Lift for a balloon by adding water vapour

Example: Surrounding air is at 25 deg C and RH=60% and P=100 kPa.
A cubic metre of the surrounding air has 1.6 kg of water evaporated into it and ends up at 95 deg C (so the system is a cubic metre of air at 25 deg C and 1.6 kg of water both heated to 95 deg C). The buoyancy of 1 cubic metre of this hot expanded air is now 0.462 kg. The RH is now 82.03% (after water added and heated). With no water added but surrounding air heated to 95 deg C the RH would be 2.25% and the buoyancy would be 0.221 kg.

Here is some computer code that will work out the details for you - I wrote it is Delphi. 

Inputs are Tair1 (temperature of surrounding air in deg C), 

RH1 (relative humidity of surrounding air (percent), 

P (atmospheric pressure in kPa, 

Tair2 (temperature to which air and water will be heated to become moist air), 

ng (number of grams of water that will be heated with 1 cubic metre of surrounding air to Tair2).

Delphi code is below:

label 1;

var

HR1,HR2,Tk1,Tk2,Tair1,RH1,RH2,P,Tair2,ng,Psatv1,Psatv2,Pv1,Pv2,Mv1,Ma1,Mv2,Ma2,

denss,densp,F,a,Dair1,Dair2,Psati,Enth1,Enth2,Enth2m1,Tav,Enthv,Enthvkg,mf1,mf2:extended;

errors1:boolean;

calcstr1,calcstr2,calcstr3,calcstr4,calcstr5,calcstr6,calcstr7,calcstr8,

calcstr9,calcstr10,calcstr11,calcstr12,calcstr13,calcstr14,calcstr15,calcstr16,

calcstr17,calcstr18,calcstr19,calcstr20:string[30];

begin

errors1:=false;

form20.hide;

form20.show;

try

Tair1:=strtofloat(form20.edit1.Text);

RH1:=strtofloat(form20.edit2.Text);

P:=strtofloat(form20.edit3.Text);

Tair2:=strtofloat(form20.edit4.Text);

ng:=strtofloat(form20.edit5.Text);

except

errors1:=true;

end;

if (errors1=true) or

(Tair1<0.1) or (Tair1>80) or (Rh1<1) or (RH1>100) or (P<50) or (P>150) or (Tair2<0.1) or (Tair2>98)

or (ng<0) or (ng>10000)

then begin

form20.canvas.textout(0,100,'CHECK ENTRIES.');

goto 1

end;

Tk1:=Tair1+273.15;

Tk2:=Tair2+273.15;

Psatv1:=0.61121*exp((18.678-Tair1/234.5)*Tair1/(257.14+Tair1));

Pv1:=Psatv1*RH1/100;

Mv1:=Pv1/(0.4615*Tk1);

Ma1:=(P-Pv1)/(0.287*Tk1);

HR1:=Mv1/Ma1;

HR2:=(Mv1+(ng/1000))/Ma1;

Psatv2:=0.61121*exp((18.678-Tair2/234.5)*Tair2/(257.14+Tair2));

RH2:=(HR2*P)/((0.622+HR2)*Psatv2);

Pv2:=Psatv2*RH2;

Mv2:=Pv2/(0.4615*Tk2);

Ma2:=(P-Pv2)/(0.287*Tk2);

Enth1:=ma1*1.005*Tair1+(ma1)*HR1*(2501.3+1.88*Tair1)+(ng/1000)*4.18*Tair1;

Enth2:=ma1*1.005*Tair2+(ma1)*HR2*(2501.3+1.88*Tair2);

Enth2m1:=Enth2-Enth1;

Dair1:=(Mv1+Ma1);

Dair2:=(Mv2+Ma2);

denss:=Dair1;

Densp:=Dair2;

F:=9.80665*(Denss-densp);

a:=9.80665*((denss/densp)-1);

Tav:=(Tair1+Tair2)/2;

mf1:=Pv1/P;

mf2:=Pv2/P;

Enthv:=2501.3-2.4432443*(Tav-0.01);

Enthvkg:=ng*Enthv/3600;

str(Mv1:12:6,calcstr1);

str(Ma1:12:6,calcstr2);

str(Dair1:12:6,calcstr3);

str(HR1:12:6,calcstr4);

str(Mv2:12:6,calcstr5);

str(Ma2:12:6,calcstr6);

str(Dair2:12:6,calcstr7);

str(HR2:12:6,calcstr8);

str(RH2*100:12:2,calcstr9);

str(F:12:3,calcstr10);

str((F/9.80665):12:3,calcstr11);

str(a:12:3,calcstr12);

str(Enth1:12:3,calcstr13);

str(Enth2:12:3,calcstr14);

str(Enth2m1:12:3,calcstr15);

str(Enth2m1*(1000/3600):12:3,calcstr16);

str(Enthv:12:3,calcstr17);

str(Enthvkg:12:3,calcstr18);

str(mf1:12:3,calcstr19);

str(mf2:12:3,calcstr20);

form20.canvas.textout(0,100,'For surrounding air, mass of vapour Mv1 is: '+calcstr1+' kg/m^3.');

form20.canvas.textout(0,130,'For surrounding air, mass of air Ma1 is: '+calcstr2+' kg/m^3.');

form20.canvas.textout(0,160,'For surrounding air, density of air Dair1 is: '+calcstr3+' kg/m^3.');

form20.canvas.textout(0,190,'For surrounding air, humidity ratio HR1 is: '+calcstr4+' kg/kg.');

form20.canvas.textout(0,230,'For parcel, Mv2 is: '+calcstr5+' kg/m^3.');

form20.canvas.textout(0,260,'For parcel, Ma2 is: '+calcstr6+' kg/m^3.');

form20.canvas.textout(0,290,'For parcel, Dair2 is: '+calcstr7+' kg/m^3.');

form20.canvas.textout(0,320,'For parcel, HR2 is: '+calcstr8+' kg/kg.');

form20.canvas.textout(0,350,'For parcel at Tair2 deg C, relative humidity RH2 after adding water is: '+calcstr9+'%.');

form20.canvas.textout(0,390,'Buoyancy force on a 1 cubic metre parcel of air is: '+calcstr10+' N ('+calcstr11+' kg force).');

form20.canvas.textout(0,420,'Initial acceleration of the parcel of air is: '+calcstr12+' m/(s.s).');

form20.canvas.textout(0,460,'For 1 m^3 Enth1 is: '+calcstr13+' kJ.      Enth2 is: '+calcstr14+' kJ.');

form20.canvas.textout(0,490,'Enthalpy difference for the two situations is Enth2 - Enth1, which is: '+calcstr15+' kJ.');

form20.canvas.textout(0,520,'Enthalpy difference starting with 1000 m^3 is 1000(Enth2 - Enth1) is: '+calcstr16+' kWh.');

form20.canvas.textout(0,550,'Enthalpy of vaporisation at Tav=(Tair1+Tair2)/2 is: '+calcstr17+' kJ/kg.');

form20.canvas.textout(0,580,'Enthalpy of vaporisation at Tav=(Tair1+Tair2)/2 for ng kg is: '+calcstr18+' kWh.');

form20.canvas.textout(0,610,'For surrounding air, mole fraction of water vapour mf1 is: '+calcstr19+' For parcel, mf2 is: '+calcstr20+'.');

1: end;

Amount of basalt to react with CO2 in air

I believe we will need to take CO2, SO2, NOx, etc, out of the air with powdered basalt and other alkaline rocks. You can join my group https://mewe.com/group/5dca2395b0d95521915ec355
Here is Delphi computer code that will calculate approximately how much basalt powder is needed The inputs are T (air temperature in deg C), RH in %, P (atmospheric pressure in kPa - at the coast P is about 101 kPa), ppm (parts per million of CO2 (probably 415 or more parts).
Code is below:
label 1;
var
Dair,Ma,Tk,Mv,Pv,HR,P,RH,T,Psatw,Pdair,PCO2,MCO2,
Dairkg,PercCV,PercVV,ppm,molH2O,molCO2,thirdc,mbthirdc:extended;
errors1:boolean;
calcstr1,calcstr2,calcstr3,calcstr4,calcstr5,calcstr6,calcstr7,
calcstr8,calcstr9,calcstr10,calcstr11,calcstr12,calcstr13:string[30];
begin
errors1:=false;
form21.hide;
form21.show;
try
T:=strtofloat(form21.edit1.Text);
RH:=strtofloat(form21.edit2.Text);
P:=strtofloat(form21.edit3.Text);
ppm:=strtofloat(form21.edit4.Text);
except
errors1:=true;
end;
if (errors1=true) or
(T<0) or (T>70) or (RH<1) or (RH>100) or (P<10) or (P>150) or
(ppm<100) or (ppm>2500)
then begin
form21.canvas.textout(0,100,'CHECK ENTRIES.');
goto 1
end;
Tk:=T+273.15;
Psatw:=0.61121*exp((18.678-T/234.5)*T/(257.14+T));
Pv:=Psatw*RH/100;
Pdair:=(P-Pv);
PCO2:=Pdair*ppm/1000000;
{HR:=0.622*Pv/(P-Pv);}
Mv:=Pv/(0.4615*Tk);
Ma:=(P-Pv)/(0.287*Tk);
MCO2:=PCO2/(0.1889*Tk);
HR:=Mv/Ma;
Dair:=Mv+Ma;
PercCV:=(PCO2/P)*100;
PercVV:=(Pv/P)*100;
molH2O:=Pv/(0.00831447*Tk);
molCO2:=PCO2/(0.00831447*Tk);
thirdc:=MCO2*1000000*(1/3);
mbthirdc:=thirdC*3.33333;
str(Psatw:13:4,calcstr1);
str(Pv:13:4,calcstr2);
str(HR:13:6,calcstr3);
str(Mv*1000:13:3,calcstr4);
str(Ma*1000:13:3,calcstr5);
str(Dair*1000:13:3,calcstr6);
str(MCO2*1000:13:3,calcstr7);
str(MCO2*1000000:13:2,calcstr8);
str(PercCV:13:3,calcstr9);
str(PercVV:13:3,calcstr10);
str(molCO2:13:4,calcstr11);
str(molH2O:13:4,calcstr12);
str(mbthirdc:13:1,calcstr13);
form21.canvas.textout(0,120,'Saturation vapour pressure over water (Psatwater) is: '+calcstr1+' kPa.');
form21.canvas.textout(0,150,'Pv (actual vapour pressure over water) is: '+calcstr2+' kPa.');
form21.canvas.textout(0,180,'HR (humidity ratio over water) is: '+calcstr3+' kg/kg.');
form21.canvas.textout(0,210,'Mass of vapour in one cubic metre is: '+calcstr4+' grams.');
form21.canvas.textout(0,240,'Mass of DRY air in one cubic metre is: '+calcstr5+' grams.');
form21.canvas.textout(0,270,'Density of air (includes water vapour) is: '+calcstr6+' grams per cubic metre.');
form21.canvas.textout(0,300,'Mass of carbon dioxide in one cubic metre is: '+calcstr7+' grams.');
form21.canvas.textout(0,330,'Mass of carbon dioxide in one cubic kilometre is: '+calcstr8+' tonnes.');
form21.canvas.textout(0,360,'Carbon dioxide (volume percent) is: '+calcstr9+'%.');
form21.canvas.textout(0,390,'Water vapour (volume percent) is: '+calcstr10+'%.');
form21.canvas.textout(0,420,'Moles of carbon dioxide in one cubic metre is: '+calcstr11+' mol.');
form21.canvas.textout(0,450,'Moles of water vapour in one cubic metre is: '+calcstr12+' mol.');
form21.canvas.textout(0,480,'Mass of basalt needed to take out a third of the CO2 in one cubic km is: '+calcstr13+' tonnes.');
1: end;

Water from air Delphi2 Code

THE CODE BELOW IS FREEWARE:
T1 is the temperature of the surrounding air in deg C, RH1 is the relative humidity in percent, P1 is the atmospheric pressure in kPa (at sea level it is about 101.325 kPa), T2 is the temperature to which the air will be cooled, V1 is the volume of air you are going to cool in cubic metres. Code below:
label 1;
var
EnthV1MWh,EnthV1kWh,EnthV1,NokgdaV1,Enthd,Psatw1,Psatw2,Tk1,Tk2,Mvpa1,Enth1,Enth2,Ma1,Mv1,HR1,HR2,
Pv1,PV2,T1,RH1,P1,P2,T2,V1,kgDryAir,EdpkgDA,EnthDV,HR1mHR2,HR1mHR2V,kgpkWh,Td,RH,Psat,Pv,Tdew:extended;
errors1:boolean;
calcstr2,calcstr1,calcstr3,calcstr4,calcstr5,calcstr6,calcstr7,
calcstr8,calcstr9,calcstr10,calcstr11,calcstr12,calcstr13,calcstr14,
calcstr15,calcstr16,calcstr17,calcstr18:string[30];
begin
errors1:=false;
form2.hide;
form2.show;
try
T1:=strtofloat(form2.edit1.Text);
RH1:=strtofloat(form2.edit2.Text);
P1:=strtofloat(form2.edit3.Text);
T2:=strtofloat(form2.edit4.Text);
V1:=strtofloat(form2.edit5.Text);
except
errors1:=true;
end;
if (errors1=true) or
(T1<0.1) or (T1>100) or (RH1<0.1) or (RH1>100) or (P1<50) or (P1>150) or (T2<0.1) or
(T2>100) or (V1<0) or (T2>=T1-0.01)
then begin
form2.canvas.textout(0,100,'CHECK ENTRIES.');
goto 1
end;
Tk1:=T1+273.15;
Tk2:=T2+273.15;

Td:=T1;
RH:=RH1;
Psat:=0.61121*exp((18.678-Td/234.5)*Td/(257.14+Td));
Pv:=(RH/100)*Psat;
Tdew:=Td;
repeat
Tdew:=Tdew-0.001;
Psat:=0.61121*exp((18.678-Tdew/234.5)*Tdew/(257.14+Tdew));
until (Psat<=Pv);
str(Tdew:12:2,calcstr18);


Psatw1:=0.61121*exp((18.678-T1/234.5)*T1/(257.14+T1));
Pv1:=Psatw1*RH1/100;
HR1:=0.622*Pv1/(P1-Pv1);
Ma1:=(P1-Pv1)/(0.287*Tk1);
Enth1:=(1.005*T1)+HR1*(2501.3+1.88*T1);
Psatw2:=0.61121*exp((18.678-T2/234.5)*T2/(257.14+T2));
HR2:=Hr1;
P2:=P1;
if (Psatw2<PV1) then begin
Pv2:=Psatw2;
HR2:=0.622*Pv2/(P2-Pv2);
form2.canvas.textout(0,90,'Water condenses out. Dew point is:'+calcstr18+' deg C');
end;
Enth2:=(1.005*T2)+HR2*(2501.3+1.88*T2);
kgDryAir:=V1*Ma1;
EdpkgDA:=(Enth2-Enth1);
EnthDV:=kgDryAir*EdpkgDA;
HR1mHR2:=HR1-HR2;
HR1mHR2V:=kgDryAir*HR1mHR2;
kgpkWh:=(-1)*HR1mHR2V/(EnthDV/3600);
str(Enth1:13:4,calcstr1);
str(Enth2:13:4,calcstr2);
str(EdpkgDA:13:6,calcstr3);
str(Ma1:13:4,calcstr4);
str(kgDryAir,calcstr5);
str(EnthDV,calcstr6);
str((EnthDV/3600),calcstr7);
str((EnthDV/3600000),calcstr8);
str((HR1mHR2*1000):13:4,calcstr9);
str(HR1mHR2V,calcstr10);
str((HR1*1000):13:4,calcstr11);
str((HR2*1000):13:4,calcstr12);
str(EnthV1MWh,calcstr16);
str(kgpkWh:13:4,calcstr17);
form2.canvas.textout(0,120,'Enthalpy1 per kg dry air is: '+calcstr1+' kJ/kg dry air.');
form2.canvas.textout(0,150,'Enthalpy2 per kg dry air is: '+calcstr2+' kJ/kg dry air.');
form2.canvas.textout(0,180,'Enthalpy difference per kg dry air is: '+calcstr3+' kJ/kg dry air.');
form2.canvas.textout(0,210,'Mass of DRY air in one cubic metre of original air is: '+calcstr4+' kg.');
form2.canvas.textout(0,240,'Number of kg of dry air in V1 is: '+calcstr5+' kg.');
form2.canvas.textout(0,270,'FOR VOLUME V1: Heat added to volume V1 of air (Enthalpy difference for volume V1) is: '+calcstr6+' kJ.');
form2.canvas.textout(0,300,'FOR VOLUME V1: Heat added to volume V1 of air is: '+calcstr7+' kWh.');
form2.canvas.textout(0,330,'FOR VOLUME V1: Heat added to volume V1 of air is: '+calcstr8+' MWh.');
form2.canvas.textout(0,370,'HR1 - HR2 g/kg is: '+calcstr9+' grams of water vapour per kg dry air (CONDENSES OUT).');
form2.canvas.textout(0,400,'(HR1 - HR2 kg/kg)x(number of kg dry air in volume V1) is: '+calcstr10+' kg of water (CONDENSES OUT).');
form2.canvas.textout(0,440,'HR1 is: '+calcstr11+' g/kg.');
form2.canvas.textout(0,470,'HR2 is: '+calcstr12+' g/kg.');
form2.canvas.textout(0,520,'SUMMARY. Total mass of water condensing out of the air is: '+calcstr10+' kg.');
form2.canvas.textout(0,550,'SUMMARY: Total heat added to the air is: '+calcstr7+' kWh.');
form2.canvas.textout(0,580,'SUMMARY: kg of water produced per kWh of heat removed from air: '+calcstr17+' kg/kWh.');
form2.canvas.textout(0,610,'SUMMARY: Dew point temperature is: '+calcstr18+' deg C.');
1: end;

Removal of methane from the atmosphere

https://en.wikipedia.org/wiki/Atmospheric_methane#Removal_processes says " Furthermore, in an attempt to absorb the methane that is already being produced from landfills, experiments in which nutrients were added to the soil to allow methanotrophs to thrive have been conducted. These nutrient supplemented landfills have been shown to act as a small scale methane sink, allowing the abundance of methanotrophs to sponge the methane from the air to use as energy, effectively reducing the landfill's emissions." Also says: "Forest soils act as good sinks for atmospheric methane because soils are optimally moist for methanotroph activity, and the movement of gases between soil and atmosphere (soil diffusivity) is high " So forests could be good.

Humidification for rain

This triangular cloth-covered framework device shown will deflect a huge volume of air into the almost saturated layer just above the sea during the course of a day and will facilitate the evaporation of spray above the surface by creating turbulence with the downwards direction of the air. Just before a sea breeze develops (breeze from sea to land) the air that is being warmed on land expands upwards and outwards, keeping the sea breeze from developing. Warm dry air from this expanded volume of air will be mixed, by this device, into the spray and almost-saturated air region just above the sea just before the sea breeze develops. A deeper moist layer will then blow in with the sea breeze, facilitating rain.
If the spray was just rising a few cm without the device and is forced a few m upwards by the wind with the device there will be a big increase in evaporation. If a few cubic km of air per day is directed down by the device you will have a huge amount of moist air. Imagine the device is 1 km long, 1/100 km high and wind blows at 10 km/hour for 20 hours. Then 2 cubic km of air per day could be forced down.
When there is a drought the land gets hotter because of no evaporation (hot deserts are hotter than tropical forests where there is evaporation). When land gets hotter the air above the land gets hotter and then relative humidity of the air drops and the vapour pressure deficit (VPD) increases a lot. From what I have read plants generally like a vapour pressure deficit (VPD) of between 0.8 kPa and 1.2 kPa. This means that the vapour pressure inside the plant minus the vapour pressure in the air should be 0.8 to 1.2 kPa. Failing this there is wilting, etc. Even if you do not get rain it might be worthwhile humidifying the air around the coast or planting more trees

Wetter or drier?

Will the world become wetter or drier overall? The article https://www.aaas.org/news/drying-atmosphere-spurs-decline-vegetation-growth says: "The findings reveal that atmospheric water vapor is expected to further wane throughout the 21st century due to rising air temperatures and a decline in the evaporation of the world's oceans."
People might find it strange that as air temperatures increase the evaporation from oceans decreases, but here is an explanation: When it comes to evaporation from the sea, most evaporation occurs when the sea is hotter than the air above it. With global warming the land heats up more than the sea and so you might tend to get air that is relatively hotter than the sea blowing to sea. When hotter air is above the sea the sea cools the air above and the relative humidity (RH) of this air increases and less evaporation results into the air because RH is higher. When The sea is warmer than the air above it the sea heats the air above it and the RH of the air decreases and more evaporation occurs because RH is low. When you have cold sea near to hot land you can get "negative evaporation" - the water vapour condenses out of the hot air above the cold sea and fog occurs.
One factor that could increase the evaporation is increased downwelling sky radiation from hotter air, but the effect of increasing RH of hotter air above cold water appears to be a clear winner in certain regions at present.
Solution: A stagnant saturated layer builds up just above the sea surface so a sheet along the coast and above the sea that deflects air downwards and gives turbulence could partially solve the problem and increase evaporation. Also see https://journals.ametsoc.org/doi/full/10.1175/JCLI3519.1 
I have been dealing with evaporation equations for a while and I take the average of 5 equations (equations from Fitzgerald, Rohwer, Engineering Toolbox, Horton, Meyer) and here are some results: Suppose wind speed is 10 km per hour, RH=80% above the sea, sea temperature is 18 deg C and atmospheric pressure is 760 mm of mercury.
If we increase the air temperature above the sea we get this. 
Tair= 0 deg C then evaporation is 13 mm per day. 
Tair=10 deg C then evaporation is 9 mm per day. 
Tair= 20 deg C then evaporation is 1 mm per day. 
Tair=30 deg C then fog forms above water.

Cooling cities near the coast

For cities near the coast one could use mirrors to shade the ground and also reflect solar energy to old floating abandoned ships. The old ships will heat up and transfer infrared radiation to the the sea surface. Infrared radiation only penetrates the sea a few mm. This could cause clouds to form to shade and cool and bring rain.  Rising moist hot air from the sea could increase air circulation and reduce air pollution in cities.

Carbon dioxide and drowsy accidents

The world is drifting towards drowsiness and accident-proneness: If one calculates, using equations related to the Keeling Curve, one finds that by 2070 the concentration of carbon dioxide in the atmosphere will be about 700 ppm (at present it is around 400 ppm). By 2095 it will be about 1000 ppm. Now when the concentration becomes 1000 ppm people start to feel drowsy and will lose concentration. 
But another problem is that when atmospheric carbon dioxide levels are high and one enters an enclosure (room and so on) the carbon dioxide levels quickly reach dangerous levels because of breathing. So if there are 10 people in a room of volume 100 cubic metres and atmospheric concentration of carbon dioxide is 400 ppm then it will take about 22 minutes for levels to reach 1000 ppm. If you start with atmospheric levels of carbon dioxide at 700 ppm then it takes only about 11 minutes for levels to reach 1000 ppm.
Fires (which produce carbon dioxide) will become far more dangerous regarding breathing.

Density of breath and surrounding air

The Delphi 2 code (Pascal) below calculates the density of breath and of the surrounding air. If you enter a hollow or lift or tank and so on and your breath is more dense than the surrounding air it will sink and you could surround yourself with air that has a lot of carbon dioxide in. This could happen during heatwave conditions.
The inputs to the program are: Ts (temperature in deg C of the surrounding air, eg 38), RHs (relative humidity of the surrounding air in %, eg 34), P (the atmospheric air pressure in kPa, eg 100), PercO (the percentage of oxygen used up in breathing, eg 25), Tbr (the temperature of the breath in deg C, eg 37). The Delph1 2 code is below:
label 1;
var
PvBr,Mas,Mvs,Pvs,Psatws,PercO,Ts,Tsk,P,RHs,Tbr,Tbrk,Psatwbr,Denssa,Densbr,a,molDaBr,
MolNBr,MolOBri,MolOBrf,molCO2Bri,molCO2Brf,MolVBr,MolArBr,MolNeBr,MolHeBr,Mol,
MolKrBr,MolXeBr,VN,VO,VCO2,VV,VArg,VNe,VHe,VKr,VXe,MNbr,MOBrf,MCO2brf,MVBr,
MArBr,MNeBr,MHeBr,MKrBr,MXeBr,MBr,VolBr,TDI,Td,RHDI,DI,Pw,Tw,Aw,Bw,RHSw,Psat,
Tdew,Pvd,RH,LCL:extended;
errors1:boolean;
calcstr2,calcstr1,calcstr3,calcstr4,calcstr5,calcstr6,calcstr7,
calcstr8,calcstr9,calcstr10,calcstr11,calcstr12,calcstr13,calcstr14,calcstr15,calcstr16,
calcstr17,calcstr18,calcstr19,calcstr20,calcstr21,calcstr22,calcstr23,calcstr24:string[30];
begin
errors1:=false;
form16.hide;
form16.show;
try
Ts:=strtofloat(form16.edit1.Text);
RHs:=strtofloat(form16.edit2.Text);
P:=strtofloat(form16.edit3.Text);
PercO:=strtofloat(form16.edit4.Text);
Tbr:=strtofloat(form16.edit5.Text);
except
errors1:=true;
end;
if (errors1=true) or
(Ts<0) or (Ts>70) or (RHs<=0) or (RHs>100) or (P<10) or (P>150) or(percO<0) or (percO>90)
or (Tbr<0) or (Tbr>70) then begin
form16.canvas.textout(0,100,'CHECK ENTRIES.');
goto 1
end;
Tsk:=Ts+273.15;
Tbrk:=Tbr+273.15;
Psatws:=0.61121*exp((18.678-Ts/234.5)*Ts/(257.14+Ts));
Psatwbr:=0.61121*exp((18.678-Tbr/234.5)*Tbr/(257.14+Tbr));
Pvs:=Psatws*RHs/100;
Mvs:=Pvs/(0.4615*Tsk);
Mas:=(P-Pvs)/(0.287*Tsk);
PvBr:=Psatwbr;
molDaBr:=100-(PvBr/P)*100;
MolNBr:=78.03/100*moldaBr;
MolOBri:=20.99/100*molDaBr;
MolOBrf:=MolOBri-(percO/100*molObri);
MolCO2bri:=(0.033/100)*moldaBr;
MolCO2Brf:=molCO2bri+(percO/100*molObri);
molVbr:=100*Pvbr/P;
molArBr:=(0.94/100)*moldabr;
molNeBr:=(0.0015/100)*moldabr;
molHeBr:=(0.000524/100)*moldabr;
molKrBr:=(0.00014/100)*moldabr;
molXeBr:=(0.000006/100)*moldabr;
MNbr:=molNbr*0.028013;
MObrf:=molOBrf*0.031999;
MCO2Brf:=molCO2Brf*0.04401;
MVBr:=molVBr*0.018015;
MArBr:=molArBr*0.039948;
MNeBr:=molNeBr*0.020183;
MHeBr:=molHeBr*0.004003;
MKrBr:=molKrBr*0.08380;
MXeBr:=molXeBr*0.13130;
Denssa:=Mvs+Mas;
MBr:=MNbr+MOBrf+MCO2Brf+MVBr+MArBr+MNeBr+MHeBr+MKrBr+MXeBr;
VolBr:=100*0.00831447*Tbrk/P;
DensBr:=MBr/VolBr;
a:=9.80665*((denssa/densbr)-1);
VN:=100*MolNBr*(0.00831447*TBrk/P)/VolBr;
VO:=100*MolOBrf*(0.00831447*TBrk/P)/VolBr;
VCO2:=100*MolCO2Brf*(0.00831447*TBrk/P)/VolBr;
VV:=100*MolVBr*(0.00831447*TBrk/P)/VolBr;
VArg:=100*MolArBr*(0.00831447*TBrk/P)/VolBr;
VNe:=100*MolNeBr*(0.00831447*TBrk/P)/VolBr;
VHe:=100*MolHeBr*(0.00831447*TBrk/P)/VolBr;
VKr:=100*MolKrBr*(0.00831447*TBrk/P)/VolBr;
VXe:=100*MolXeBr*(0.00831447*TBrk/P)/VolBr;
str(Denssa:13:3,calcstr1);
str(DensBr:13:3,calcstr2);
str(a:13:2,calcstr3);
str(VN:13:4,calcstr4);
str(VO:13:4,calcstr5);
str(VCO2:13:4,calcstr6);
str(VV:13:4,calcstr7);
str(VArg:13:4,calcstr8);
str(VNe:13:4,calcstr9);
str(VHe:13:4,calcstr10);
str(VKr:13:4,calcstr11);
str(VXe:13:4,calcstr12);

form16.canvas.textout(0,100,'SURROUNDING AIR: DENSITY is: '+calcstr1+' kg/m^3.');
form16.canvas.textout(0,130,'BREATH:          DENSITY is: '+calcstr2+' kg/m^3.');
form16.canvas.textout(0,170,'BREATH: ACCELERATION (negative is down and positive is up) is: '+calcstr3+' m/s^2.');
form16.canvas.textout(0,210,'Breath % vol N2: '+calcstr4+'%.');
form16.canvas.textout(0,240,'Breath % vol O2: '+calcstr5+'%.');
form16.canvas.textout(0,270,'Breath % vol CO2: '+calcstr6+'%.');
form16.canvas.textout(0,300,'Breath % vol H20(g): '+calcstr7+'%.');
form16.canvas.textout(0,330,'Breath % vol Ar: '+calcstr8+'%.');
form16.canvas.textout(0,360,'Breath % vol Ne: '+calcstr9+'%.');
form16.canvas.textout(0,390,'Breath % vol He: '+calcstr10+'%.');
form16.canvas.textout(0,420,'Breath % vol Kr: '+calcstr11+'%.');
form16.canvas.textout(0,450,'Breath % vol Xe: '+calcstr12+'%.');
Td:=Ts;
TDI:=Td;
RHDI:=RHs;
DI:= (2*TDI) + (RHDI/100*TDI)+24;
Pw:=P/101.325;
Tw:=Td;
repeat
Tw:=Tw-0.001;
 Aw:=611.2*exp(17.502*Tw/(240.97+Tw))-66.8745*(1+0.00115*Tw)*Pw*(Td-Tw);
Bw:=6.112*exp(17.502*Td/(240.97+Td));
RHSw:=Aw/Bw;
until (RHSw<=rhs);
RH:=RHs;
Tdew:=Td;
Psat:=0.61121*exp((18.678-Td/234.5)*Td/(257.14+Td));
Pvd:=(RH/100)*Psat;
repeat
Tdew:=Tdew-0.001;
Psat:=0.61121*exp((18.678-Tdew/234.5)*Tdew/(257.14+Tdew));
until (Psat<=Pvd);
LCL:=125*(Td-Tdew);
str(DI:13:1,calcstr13);
str(Tw:13:2,calcstr14);
str(Tdew:13:2,calcstr15);
str(LCL:13:2,calcstr16);
form16.canvas.textout(0,490,'Discomfort Index: '+calcstr13+' (90 to 100 very uncomfortable, 100 to 110 extremely uncomfortable, >110 hazardous).');
form16.canvas.textout(0,520,'Wet Bulb T: '+calcstr14+' deg C');
form16.canvas.textout(0,550,'Dew point: '+calcstr15+' deg C.');
form16.canvas.textout(0,580,'LCL: '+calcstr16+' m.');

1: end;

Rain from dry and wet air layers

When wind blows up a mountain the air cools at about 9.8 deg C for every 1000 m rise (the dry adiabatic lapse rate for air that is not saturated). However, if the air is moist, one might have pockets where condensation occurs and this releases heat of condensation, keeping the air warmer. 
What sometimes happens is that one has a layer of moist air under a layer of drier air and as it blows up the mountain the top dry layer cools at 9.8 deg C for every 1000 m rise and the moist layer below might cool at an average of, say, 8 deg C per 1000m m rise.
So the bottom moist layer is continuously getting warmer relative to the top dry layer. This causes the bottom layer to continuously get less dense than the top layer and so the bottom rises by convection as well as by being blown up the mountain. The bottom layer could rise high enough by this method for rain to form, even if the mountains are low.
It does not take much to heat moist air, but it does take a lot of heat to evaporate water to make air moist. So if one could heat the air just above the sea, where it is moist, by infrared radiation (evaporation of spray would also occur), then one could have more rain even with low mountains. One could do this using solar power to heat dark objects on land and then reflect the infrared heat from the dark objects to the air above the sea surface. When the wind blows to land rain could result even with low mountains.
From
https://journals.ametsoc.org/doi/pdf/10.1175/1520-0442%281991%29004<1023%3AHPOTO>2.0.CO%3B2 
it appears that the RH is usually greatest at about 940 mb (about 600 m above the ocean). If one could increase RH to about 90% in the first 500 m or so above the ocean the above method could work well. 

Delphi Code Example

START
form5.canvas.font.size:=10;
form5.canvas.font.style:=[fsbold];
form5.canvas.font.color:=clblack;
form5.edit1.text:='';
form5.edit2.text:='';
form5.edit3.text:='';
form5.show;

PROCEDURE
label 1;
var
TdewC,T1C,T2C,Ht,LCL,RHd,Tdew,td,T1,T2,densair,m,T,Rh,Rh1,Rh2,P,V,Tsat,Pa,Pv,
Pvd,Psat,omega,tdp,pdp,Va,Vv,Pvs,Pas,omegas,Mv,Ma,Psat1,Psat2,
relh,Tw,LCLC,HTC,LCLT1:extended;
errors1:boolean;
calcstr8,calcstr2,calcstr1,calcstr3,calcstr4,calcstr5,calcstr6,calcstr7:string[30];
begin
errors1:=false;
form5.hide;
form5.show;
Memo1.clear;
Memo1.font.size:=12;
Memo1.font.style:=[fsbold];
Memo1.font.color:=clblack;
try
T1:=strtofloat(form5.edit1.Text);
Rh1:=strtofloat(form5.edit2.Text);
T2:=strtofloat(form5.edit3.Text);
except
errors1:=true;
end;
if (errors1=true) or
(t1<0.01) or (t1>69) or (rh1<0) or (rh1>100) or
(t2<0.1) or (t2>70) or (T2<T1+0.02)
then begin
form5.Memo1.Lines.Add('CHECK ENTRIES.');
goto 1
end;
T1C:=T1;
ht:=1000*(T2-T1)/3.3;
RH1:=RH1/100;
Psat1:=0.61121*exp((18.678-T1/234.5)*T1/(257.14+T1));
Pv:=Psat1*RH1;
Psat2:=0.61121*exp((18.678-T2/234.5)*T2/(257.14+T2));
Rh2:=Pv/Psat2;  {ie partial pressure of water vapour/psat2}
str(rh2*100:9:2,calcstr1);
form5.Memo1.lines.Add('Final relative humidity (at T2) is: '+calcstr1+'%.');
Tdew:=T1;
repeat
Tdew:=Tdew-0.001;
Psat:=0.61121*exp((18.678-Tdew/234.5)*Tdew/(257.14+Tdew));
until (Psat<=Pv);
str(Tdew:12:2,calcstr2);
form5.Memo1.Lines.Add('Dew point is: '+calcstr2+' deg C.');
1: end;

Shortened Water from Air code

T1 is the temperature of the surrounding air in deg C, RH1 is the relative humidity in percent, P1 is the atmospheric pressure in kPa (at sea level it is about 101.325 kPa), T2 is the temperature to which the air will be cooled, V1 is the volume of air you are going to cool in cubic metres. Code below:

label 1;
var
EnthV1MWh,EnthV1kWh,EnthV1,NokgdaV1,Enthd,Psatw1,Psatw2,Tk1,Tk2,Mvpa1,Enth1,Enth2,Ma1,Mv1,HR1,HR2,
Pv1,PV2,T1,RH1,P1,P2,T2,V1,kgDryAir,EdpkgDA,EnthDV,HR1mHR2,HR1mHR2V,kgpkWh,Td,RH,Psat,Pv,Tdew:extended;
errors1:boolean;
calcstr2,calcstr1,calcstr3,calcstr4,calcstr5,calcstr6,calcstr7,
calcstr8,calcstr9,calcstr10,calcstr11,calcstr12,calcstr13,calcstr14,
calcstr15,calcstr16,calcstr17,calcstr18:string[30];
begin
errors1:=false;
form3.hide;
form3.show;
try
T1:=strtofloat(form3.edit1.Text);
RH1:=strtofloat(form3.edit2.Text);
P1:=strtofloat(form3.edit3.Text);
T2:=strtofloat(form3.edit4.Text);
V1:=strtofloat(form3.edit5.Text);
except
errors1:=true;
end;
if (errors1=true) or
(T1<1) or (T1>60) or (RH1<=0) or (RH1>100) or (P1<20) or (P1>120) or (T2<1) or
(T2>60) or (V1<0)
then begin
form3.canvas.textout(0,100,'CHECK ENTRIES.');
goto 1
end;
Tk1:=T1+273.15;
Tk2:=T2+273.15;

Td:=T1;
RH:=RH1;
Psat:=0.61121*exp((18.678-Td/234.5)*Td/(257.14+Td));
Pv:=(RH/100)*Psat;
Tdew:=Td;
repeat
Tdew:=Tdew-0.001;
Psat:=0.61121*exp((18.678-Tdew/234.5)*Tdew/(257.14+Tdew));
until (Psat<=Pv);
str(Tdew:12:2,calcstr18);


Psatw1:=0.61121*exp((18.678-T1/234.5)*T1/(257.14+T1));
Pv1:=Psatw1*RH1/100;
HR1:=0.622*Pv1/(P1-Pv1);
Ma1:=(P1-Pv1)/(0.287*Tk1);
Enth1:=(1.005*T1)+HR1*(2501.3+1.82*T1);
Psatw2:=0.61121*exp((18.678-T2/234.5)*T2/(257.14+T2));
HR2:=Hr1;
P2:=P1;
if (Psatw2<PV1) then begin
Pv2:=Psatw2;
HR2:=0.622*Pv2/(P2-Pv2);
form3.canvas.textout(0,120,'Water condenses out. Dew point is:'+calcstr18+' deg C');
end;
Enth2:=(1.005*T2)+HR2*(2501.3+1.82*T2);
kgDryAir:=V1*Ma1;
EdpkgDA:=(Enth2-Enth1);
EnthDV:=kgDryAir*EdpkgDA;
HR1mHR2:=HR1-HR2;
HR1mHR2V:=kgDryAir*HR1mHR2;
kgpkWh:=(-1)*HR1mHR2V/(EnthDV/3600);
str((EnthDV/3600),calcstr7);
str(HR1mHR2V,calcstr10);
str(kgpkWh:13:4,calcstr17);
form3.canvas.textout(0,150,'Total mass of water condensing out of the air is: '+calcstr10+' kg.');
form3.canvas.textout(0,180,'Total heat added to the air is: '+calcstr7+' kWh.');
form3.canvas.textout(0,210,'Number of kg of water produced per kWh of heat removed from air: '+calcstr17+' kg/kWh.');
form3.canvas.textout(0,240,'Dew point temperature is: '+calcstr18+' deg C.');
1: end;

Water from pipe from sea to land

Sometimes, in dry areas, the sea nearby is warmer than the land at night. If one has a pipe with inlet over the sea and outlet over the land it is possible the warmer less dense sea air will rise in the pipe and water will condense out on the land side. At the same temperature moist air has a lower density than dry air. Here is computer code that will help one determine whether water will condense out (RH becomes greater than 100% - if that were possible).
T is in deg C, P is in kPa, RH is a percentage.
The inputs are initial temperature Ti above sea surface, initial relative humidity (RH) above sea surface, initial atmospheric pressure just above the sea surface (P1) and final temperature Tf (above land of desert), and P2 (at altitude of land of desert). Delphi computer code below:

label 1;
var
Tdewi,Tdewf,Pvf,LCL1,LCL2,Pvi,Psatwi,Psatwf,RHf,HR,RH,p1,p2,Ti,Tf,Hinc,q,lnx,x,
Psat,h,p:extended;
errors1:boolean;
calcstr6,calcstr5,calcstr4,calcstr2,calcstr1,calcstr3:string[30];
begin
errors1:=false;
form47.hide;
form47.show;
try
Ti:=strtofloat(form47.edit1.Text);
RH:=strtofloat(form47.edit2.Text);
P1:=strtofloat(form47.edit3.Text);
Tf:=strtofloat(form47.edit4.Text);
P2:=strtofloat(form47.edit5.Text);
except
errors1:=true;
end;
if (errors1=true) or (Ti<0) or (Ti>100) or (RH>100) or (RH<0) or (P1<2) or (P1>500) or (Tf<0)
or (Tf>100) or (P2<2) or (P2>500)
then begin
form47.canvas.textout(0,100,'CHECK ENTRIES.');
goto 1
end;
Psatwi:=0.61121*exp((18.678-Ti/234.5)*Ti/(257.14+Ti));
Pvi:=Psatwi*(RH/100);
HR:=0.622*Pvi/(P1-Pvi);
Psatwf:=0.61121*exp((18.678-Tf/234.5)*Tf/(257.14+Tf));
RHf:=(HR*P2)/((0.622+HR)*Psatwf);
Pvf:=RHf*Psatwf;

Tdewi:=Ti;
repeat
Tdewi:=Tdewi-0.001;
Psat:=0.61121*exp((18.678-Tdewi/234.5)*Tdewi/(257.14+Tdewi));
until (Psat<=Pvi);
str(Tdewi:12:2,calcstr3);

Tdewf:=Tf;
repeat
Tdewf:=Tdewf-0.001;
Psat:=0.61121*exp((18.678-Tdewf/234.5)*Tdewf/(257.14+Tdewf));
until (Psat<=Pvf);
str(Tdewf:12:2,calcstr4);

LCL1:=125*(Ti-Tdewi);
LCL2:=125*(Tf-Tdewf);
str(LCL1:15:2,calcstr5);
str(LCL2:15:2,calcstr6);

str((100*RHf):15:2,calcstr1);
str(HR:15:5,calcstr2);
If(RHf<1) then form47.canvas.textout(0,120,'Relative humidity (RH) AFTER PARCEL HAS MOVED is: '+calcstr1+'%.');
If (RHf<1) then form47.canvas.textout(0,150,'Humidity ratio (HR) remains the same as the air parcel moves: '+calcstr2+' kg/kg.');
If(RHf>1) then form47.canvas.textout(0,120,'Air has become SATURATED and FICTITIOUS relative humidity (RH) AFTER PARCEL HAS MOVED is: '+calcstr1+'%.');
form47.canvas.textout(0,190,'Dew point for initial Ti and P1 situation is: '+calcstr3+' deg C.');
if (RHf<1) then form47.canvas.textout(0,220,'Dew point for final Tf and P2 situation is '+calcstr4+' deg C.');
form47.canvas.textout(0,260,'LCL1 for initial Ti and P1 situation is: '+calcstr5+' m.');
if (RHf<1) then form47.canvas.textout(0,290,'LCL2 for final Tf and P2 situation is '+calcstr6+' m.');

1: end;

To get some plants to cover dry ground and reduce dust one could use a dew drip system and grow rows of plants, say with 20 m between rows. Some plants could absorb toxic matter. 
When cold or warmth needs to be transported a long way secondary refrigerants are used. Water is often used as a secondary refrigerant as it holds a lot of heat per kg. So why not cool water using the usual sort of refrigerant and run the cold water through long pipes with spikes on pointing downwards? Dew will form and run down the spikes giving drip irrigation. One could also use bends in the pipe to allow dew to drip down. To prevent frost run hot water through the pipes. The system could be powered by a wind turbine.
To get an idea of how much water one could get from the air you can use my Delphi computer code at 
https://www.facebook.com/groups/WaterFromAir/  or from
 https://airartist.blogspot.com/2019/02/i-have-finished-delphi-computer-program.html

Delphi code for water from air.

I have finished a Delphi computer program that will tell people if water will condense out of the air when the air is cooled. One could use wind power. Anyone or any institution may use it in any legal context. It also tells how much water will condense out and it tells one how much heat (in kJ and kWh) must be extracted from the air and so one can get an idea of the power requirements.
One could use an air conditioning unit to cool the air to the best temperature and there are do-it-yourself (DIY) sites that tell one how to do this -see https://www.instructables.com/id/DIY-Atmospheric-Water-Generator/
 Inputs are T1 (temperature of atmospheric air in deg C before cooling), RH1 (relative humidity in % of atmospheric air before cooling), P1 (atmospheric pressure in kPa which is assumed to remain the same during cooling), T2 (temperature in deg C to which air will be cooled), V1 (the volume in cubic metres  of air at T1 and RH1 and P1 that is to be cooled).
Note: It is a convention to say "enthalpy per kg of dry air," meaning the enthalpy related to the kg of dry air and the water vapour associated with that kg of dry air.
See code below:

label 1;

var

EnthV1MWh,EnthV1kWh,EnthV1,NokgdaV1,Enthd,Psatw1,Psatw2,Tk1,Tk2,Mvpa1,Enth1,Enth2,Ma1,Mv1,HR1,HR2,

Pv1,PV2,T1,RH1,P1,P2,T2,V1,kgDryAir,EdpkgDA,EnthDV,HR1mHR2,HR1mHR2V,kgpkWh,Td,RH,Psat,Pv,Tdew:extended;

errors1:boolean;

calcstr2,calcstr1,calcstr3,calcstr4,calcstr5,calcstr6,calcstr7,

calcstr8,calcstr9,calcstr10,calcstr11,calcstr12,calcstr13,calcstr14,

calcstr15,calcstr16,calcstr17,calcstr18:string[30];

begin

errors1:=false;

form2.hide;

form2.show;

try

T1:=strtofloat(form2.edit1.Text);

RH1:=strtofloat(form2.edit2.Text);

P1:=strtofloat(form2.edit3.Text);

T2:=strtofloat(form2.edit4.Text);

V1:=strtofloat(form2.edit5.Text);

except

errors1:=true;

end;

if (errors1=true) or

(T1<0.1) or (T1>100) or (RH1<0.1) or (RH1>99) or (P1<50) or (P1>150) or (T2<0.1) or

(T2>100) or (V1<0) or (T2>=T1)

then begin

form2.canvas.textout(0,100,'CHECK ENTRIES.');

goto 1

end;

Tk1:=T1+273.15;

Tk2:=T2+273.15;

Td:=T1;

RH:=RH1;

Psat:=0.61121*exp((18.678-Td/234.5)*Td/(257.14+Td));

Pv:=(RH/100)*Psat;

Tdew:=Td;

repeat

Tdew:=Tdew-0.001;

Psat:=0.61121*exp((18.678-Tdew/234.5)*Tdew/(257.14+Tdew));

until (Psat<=Pv);

str(Tdew:12:2,calcstr18);

Psatw1:=0.61121*exp((18.678-T1/234.5)*T1/(257.14+T1));

Pv1:=Psatw1*RH1/100;

HR1:=0.622*Pv1/(P1-Pv1);

Ma1:=(P1-Pv1)/(0.287*Tk1);

Enth1:=(1.005*T1)+HR1*(2501.3+1.88*T1);

Psatw2:=0.61121*exp((18.678-T2/234.5)*T2/(257.14+T2));

HR2:=Hr1;

P2:=P1;

if (Psatw2<PV1) then begin

Pv2:=Psatw2;

HR2:=0.622*Pv2/(P2-Pv2);

form2.canvas.textout(0,90,'Water condenses out. Dew point is:'+calcstr18+' deg C');

end;

Enth2:=(1.005*T2)+HR2*(2501.3+1.88*T2);

kgDryAir:=V1*Ma1;

EdpkgDA:=(Enth2-Enth1);

EnthDV:=kgDryAir*EdpkgDA;

HR1mHR2:=HR1-HR2;

HR1mHR2V:=kgDryAir*HR1mHR2;

kgpkWh:=(-1)*HR1mHR2V/(EnthDV/3600);

str(Enth1:13:4,calcstr1);

str(Enth2:13:4,calcstr2);

str(EdpkgDA:13:6,calcstr3);

str(Ma1:13:4,calcstr4);

str(kgDryAir,calcstr5);

str(EnthDV,calcstr6);

str((EnthDV/3600),calcstr7);

str((EnthDV/3600000),calcstr8);

str((HR1mHR2*1000):13:4,calcstr9);

str(HR1mHR2V,calcstr10);

str((HR1*1000):13:4,calcstr11);

str((HR2*1000):13:4,calcstr12);

str(EnthV1MWh,calcstr16);

str(kgpkWh:13:4,calcstr17);

form2.canvas.textout(0,120,'Enthalpy1 per kg dry air is: '+calcstr1+' kJ/kg dry air.');

form2.canvas.textout(0,150,'Enthalpy2 per kg dry air is: '+calcstr2+' kJ/kg dry air.');

form2.canvas.textout(0,180,'Enthalpy difference per kg dry air is: '+calcstr3+' kJ/kg dry air.');

form2.canvas.textout(0,210,'Mass of DRY air in one cubic metre of original air is: '+calcstr4+' kg.');

form2.canvas.textout(0,240,'Number of kg of dry air in V1 is: '+calcstr5+' kg.');

form2.canvas.textout(0,270,'FOR VOLUME V1: Heat added to volume V1 of air (Enthalpy difference for volume V1) is: '+calcstr6+' kJ.');

form2.canvas.textout(0,300,'FOR VOLUME V1: Heat added to volume V1 of air is: '+calcstr7+' kWh.');

form2.canvas.textout(0,330,'FOR VOLUME V1: Heat added to volume V1 of air is: '+calcstr8+' MWh.');

form2.canvas.textout(0,370,'HR1 - HR2 g/kg is: '+calcstr9+' grams of water vapour per kg dry air (CONDENSES OUT).');

form2.canvas.textout(0,400,'(HR1 - HR2 kg/kg)x(number of kg dry air in volume V1) is: '+calcstr10+' kg of water (CONDENSES OUT).');

form2.canvas.textout(0,440,'HR1 is: '+calcstr11+' g/kg.');

form2.canvas.textout(0,470,'HR2 is: '+calcstr12+' g/kg.');

form2.canvas.textout(0,520,'SUMMARY. Total mass of water condensing out of the air is: '+calcstr10+' kg.');

form2.canvas.textout(0,550,'SUMMARY: Total heat added to the air is: '+calcstr7+' kWh.');

form2.canvas.textout(0,580,'SUMMARY: kg of water produced per kWh of heat removed from air: '+calcstr17+' kg/kWh.');

form2.canvas.textout(0,610,'SUMMARY: Dew point temperature is: '+calcstr18+' deg C.');

1: end;