Calculate humidity from dry/Wet bulb temp

[dhkdeoen]

Homework Statement

Dry-bulb Temperature 46'C
Wet-bulb Temperature 34'C
P=101.3kPa

with these, I have to 'calculate' Specific Humidity(w), Relative humidity(φ), Specific Enthalpy(h)

Homework Equations

Specific Humidity
w=0.622*Pv/Pa
when Pa=P-Pv

Relative Humidity
φ=w*P/[(0.622+w)*Pg]

Specific Enthalpy
h=ha+w*hg

for this problem, I used approximated values based on temperature for ha.
so
ha=1.005[kJ/KgC]*T[C]
ha[kJ/KG]

The Attempt at a Solution

P is 101.3
T is 46'C

from Table A-4, Pv is 10.14kPa at 46'C

w=(0.622*10.14)/(101.3-10.14)=0.069
6.9%

φ=0.069*101.3/[(0.622+0.069)*10.14)=0.9976
99.76%
this is when I start to feel a bit weirdh=1.005*46+0.069*2584.18
=224.54kJ/kgThen I checked my Psychrometric Chart, and found out this makes no sense. Where did I go wrong? or am I reading the the chart wrong?

 

[rude man]

Since your wet bulb temp is close to freezing the curves on the psychrometric chart are at one end so it's a bit hard to get accurate readings, but I got
rel humidity ~ 33%
enthalpy ~ 12.5 BTU per lb. dry air. Sorry these units are not SI.
humidity mass ratio ~ .004 moisture/dry air.
Don't know about your equations but I can tell from their simplicity that they are rough approximations to the chart numbers.

 

[SteamKing]

Since your wet bulb temp is close to freezing the curves on the psychrometric chart are at one end so it's a bit hard to get accurate readings, but I got
rel humidity ~ 33%
enthalpy ~ 12.5 BTU per lb. dry air. Sorry these units are not SI.
humidity mass ratio ~ .004 moisture/dry air.
Don't know about your equations but I can tell from their simplicity that they are rough approximations to the chart numbers.

Wet bulb temperature is 34° C. That's not close to freezing. In fact, it's rather balmy. I think you got your temperature units mixed up.

 

[SteamKing]

Homework Statement

Dry-bulb Temperature 46'C
Wet-bulb Temperature 34'C
P=101.3kPa

with these, I have to 'calculate' Specific Humidity(w), Relative humidity(φ), Specific Enthalpy(h)

Homework Equations

Specific Humidity
w=0.622*Pv/Pa
when Pa=P-Pv

Relative Humidity
φ=w*P/[(0.622+w)*Pg]

Specific Enthalpy
h=ha+w*hg

for this problem, I used approximated values based on temperature for ha.
so
ha=1.005[kJ/KgC]*T[C]
ha[kJ/KG]

The Attempt at a Solution

P is 101.3
T is 46'C

from Table A-4, Pv is 10.14kPa at 46'C

w=(0.622*10.14)/(101.3-10.14)=0.069
6.9%

φ=0.069*101.3/[(0.622+0.069)*10.14)=0.9976
99.76%
this is when I start to feel a bit weirdh=1.005*46+0.069*2584.18
=224.54kJ/kgThen I checked my Psychrometric Chart, and found out this makes no sense. Where did I go wrong? or am I reading the the chart wrong?

It's not clear what Table A-4 is. Can you link to it?

Here is a psychrometric chart in SI Units:

http://www.uigi.com/UIGI_SI.PDF

Your dry bulb/wet bulb temps are almost off this chart.

I would say the enthalpy is about 122 kJ/kg of dry air and the relative humidity is about 45%

 

[rude man]

Wet bulb temperature is 34° C. That's not close to freezing. In fact, it's rather balmy. I think you got your temperature units mixed up.

Well you're absolutely right. I'll take another shot at it soon.

 

[rude man]

With the correct temperatures I got about 38% rel. humidity, 0.025 kg moisture /kg dry air, and H ~ 56 BTU//lb dry air = 129 kJ/kg. On my chart I had to estimate enthalpy a bit.