chemical (mole fraction) gradients, temperature gradients and gravity. But temperature and gravity are negligible.

· MR: Forward to discussion about forced/free convection in last meeting, found section in page Transport Phenomenon (2^nd edition) by Bird and Lightford on page 317 which addresses this issue. Need to read this.

· Continuous operation of solar ammonia absorption ice maker.

o MR: instead of evaporating water (which may be a scarce resource), why not use evaporation of ammonia which already occurs in cooled chamber to cool the generator.

o QS: will consider this option bearing in mind we don’t want to waste all the heat extraction from evaporation of ammonia in cooling generator so much so that primary objective of cooling chamber is overseen.

· Research by Marzan Rahman:

o MR: confirm evaporation is defined as specific case of vapourisation (see wiki).

o QS: is it preferable to add cold water to DF as suggested by M Rahman? MR: cold water would absorb more heat but would take longer to start vapourising. Need to experiment.

· 2^nd law of thermodynamics and net heat transfer

o Agreed that cooling of DF across against temperature gradient does not break 2^nd law of thermodynamics as work driving process is provided by heat removal by evaporation

o Agreed that net heat transfer in our 1D model of DF should be constant but not zero.

Actions

· QS, MR: to understand derivation of mass transfer driving force.

· QS: to run experiment o investigate the effect of temperature of water added

Next meeting Thursday 22/7/10 at 9pm on Skype (to join add skype ID: qasidsafeer)

response to your previous post, how about using the ice produced to cool down the pipe? rather than water, surely water would be in short supply as well.

also ammonia gas is not produced in anaerobic digestors, its methane co2 other annoying by products.

gotta find another method of creating ammonia.

________

muddassar

Continuous operation of solar ice maker

Dan found a brilliant article about a cheap and simple solar cooling system which can reach temperature low enough to freeze water (solar ice maker)

The only problem is that the system does not operate continuously - first it has to recharge with ammonia then it can evaporate it of to induce cooling.

As suggested in the article this could be solved by having several systems all operating simultaneously on staggered cycles - so as one recharges there are others which are still working. So the quicker a cycle can be completed the more efficient it would be (instead of each unit doing one cycle per day it could do N per day)

According to the article once the generator has gathered enough heat (which takes 3 hours), it is covered to from the sun to cool down to complete the cycle. Now it dosnt say how long it takes to cool down by just covering it from the sun but I reckon it would be much faster to cool it by evaporating water on its surface. This could be done by sticking a pipe with tiny holes to pour little drops of water (which would evaporate on its surface in the sun) when it need to cool down (as shown in diagram below).

If this works, it could have incredible implications for cold storage and air conditioning in developing countries.

2nd law of thermodynamics and net heat transfer

Just received a very interesting email regarding the 2nd law of thermodynamics and net heat transfer with respect to the Desert Fridge:

Foaad: My initial impression is that the assumption of net heat transfer in steady state is incorrect as it would imply constant cooling without limit. Also it seems to violate the 2nd law of thermodynamics that heat transfer from lower to higher temperature is not possible without input of work. I think by making net heat transfer zero, we can calculate the minimum temperature possible based upon outside temperature and humidity. In such a case any heat transfer caused by the temperature difference would be resulting in evaporation only.

Qasid: I agree that there would initially be an unsteady state heat transfer process occurring in the DF as water begins to evaporate and draw away heat from air inside the inside pot and thus gradually reducing the temperature of the air inside the inner pot and thus setting up an increasing temperature gradient and therefore driving an increasing rate of heat transfer from the constantly hot surrounding air to the cooling air insider the inner pot. Ceteris paribus, this system would eventually reach a balance between these two competing heat transfer processes (reference) and therefore we would not have infinite constant cooling

Regarding the second law of dynamics, it is proposed that the refrigeration process is maintained by the heat carried by the molecules of evaporating water (reference:Dr Ataul Habib Khalid). It is the same method whereby we can cool our bodies even when the surrounding temperature is hotter than our core body temperature (reference).

If we assume the net rate of heat transfer is zero then, by newtons law of cooling, this suggest there is no temperature difference between the surroundings and inside Desert Fridge.

"Cooling by Evaporation in a Porous Pot" by Marzan Rahman

Reading through an undergraduate research project conducted by Marzan Rahman,(a student of Dr Nadeem Malik at Queen Mary University London) entitled Cooling by evaporation in a porous pot

Here are a few points which I found interesting about this project

· Mentions evaporation is a special case of vapourisation where which only occurs at the surface of the liquid;

o However the reference given for this is a wiki page – is there any proper reference for this?

· Uses wet cloth as lid for Desert Fridge;

o HT through the upper surface (including the exposed sand) is not acconted for in the 1D model of the DF we are developing - is negligible or must we consider this later?

· Assumes the purpose of the sand layer is for insulation.

o If this is the case since we have heat transfer processes occurring in both direction (we want to draw heat out as well), so is insulation a good thing?

· Assumes it is necessary to add cold water

o Why? I guess it would help as a buffer as the water would draw heat as it increases in temperature, but would the lower temperature reduce the rate of vapourisation thus reducing the performance as well?

· Claims convection draws heat through the porous layers

o Wouldn’t the mass flow rate of water b through the porous layers be far too small for there to be any significant convection as opposed to simple conduction?

· States: “evaporated water molecule diffuse through the porous clay pot wall in inside the inner pot… the concentration of water is greater than the air molecules present inside the inner pot. So the evaporated water molecule diffuses inside the inner pot and drive away the hot air molecule.

o Not sure I understand this idea. Is it suggested that water evaporated from the outside surface somehow diffuses though to the inner chamber? Would it not all evaporate directly into the surrounding air?

o In any case how would this process” drive away the hot air molecule”?

....... to be continued...

Steel vs aluminium experiment (new results!)

Just started an experiment to compare the use of steel to aluminium as a material for the inner pot of the DF.

Idea was suggested by prof. Worsely (DF research meeting 5/7/10) as steel has a higher emissivity than aluminium and therefore would enable more efficient heat removal from inner chamber.
However we know there is also heat transfer occurring in the opposite direction (i.e from the outside-in) therefore would this not increase the rate of heat transfer in this direction as well?
So we decided to run this experiment to find out..



2 Desert Fridges (DFs) set up in oven kept at approx 30 oC.

One DF with aluminium inner pot (coka cola can) and other with steel inner pot (pepsi can). Used a magnet to verify which is steel/aluminium.

Porous clay (terracotta) flower pots used for outer pots both DFs. Sand from Swansea bay.
100ml water added to sand in each DF.

Logging temperatures inside each DF and ambient oven temp simultaneously for 24 hours (3 min log interval) using thermocouples (AU:T1 , AU:T2, AWE:T1) immersed in water (to give better contact and increase thermal mass to reduce fluctuations)

Results

The experiment was conduced as set out above, starting on 16/7/10 and ran for more than a day. Below is a graph of the results which has been plotted from the measurement recorded by data logging equipment.
Oven temperature: Initially increases as door was left open for some short period, while the thermocouples were put in place which caused the temperature to drop from the level it was originally being maintained at 29.8oC (air temp). When experiment started, door closed and temperature returned to set level (not quite - actually only went up 26-27). Also, it may have taken time for the water which the thermocouples were immersed in to heat up. Not sure why oven didnt maintain constant temperature - maybe rubbish thermostat control.

Aluminium: very poor performance (temperature difference of little more than a single oC) but consistently followed. Dont know why the results were so much worse than the previous experiment conducted using an aluminum inner pot (which gave a temperature difference of 10oC)

Steel: Definitely performed much better than aluminum by any measure (max 6oC cooler than ambient), except consistency - why is there such a massive swing in temperature?!

Maybe the tip of the thermocouple was wetted but not properly immersed in the water in the can. So the little residual water evaporated thus causing the initial cooling. Then once this residual water had completely dried it began to heat up with the ambient oven temperature. Then the main evaporate cooling process may have kicked in which caused the next phase of cooling. Then the water in the sand and outer clay pot dried out so the temperature began to increase towards the surrounding oven temperature again. Just a guess but decided to re-run this experiment to check if these were consistent results or just down to some experimental error as suggested.

Humidity was recorded at beginning and end of experiment at 56%.

Next experiment
Just started (18/7/10 at 15:41) next run where the oven temperature has been increased (to approx 350 C. ). The outer pots used in the first run will be switched (i.e the outer pot used for he alumimum can in the first run will be used for the steel can in the second run and visa-versa) as I noticed on of the pots seemed more porous than the other. The position of the DFs in the oven (LHS/RHS) have also been switched. Theoretically the switch should have no effect on the results as the purpose is to prove that the only variable is the material of the inner pot.

Test Cell

Discussing design of test cell (see diagram below) with Dan today.

We could use this to efficiently test the effect using different materials of different thickness for the inside and outside pot material and the sand layer on the performance of the Desert Fridge by using thermocouples with data logging equipment to record the temperatures inside the small quantity of water insider each of the Desert Fridge test cell units.

Using this arrangement we could simultaneously compare multiple (maybe 6) configurations at a time. We would also have one thermocouple logging the temperature of the surrounding air inside the environmental chamber.

Inside temperature

The ultimate aim is to determine the steady state inside temperature of the Desert Fridge for a given set of external (outside temperature, humidity etc) and internal (material conductivity, thickness etc) parameters.
Once we have know the steady state mass flux through the Desert Fridge (n.) we can take this as the mass of water vapourising (D. K. Edwards, 1978, p. 145) and therefore determine the heat transfer due to vaporization by:

Where:
q._evap = heat flux by evaporation (W/m2)
n.= mass flux of vapourising water (Kg/m2.s)
L = latent heat of vapourisation which, for water, is given by the following function of temperature (evaluated at the wet bulb temperature in our case I think):

Lwater(T) = 0.0000614342T3 - 0.00158927T2 + 2.36418T - 2500.79 (Yau, 1989)

As the water vapourises at the surface of the outer pot it will draw heat () away from the air inside the inner pot (it may also draw some part of this heat from the outside air by convection and radiation but lets keep it simple for now) which will cause the temperature of the inside the inner pot to decrease thus setting up a temperature gradient from the outside to the inside which will drive a heat flux in the reverse direction.
Eventually the heat fluxes in the opposite directions will reach equilibrium at which point, the surface of the outer pot maintained at the wet bulb temperature of the outside air (Kreith) (which can be determined from a psychometric chart for a given dry bulb and humidity) as shown in the diagram below:

The heat transfer in the reverse direction (q_back) can be determined using the standard method for heat transfer through a composite slab (Wong, 1997, p. 14). This requires the overall heat transfer coefficient which in this case is given by;

Where:
U = overall heat transfer coefficient (W/m2)
h = film (convective) heat transfer coefficient (W/m2)
k = conductive heat transfer coefficient (W/m)
x= thickness (m)

Then the heat flux due to the temperature difference between the outside (T_h) and inside (T_c) air is given by Newton’s law of heating i.e.

Now, although q_evap and q_back will be in steady state equilibrium, this does NOT mean they must be equal (this would imply no temperature difference), it means that the net heat flux will be constant which is given by:

Equation 1

To find a useful expression for q_net we use the same theory for heat transfer through compound walls as used above but here the heat transfer is from the inside air to the vapourising water at the outer surface (at T_wb) opposed to the outside air (at T_h) as before – so we can exclude the h_2 term (which is the convective heat transfer coefficient from outer surface to outside air) so we have:

Where;

Now substituting the expressions derived for q_net, q_evap and q_back, back into equation 1 gives:

Rearranging this equation to make T_c the subject gives:

Bingo!... only problem we still need to know mass flux

Works Cited

D. K. Edwards, V. E. (1978). Transfer Processes: An Introduction to Diffusion, Convection and Radiation. Hemisphere Publishing Corporation.
Yau, R. &. (1989). A Short Course in Cloud Physics. Pergamon press.

Mass transfer through Desert Fridge... Fick's law?

Tried to solve the problem of mass transfer through the Desert Fridge problem by assuming steady diffusion (there will actually also be convection but lets keep it simple for now) and using the approach suggested by Taylor on page 131 of Multicomponent Mass Transfer (he actually looks at mass transfer from gas to liquid – I have rewritten his approach in reverse) as follow:

The transfer of a liquid to a gas may be regarded as taken place in three phases:

1) Diffusion from the bulk liquid (water contained in the porous sand and clay) to the liquid surface (at the surface of the outer pot).

2) Dissolving of the liquid into the gas (to give a water vapour – air mixture in our case).

3) Diffusion from the liquid surface to the gas (warm air of the surroundings).

Now according to section 3.2.3 Concentrations at interfaces in Transfer Processes by D.K Edwards the mass fractions of water can determine across the air-water interface as follows:

Define imaginary surfaces infinitely close to, and at either side of the interface as shown in the diagram below:
Where:
x= mole fraction
u = liquid phase
s = gas phase (vapour-air mixture)
e= bulk gas phase?

• Mass fraction of water at u (m_H2O_u) = 1
• For (m_H2O_s):

1. Look up saturation pressure of steam at the temperature of the water (p_sat_H20_s) in thermodynamics steam tables

2. Then mole fraction is given by:

So in our case we have:

3. Then mass fraction is then given by:

Where:
m= mass fractions
M = molecular weights (M_air= 29g/mol , M_H2O=18)
x = mole fractions
subscript i is water in our case and j is water and air.

So we have:

Using the definition of mass fractions (i.e sum of mass fractions = 1) we can further simplify giving:


(Nb. this is where the first equation in this sweat sweat cooling example must come from)

• And finally for mass fraction of water in the bulk air (m_h2O_e) we can directly deduce this as the mass fraction of water is the humidity of the air (kg water vapour/kg air) (I think) which can be found using a psychometric chart for a known dry bulb temperature and relative humidity (can this be done analytically or can data from psychometric chart be used in matlab?)

And it turns out we have exactly the same mass concentration profile through a porous sweat cooling wall as shown in diagram below (this was copied from some book which I cant find now) :

Now to solve for the mass transfer rate iv tried using:

a) Fick's law:
where:
j = diffusive mass flux
p = fluid density
D = diffusivity coefficient
dm/dz = mass concentration gradient (m =mass fraction, z= diffusion length)

The density of water vapor will be a function of temperature . Now do we use the temperature of the water at the surface of outer pot or the surrounding air here?

The diffusivity coefficient can be taken as approximately 22 m2/s (or does it vary significantly with temperature?).

Finally, although we have can derive the mass fractions at the various phases (see above) I have no idea how to determine the mass concentration gradient as this will require some knowledge of the diffusion length.
For diffusion through the porous layers, it may be their thickness, but what would it be for diffusion from the surface of the outer pot to the surrounding air?

Stuck here so try another method:

b) Mass transfer driving force…