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clurt
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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
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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