Enthalpies of saturated water and steam at sub-zero temperatures.

[athomas236]

Does anyone know of a reference or website that provides the enthalpies of saturated water and steam at sub-zero temperatures.

Regards,

athomas236

 

[25362]

The ChE manual, Perry VI, table 3-300, shows these and other properties for ice and vapor down to -160^oF.

 

[athomas236]

Thanks 25362,

Unfortunately I have 7th edition of Perry which does not have the information but your response prompted me to look in Kemps Handbook which has the same information as Perry but down to -40C.

Neither Perry or Kemps seem to have enthalpies for saturated water. I suppose I could estimate the enthalpy for water based on specific heat of 4.2kJ/kg K.

Regards,

athomas236

 

[athomas236]

Sorry 25362,

Just realised that below 0C there are only two phases vapour and solid ice.

Best regards,

athomas236

 

[25362]

Indeed, water in liquid form at temperatures below the thermodynamic equilibrium temperature for phase transformation is known as "supercooled" and is considered in a thermodynamic metastable state.

Freezing may not occur until pure water is subcooled to several degrees below the equilibrium temperature. The transformation of water into ice must be intiated by a microscopic ice cluster named a "nucleus", itself considered a probabilistic event.

Experiments have shown that water under atmospheric conditions can be subcooled to about -45^oC before homogeneous nucleation occurs. For this reason this temperature has been labeled the homogeneous nucleation temperature of water, corresponding to a critical cluster of some 25 molecules with a radius of ~0.4 nm.

Introducing an impurity, such as a piece of ice, in subcooled water triggers immediate ice propagation by heterogeneous nucleation.

 

[athomas236]

Thanks 25362

athomas236

 

[kenvlach]

Perry's 7th edition has data down to -160 F (Table 2-350) and in SI units down to 150 K (Table 2-352). Also some data along the melting line, down to -8.803 C @ 1000 bar (Table 2-357).

More water info than you may care to wade through:

http://www.iapws.org/

NIST has lots of data:

http://srdata.nist.gov/gateway/gateway?substance=water&subft=Submit&rddesc=desc

An interesting paper:
'International Equations for the Pressure along the Melting and along the Sublimation Curve of Ordinary Water Substance'

http://www.nist.gov/srd/PDFfiles/jpcrd477.pdf

 

[athomas236]

kenlach,

Thanks for all the useful information. I really will have to get my eyes tested.

Regards,

athomas236

 

[25362]

The vapor pressure of supercooled water is higher than that for ice at equal temperatures.
Goggle has plenty of information on this subject. For example:

http://cires.colorado.edu/~voemel/vp.html

​

 

[athomas236]

No wI have the water and ice properties, this is my attempt at calculating what happens when water at 95C and 1bara is sprayed into air at minus 40C.

This is basic data:
Enthalpy of water at 95C & 1bara = 398.03
Entahlpy of ice at minus 40C = minus 411.97
Enthalpy of vapour at minus 40C = 2426.99

All enthalpies are kJ/kg, those for ice and vapour are from Perry version 7, table 2-352 and that for water at 95C is from steam tables.

By heat balance with X as from of ice

398.03 = -411.97*X + (1-X)*2426.99

Which simplifies to

(398.03+411.97+2426.99)*X = 2426.99

Which gives X = 0.727
ie 72.7% of water at 95C ends up as ice at minus 40C, the remainder is vapour.

Does any one have any comments on this calculation? My problem is that other posts on this site suggest that there will be little ice.

Regards,

athomas236

 

[25362]

Some thoughts:

- Heat transfer takes time. Thus, if the spraying is done from a flying airplane it may reach the point of ice formation and ice cooling, beside some water supercooling.

- Upward air buoyancy may affect the rate of ice crystallites falling down.

- The heat balance should be: heat gained by air = heat lost by water. Similar to a cooling tower.

- Air at -40 Celsius wouldn't be able, by simple heat exchange, to cool the water to the same temperature (approach [→] 0 degrees ?).

- Air picks up heat and humidity and rises by buoyancy at a certain higher temperature.

- The heat balance should be: heat gained by air = heat lost by water. Similar to what happens in a cooling tower.

- If ice nucleation starts at zero Celsius, at some point heat transfer would take place between ice and air.

- As published elsewhere high tropospheric cirrus clouds contain both ice and supercooled water at -40 Celsius.

Kindly comment.

 

[athomas236]

25362,

Thanks for your response. I now realise that my approach is simplistic since it only considers what happens when water at 95C is adiabatically reduced to minus 40C.

The cooling tower analogy is useful and gives a more meaningful heat balance so that heat lost by water = heat gained by air = heat transfered from water to air. However, it poses problems with obtaining a solution.

One problem is to estimate the amount of air that will be involved in the process.

For natural draft cooling towers this is relatively simple because the the difference in static head between a water air mixture at its molecular weight and temperature inside the tower and that of the air at its molecular weight and temperature outside the tower can be balanced against the friction loss caused by the flow of air through the tower.

I do not know how to make a similar calculation in free air because there is no obvious friction loss that is equivalent to that in the tower.

The other problem is to determine the heat transfer coefficient between air and water. Again for cooling towers the supplier would fix a base value based on his empirical data and this base value would be scaled for other operating conditions.

For the case I am considering I do not yet know how to estimate the heat transfer coefficient.

Regards,

athomas236

 

[25362]

This is a case where heat and mass flow in the same direction from water to air.

In towers with countercurrent flow, overall heat transfer coefficients, U, kW/m³.^oC for sprays vary between 1.8 and 5.6, and are estimated from:

U = 1.2 F_g^0.8 F_l^0.4 / h^0.5​

h = contact height, m
F_g = air mass flowrate, kg/m².s
F_l = liquid mass flowrate, kg/m².s

You may find some help in Donald R. Woods' Process Design and Engineering Practice, PTR Prentice Hall