Evaporate water using compressed air I want to evaporate water by forming air bubbles which are formed by flow of compresses air through perforations. 
What is the energy (in KWh) input to compress 1 cubic meter volume of atmospheric air to 50 bar ? What will be the temparature of the compressed air if ambient temperature is 30 degree celcius ?
If I compress atmospheric air to 50 bar then to evaporate 1 litre of water how much volume of air has to be compressed and supplied ?
For any missing constants please assume and let me know.
Is this process efficient or inefficient when compared to direct boiling of water ?
How much energy (KWh) is needed for direct evaporation of water (lets say a heating coil in a vessel) ?
Thanks.
 A: If you assume adiabatic compression, the equations are
$$PV^\gamma = K \text{ (constant)}\tag1$$
Where for air (mostly a diatomic gas), $\gamma=\frac75$. This means that the final volume $V_f$ when the pressure is 50 bar (50x atmospheric) is
$$V_f = V_i \left(\frac{P_i}{P_f}\right)^\frac{1}{\gamma} = 61.2 L$$
The temperature of the gas is given by the ideal gas law:
$$PV = nRT\\
T_f = T_i\frac{P_i V_i}{P_f V_f} = 928 K = $$
and we find $T_f = 655 °C$
The work done is then given by
$$W = K\frac{V_f^{1-\gamma} - V_i^{1-\gamma}}{1-\gamma}\tag2$$
Substituting the initial conditions into (1) we eliminate $K$ from (2) and get
$$W = P_i V_i^\gamma\frac{V_f^{1-\gamma} - V_i^{1-\gamma}}{1-\gamma}=-5.1 \mathrm{\;bar\; m^3} = 0.14 \mathrm{\;kWh}\tag2$$
for the power required to compress 1 cubic meter of ambient air to a pressure of 50 bar.
The resulting hot air could be used to heat water - but recognize that the air expanding into the water will undergo adiabatic cooling, so you would have to keep the water at the higher pressure in order to take advantage of the heat in the air, and then you end up with steam at this higher pressure as well...
And in this process, all we did was use mechanical energy to generate heat; and while the next step of the process (using small bubbles of hot air to evaporate water) is probably quite efficient (in terms of heat transfer), the first step (convert electrical energy to hot air by compression) is probably not - and you end up with steam at a pressure which might not be useful for you.
So unless you have a source of mechanical energy (say, a windmill) and you need the steam at a high pressure, I think this is not a good method for evaporating water. If you put an electrical heating element in direct contact with the water, all the electrical power is directly converted into heat, and thus into steam. I think it would be a better approach.
In general, a heating element in water would need to heat the water (from 30C to 100 C), then supply additional heat to convert the water to steam. The heat capacity of water is about 4.2 kJ/kg/°C, and the latent heat of evaporation is 2260 J/g. This means the total energy needed to boil one liter of water initially at 30C is 
$$E = (100-30)*4.2 + 2260 = 2.5 MJ = 0.71 kWh$$
Note most of this energy is to actually evaporate the water, not to heat it up. It also tells us that if the evaporating is done in order to distill the water, we can get a LOT more energy efficient by condensing the steam using the incoming (dirty) water - the water will be heated "for free" by the steam, and the steam will turn back into (clean) water.
All this quickly turns into engineering... I recommend that if you have follow up questions, you direct them to engineering.stackexchange.com .
