Calculate the moisture content of air

[clurt]

Homework Statement

(Simplified)
During tests at cliffside power station, the following data was recorded from the cooling tower.
A cooling tower has a 9 cell draft design, each diameter is 10m.
Data
D_o = 10m
V_exit = 14.4 m/s
T_exit(15 celsius) = 288.15K

Ambient conditions : T(15 celsius) = 288.15K , Φ = 30%
Flowrate = 7111.11 kg/s
T_entry(22 celsius) = 295.15K
1% of water lost to evaporation

>Calculate the moisture content(ω) of the air leaving the tower (kg/kg)?

Homework Equations

ω = m_v/m_a
ω = v_a/v_v

ω = 0.622 * (Pv)/(P - Pv)

Φ = (ωPa)/(0.622Ps)

The Attempt at a Solution

[/B]
Assumptions
P_atm = 101.3 kPa
Mass is conserved through system

Found:
A_o = 78.54 m^2
A_tot = 706.86 m^2
Psat (leaving tower, 15) = 1.7057 kPa in steam tables

Therefore Pv = 0.3*1.7057 = 0.51171 ??

using
"ω = 0.622 * (Pv)/(P - Pv)" I need to find P which is the exit pressure? So with the velocity and mass flow I have to find the exit pressure somehow.

A little lost at the moment, would love some direction.

Thanks

 

[BvU]

No quick responses, so let me try to help (not an expert, just a physicist).
You're not the only one who is a little lost in this exercise. Is it a subsection of a bigger one? There really is a lot of info.
I assume there are 9 cells of 10 m in diameter, so the volume flow at the exit is $9 \, \pi D_0^2/4 \times 10$ m/s $= 7070$ m³/s . This air contains the water that came in with the ambient air plus 1% of the 7111.11 kg/s (where this six-digit suggested accuracy comes from is a mystery to me) of the water to be cooled.

So:
find kg/s of water coming in with the ambient air
add kg/s of water evaporated
find exhaust temperature of moist air (hard part - see below)
apply your relevant equations

(ps I don't understand or believe $\omega = {v_a\over v_v}$ )
----

Probably overkill:

Basically the cooling tower is a co-current heat exchanger with a phase change and mixing of part of one inlet to the other outlet. So to do this correctly you need to solve a coupled mass- and heat balance problem. Hefty.

I don't think you can expect the moist air exit temperature to be equal to the ambient temperature (*), but I expect it to be a good approximation to let the exit pressure be one atmosphere.

For the heat balance you'll need an estimate for the heat capacity of moist air, the heat of evaporation, the density of moist air etc. (links just to give an impression).

(*) it is warmer because it has to rise up and there is a driving force for heat exchange for a finite-size device. Another reason I see in a rough heat balance: power for cooling the water 7 C is less than heat of evaporation plus heat to warm up even the dry air 7 C.

 

[clurt]

No quick responses, so let me try to help (not an expert, just a physicist).
You're not the only one who is a little lost in this exercise. Is it a subsection of a bigger one? There really is a lot of info.
I assume there are 9 cells of 10 m in diameter, so the volume flow at the exit is $9 \, \pi D_0^2/4 \times 10$ m/s $= 7070$ m³/s . This air contains the water that came in with the ambient air plus 1% of the 7111.11 kg/s (where this six-digit suggested accuracy comes from is a mystery to me) of the water to be cooled.

So:
find kg/s of water coming in with the ambient air
add kg/s of water evaporated
find exhaust temperature of moist air (hard part - see below)
apply your relevant equations

(ps I don't understand or believe $\omega = {v_a\over v_v}$ )
----

Probably overkill:

Basically the cooling tower is a co-current heat exchanger with a phase change and mixing of part of one inlet to the other outlet. So to do this correctly you need to solve a coupled mass- and heat balance problem. Hefty.

I don't think you can expect the moist air exit temperature to be equal to the ambient temperature (*), but I expect it to be a good approximation to let the exit pressure be one atmosphere.

For the heat balance you'll need an estimate for the heat capacity of moist air, the heat of evaporation, the density of moist air etc. (links just to give an impression).

(*) it is warmer because it has to rise up and there is a driving force for heat exchange for a finite-size device. Another reason I see in a rough heat balance: power for cooling the water 7 C is less than heat of evaporation plus heat to warm up even the dry air 7 C.

Really appreciate the response.

Intuitively I did something similar. 1% of 7111.11 is 71.11 kg.

Therefore (71.11 + density of water in ambient conditions*V) / (Density of dry air * V)

> (71.11 + 0.00321* 7300 m^3)/(1.201*7300m^3)

= approx 0.00107 kg/kg airI figured if 71.11kg is evaporated every second and it travels at 10.4 m/s the volume to work with is 10.4 * Area

Also here is the full context of the question. (scared if I post it people think I am just outsourcing my homework)

During tests at Cliffside Power Station in North Carolina, the following data was recorded from the cooling tower. The cooling tower was a 9-cell forced-draft design, with each cell having an exit diameter of 10m. Exit velocity of air from the tower was 10.4m/s. Ambient conditions were 15oC, Φ=30%. Condenser cooling water entered the tower at 22oC and exited at 15oC, with a flowrate of 25.6x106 kg/hr. 1% of the water was lost to evaporation. Clearly stating any assumptions that you make, calculate the moisture content of the air leaving the tower in kg/kg. Show that the temperature of the air leaving the tower must have been about 19oC