Chapter 15 : Psychrometrics

Example 15.1 Page No : 618

In [2]:
import math 

# Variables
Ps = 0.033363; 
P = 1.0132;

# Calculation and Results
W2 = (0.622*Ps)/(P-Ps);
hfg2 = 2439.9; hf2 = 109.1; cpa = 1.005;
hg = 2559.9; hw1 = hg;
T2 = 25+273; T1 = 32+273;
W1 = (cpa*(T2-T1)+(W2*hfg2))/(hw1-hf2);
Pw = ((W1/0.622)*P)/(1+(W1/0.622));
print "Specific humidity is ",round(W1,4),"kg vap./kg dry air"
print "Partial pressure of water vapour is",round(Pw,2),"bar"
print "Dew point temperature is",24.1,"degree" 			# saturation temperature at 0.03 bar

Psat = 0.048; 			# at 32 degree
fi = Pw/Psat;
print "Relative humidity is",round((fi*100),1),"%"
mu = (Pw/Ps)*((P-Ps)/(P-Pw));
print "Degree of saturation is %.3f"%mu
Pa = P-Pw;
Ra = 0.287; Tab = T1;
rho_a = (Pa*100)/(Ra*Tab);
print "Density of dry air is",round(rho_a,2),"kg/m3"
rho_w = W1*rho_a;
print "Density of water vapour is",round(rho_w,3),"kg/m3"
ta = 32.; 
tdb = 32.; 
tdp = 24.1;
h = cpa*ta + W1*(hg+1.88*(tdb-tdp));
print "Enthalpy of the mixture is",round(h,2),"kJ/kg"

# note: rounding off error is there.

Specific humidity is  0.0182 kg vap./kg dry air
Partial pressure of water vapour is 0.03 bar
Dew point temperature is 24.1 degree
Relative humidity is 60.1 %
Degree of saturation is 0.860
Density of dry air is 1.12 kg/m3
Density of water vapour is 0.02 kg/m3
Enthalpy of the mixture is 79.06 kJ/kg

Example 15.2 Page No : 620

In [14]:
# Variables
Ps = 2.339; P = 100;
W2 = (0.622*Ps)/(P-Ps);
hfg2 = 2454.1; hf2 = 83.96; cpa = 1.005;
hw1 = 2556.3;
T2 = 20.; 
T1 = 30.;

# Calculation
W1 = (cpa*(T2-T1)+(W2*hfg2))/(hw1-hf2);
Pw1 = ((W1/0.622)*P)/(1+(W1/0.622));
Ps1 = 4.246;
fi = (Pw1/Ps1);

# Results
print "Relative humidity is",round((fi*100),1),"%"
print "Humidity ratio of inlet mixture is",round(W1,4),"kg vap./kg dry air"

Relative humidity is 39.9 %
Humidity ratio of inlet mixture is 0.0107 kg vap./kg dry air

Example 15.3 Page No : 621

In [16]:
# Variables
Psat = 2.339;
fi3 = 0.50;
P = 101.3; 
cp = 1.005;

# Calculation
Pw3 = fi3*Psat;
Pa3 = P-Pw3;
W3 = 0.622*(Pw3/Pa3);
Psa1_1 = 0.7156;
Pw1 = 0.7156;
Pa1 = P-Pw1;
W1 = 0.622*(Pw1/Pa1); W2 = W1;
T3 = 293.; 
Ra = 0.287; 
Pa3 = 100.13;
va3 = (Ra*T3)/Pa3;
SW = (W3-W1)/va3;
t3 = 20.; 
tsat = 9.65; 
hg = 2518.; 
h4 = 10.;
t2 = ( W3*(hg+1.884*(t3-tsat))-W2*(hg-1.884*tsat) + cp*t3 - (W3-W2)*h4 )/ (cp+W2*1.884)

# Results
print "Mass of spray water required is",round(SW,5),"kg moisture/m3"
print "Temperature to which air must be heated is",round(t2,1),"degree"

Mass of spray water required is 0.00338 kg moisture/m3
Temperature to which air must be heated is 27.1 degree

Example 15.4 Page No : 622

In [18]:
# Variables
h1 = 82.; 
h2 = 52.; 
h3 = 47.; 
h4 = 40.;
W1 = 0.020; 
W2 = 0.0115; 
W3 = W2;
v1 = 0.887;
v = 3.33; 			# amount of free sir circulated

# Calculation
G = v/v1;
CC = (G*(h1-h3)*3600.)/14000; 			# in tonns
R = G*(W1-W3);

# Results
print "Capacity of the cooling coil in tonnes",round(CC,2),"tonnes"
print "Rate of water vapour removed is",round(R,4),"kg/s"

Capacity of the cooling coil in tonnes 33.79 tonnes
Rate of water vapour removed is 0.0319 kg/s

Example 15.5 Page No : 623

In [19]:
# Variables
W1 = 0.0058
W2 = 0.0187; 
h1 = 35.
h2 = 90.

# Calculation
G12 = 1./2; 			# G12 = G1./G2 
W3 = (W2+G12*W1)/(1+G12);
h3 = (2./3)*h2 + (1./3)*h1;

# Results
print ("Final condition of air is given by")
print "W3 = ",W3,"kg vap./kg dry air"
print "h3 = ",round(h3,2),"kJ/kg dry air"

Final condition of air is given by
W3 =  0.0144 kg vap./kg dry air
h3 =  71.67 kJ/kg dry air

Example 15.6 Page no : 624

In [3]:
# variables
T2 = 38.5           # c
o3 =11.2             # c

# calculations
h1h3 = 60.5 - 42.
w1w2  = 0.0153 - 0.0083  

# results
print "T2 = %.1f C"%T2
print "h1 - h3 = %.1f kJ/kg dry air"%h1h3
print "W1 - W2 = %.4f kg vap/kg dry air"%w1w2

T2 = 38.5 C
h1 - h3 = 18.5 kJ/kg dry air
W1 - W2 = 0.0070 kg vap/kg dry air

Example 15.7 Page No : 624

In [21]:
# Variables
h1 = 57.
h2 = h1;
h3 = 42.;
W1 = 0.0065; 
W2 = 0.0088; 
W3 = W2;
t2 = 34.5; 
v1 = 0.896;
n = 1500.; 			# seating capacity of hall
a = 0.3; 			# amount of out door air suplied

# Calculation
G = (n*a)/0.896 ; 			# Amount of dry air suplied
CC = (G*(h2-h3)*60)/14000; 			# in tonns
R = G*(W2-W1)*60;

# Results
print "Capacity of the cooling coil in tonnes",round(CC,2),"tonnes"
print "Capacity of humidifier",round(R,1),"kg/h"

Capacity of the cooling coil in tonnes 32.29 tonnes
Capacity of humidifier 69.3 kg/h

Example 15.8 Page No : 626

In [24]:
# Variables
twb1 = 15.2; 
twb2 = 26.7; 
tw3 = 30.;
h1 = 43.; 
h2 = 83.5; 
hw = 84.; 
mw = 1.15;
W1 = 0.0088; 
W2 = 0.0213;
hw3 = 125.8; 
hm = 84;
G = 1.;

# Calculation
hw34 = (G/mw)*((h2-h1)-(W2-W1)*hw);  			# hw3-hw4
tw4 = tw3-(hw34/4.19);
A = tw4-twb1;
R = tw3-tw4;
x = G*(W2-W1);

# Results
print "Temperature of water leaving the tower is",round(tw4,1),"degree"
print "Fraction of water evoporated is",x,"kg/kg dry sir"
print "Range of cooling water is",round(R,1),"degree"
print "Approach of cooling water is",round(A,1),"degree"

Temperature of water leaving the tower is 21.8 degree
Fraction of water evoporated is 0.0125 kg/kg dry sir
Range of cooling water is 8.2 degree
Approach of cooling water is 6.6 degree

Example 15.9 Page No : 627

In [26]:
# Variables
Psat1 = 0.01705; 
hg1 = 2528.9; 			# at 15 degree
Psat2 = 0.05628; 
hg2 = 2565.3; 			# At 35 degree
fi1 = 0.55;
Pw1 = fi1*Psat1;
fi2 = 1;
Pw2 = fi2*Psat2;
P = 1.;

# Calculation
W1 = (0.622*Pw1)/(P-Pw1);
W2 = (0.622*Pw2)/(P-Pw2);
MW = W2-W1;
t2 = 35; t1 = 15; 
m_dot = 2.78;
cpa = 1.005;
h43 = 35*4.187; 			# h4-h3
h5 = 14*4.187;
m_dot_w = (-(W2-W1)*h5 - W1*hg1 + W2*hg2 + cpa*(t2-t1))/(h43) ;
R = m_dot/m_dot_w ;
MW = (W2-W1)*R;
RWA = R*(1+W1);
R = 0.287; T = 288; 
V_dot = (RWA*R*T)/(P*1e02) ; 			# Pressure is in kilo Pascal

# Results
print "Make up water flow rate is",round(MW,3),"kg/s"
print "Volume flow rate of air is",round(V_dot,3),"m3/s"

Make up water flow rate is 0.129 kg/s
Volume flow rate of air is 3.437 m3/s