Chapter 10 : Psychrometrics¶

Example 10.1 Page no : 464¶

In :
# Variables
t_db = 293.; 			#K
W = 0.0095; 			#kg/kg of dry air
p_t = 1.0132;

# Calculations and Results
print ("(i) Partial pressure of vapour")
p_v = p_t*W/(W+0.622);
print ("p_v = %.3f")% (p_v), ("bar")

p_vs = 0.0234; 			#bar; From steam tables corresponding to 20 0C
phi = p_v/p_vs;
print ("(ii)relative hmidity  = %.3f")% (phi)

print ("(iii) Dew point temperature")
t_dp = 13 + (14-13)/(0.01598 - 0.0150)*(0.01524-0.0150); 			#From stea table by interpolation
print ("t_dp = %.3f")% (t_dp), ("0C")
(i) Partial pressure of vapour
p_v = 0.015 bar
(ii)relative hmidity  = 0.651
(iii) Dew point temperature
t_dp = 13.245 0C

Example 10.2 Page no : 465¶

In :
# Variables
t_db = 290.; 			#K
phi = 0.6; 			#relative humidity
p_t = 1.01325; 			#bar
p_vs = 0.0194; 			#bar

# Calculations
p_v = phi*p_vs;
W = 0.622*p_v/(p_t - p_v);
t_dp = 9 + (10-9)*(0.01164-0.01150)/(0.01230 - 0.01150); 			#By interpolation from steam tables

# Results
print ("Specific Humidity = %.4f")% (W), ("kg/kg of dry air")

print ("dew point temperature  = "), (t_dp), ("0C")
Specific Humidity = 0.0072 kg/kg of dry air
dew point temperature  =  9.175 0C

Example 10.3 Page no : 465¶

In :
# Variables
phi = 0.55;
p_vs = 0.0425; 			#bar
p_t = 1.0132; 			#bar

# Calculations
p_v = phi*p_vs;
W = 0.622*p_v/(p_t-p_v);

#Specific humidity after removing o.oo4 kg of water vapour
Wnew = W-0.004;
p_v = p_t*Wnew/(Wnew+0.622);
p_vs = 0.0234; 			#bar

# Results
print ("(i) Relative humidity"),
phi = p_v/p_vs;
print ("phi = %.3f")%(phi)

print ("(ii) Dew point temperature")

print ("Corresponding to 0.0171 bar, from steam tables")
t_dp = 15.; 			#0C
print ("t_dp = %.3f")% (t_dp), ("0C")
(i) Relative humidity phi = 0.732
(ii) Dew point temperature
Corresponding to 0.0171 bar, from steam tables
t_dp = 15.000 0C

Example 10.4 Page no : 466¶

In :
# Variables
t_db = 35.; 			#0C
t_wb = 25.; 			#0C
p_t = 1.0132; 			#bar

#Corresponding to 25 0C in steam tables
p_vs_wb = 0.0317; 			#bar

# Calculations
p_v = p_vs_wb - (p_t - p_vs_wb)*(t_db - t_wb)/(1527.4 - 1.3*t_wb);

print ("(i) Specific humidity")
W = 0.622*p_v/(p_t-p_v);
print ("W = %.3f")% (W), ("kg/kg of dry air")

print ("(ii) Relative humidity")

#Corresponding to 35 0C, from steam tables
p_vs = 0.0563;

phi = p_v/p_vs;
print ("phi %.3f")% (phi)

print ("(iii) Vapour density")
R_v = 8314.3/18;
T_v = 308.; 			#K

rho_v = p_v*10**5/(R_v*T_v);
print ("rho_v = %.3f")% (rho_v),("kg/m**3")

print ("(iv) Dew point temperature")
t_dp = 21 + (22-21)*(0.0252-0.0249)/(0.0264-0.0249);
print ("t_dp %.3f")%(t_dp), ("0C")

print ("(v) Enthalpy of mixture per kg of dry air")
cp = 1.005;
h_g = 2565.3; 			#kJ/kg; corresponding to 35 0C
h_vapour = h_g + 1.88*(t_db - t_dp);

h = cp*t_db + W*h_vapour;
print ("h = %.3f")% (h), ("kJ/kg of dry air")
(i) Specific humidity
W = 0.016 kg/kg of dry air
(ii) Relative humidity
phi 0.446
(iii) Vapour density
rho_v = 0.018 kg/m**3
(iv) Dew point temperature
t_dp 21.200 0C
(v) Enthalpy of mixture per kg of dry air
h = 76.175 kJ/kg of dry air

Example 10.5 Page no : 467¶

In :
#For the air at 35 0C DBT and 60% R.H.
p_vs = 0.0563; 			#bar; Corresponding to 35 0C from stem tables

phi = 0.6;
p_t = 1.0132; 			#bar
cp = 1.005;
t_db = 35.; 			#0C
h_g = 2565.5; 			#kJ/kg
m1 = 1.; 			#kg
m2 = 2.; 			#kg
m = m1+m2;

# Calculations
p_v = phi*p_vs;
W1 = 0.622*p_v/(p_t-p_v);

#Corresponding to 0.0388 bar, from steam tables
t_dp = 26.+(27.-26)*(0.0338-0.0336)/(0.0356-0.0336);

h_vapour = h_g + 1.88*(t_db - t_dp);
h1 = cp*t_db+W1*h_vapour;

#For the air at 20°C DBT and 13°C dew point temperature :
p_v = 0.0150; 			#bar

W2 = 0.622*p_v/(p_t-p_v);
t_db = 20.; 			#0C
t_dp = 13.;
h_g = 2538.1; 			#kJ/kg
h_vapour = h_g + 1.88*(t_db - t_dp);
h2 = cp*t_db+W2*h_vapour;

#let enthalpy per kg of moist air be h
h = ((m1*h1/(1+W1)) + (m2*h2/(1+W2)))/m;
#Let Mass of vapour/kg of moist air be M
M = (m1*W1/(1+W1) + m2*W2/(1+W2))/m;
#Let specific humidity be denoted by SH
SH = M/(1-M);

# Results
print ("Specific humidity  = %.3f")% (SH), ("kg/kg of dry air")
Specific humidity  = 0.013 kg/kg of dry air

Example 10.6 Page no : 468¶

In :
# Variables
#For air at 20 0C and 75% R.H
p_vs = 0.0234; 			#bar
phi = 0.75;
p_t = 1.0132;
cp = 1.005;
t_db = 20.; 			#0C

# Calculations
p_v = phi*p_vs;
t_dp = 15 + (16-15)*(0.01755-0.017)/(0.0182-0.017);
W = 0.622*p_v/(p_t-p_v);

h_g = 2538.1 			#kJ/kg
h_vapour = h_g + 1.88*(t_db - t_dp);
h1 = cp*t_db + W*h_vapour;

# Calculations and Results
print ("(i) Relative humidity of heated air :")

#For air at 30°C DBT
p_vs = 0.0425; 			#bar; corresponding to 30 0C
phi = p_v/p_vs;
print ("Relative humidity = %.3f")% (phi*100), ("%")

print ("(ii) Heat added to air per minute")
h_g = 2556.3; 			#kJkg
t_db = 30.;
h2 = cp*t_db+W*h_vapour;
V = 90.; 			#m**3
R = 287.;
T = 293.; 			#K

m = (p_t-p_v)*V*10**5/R/T;

Amt = m*(h2-h1);
print ("Amount of heat added per minute = %.3f")% (Amt),("kJ")
(i) Relative humidity of heated air :
Relative humidity = 41.294 %
(ii) Heat added to air per minute
Amount of heat added per minute = 1070.942 kJ

Example 10.7 Page no : 469¶

In :
#For air at 35 0C DBT and 50% RH
p_vs = 0.0563; 			#bar; At 35 0C, from steam tables
phi = 0.5;
p_t = 1.0132;
t_db1 = 35.; 			#0C
t_dp1 = 23.; 			#0C
cp = 1.005;
R = 287.;

# Calculations
p_v = phi*p_vs;
W1 = 0.622*p_v/(p_t-p_v);
h_g1 = 2565.3; 			#kJ/kg

h_vapour = h_g1 + 1.88*(t_db1 - t_dp1);
h1 = cp*t_db1+W1*h_vapour;

# Results
print ("(i) R.H. of cooled air")
p_vs = 0.0317;
phi = p_v/p_vs;
print ("RH of cooled air = %.3f")% (phi*100), ("%")

print ("(ii) Heat removed from air")
h_g2 = 2547.2; 			#kJ/kg
t_db2 = 25.; 			#0C
t_dp2 = 23.; 			#0C
W2 = W1;
T = 308.; 			#K
V = 40.; 			#m**3

h_vapour = h_g2 + 1.88*(t_db2 - t_dp2);
h2 = cp*t_db2+W2*h_vapour;
m = (p_t-p_v)*10**5*V/R/T;

#Let Heat removed be denoted by H
H = m*(h1-h2);
print ("Heat removed  = %.3f")% (H), ("kJ")
(i) R.H. of cooled air
RH of cooled air = 88.801 %
(ii) Heat removed from air
Heat removed  = 477.209 kJ

Example 10.8 Page no : 470¶

In :
#For the air at 35°C DBT and 50% R.H.

# Variables
p_vs = 0.0563; 			#bar; At 35 0C, from steam tables
phi = 0.5;
p_v = phi*p_vs;
p_t = 1.0132; 			#bar

t_dp1 = 23.; 			#0C
t_db1 = 35.; 			#0C
W1 = 0.622*p_v/(p_t-p_v);
h_g1 = 2565.3; 			#kJ/kg
R = 287.;
cp = 1.005;

# Calculations and Results
h_vapour = h_g1 + 1.88*(t_db1 - t_dp1);
h1 = cp*t_db1+W1*h_vapour;

print ("(i) Relative humidity of out coming air and its wet bulb temperature.")

print ("Relative humidity of exit air is 100 per cent.")

t_wb = 20; 			#0C
print ("Wet bulb temperture = %.3f")%(t_wb), ("0C")

p_v = 0.0234; 			#bar
p_vs = p_v;
t_db2 = 20; 			#0C
h_g2 = 2538.1; 			#kJ/kg
t_dp2 = t_db2;

W2 = 0.622*p_v/(p_t-p_v);
h_vapour = h_g2 + 1.88*(t_db2 - t_dp2);
h2 = cp*t_db2+W2*h_vapour;

T = 308.; 			#K
V = 120.; 			#m**3

W = W1-W2; 			#Weight of water vvapour removed per kg of dry air
h = h1-h2; 			#Heat removed per kg of dry air
m = (p_t-p_v)*10**5*V/R/T;

print ("(ii) Capacity of the cooling coil in tonnes of refrigeration")
C = m*(h1-h2)*60/14000;
print ("Capacity  = %.3f")% (C), ("TR")

Amt = m*(W1-W2)*60;
print ("(iii)Amount of water removed per hour = %.3f")% (Amt), ("kg/h")
(i) Relative humidity of out coming air and its wet bulb temperature.
Relative humidity of exit air is 100 per cent.
Wet bulb temperture = 20.000 0C
(ii) Capacity of the cooling coil in tonnes of refrigeration
Capacity  = 13.678 TR
(iii)Amount of water removed per hour = 24.753 kg/h

Example 10.9 Page no : 471¶

In :
# Variables
p_vs = 0.0563; 			#bar
phi = 0.2;
p_v = phi*p_vs;
p_t = 1.0132; 			#bar

# Calculations and Results
W1 = 0.622*p_v/(p_t-p_v);

print ("(i) Dew point temperature")
#
t_dp = 8+(9.-8)*(0.01126-0.01072)/(0.01150-0.01072);
print ("dew point temperature = %.3f")% (t_dp), ("0C")

print ("(ii) Relative humidity of the exit air :")
p_vs_wb = 0.0170; 			#bar
p_vs = 0.0234; 			#bar
t_db = 20.; 			#0C
t_wb = 15.; 			#0C

p_v = p_vs_wb - (p_t-p_vs_wb)*(t_db-t_wb)/(1527.4-1.3*t_wb);
W2 = 0.622*p_v/(p_t-p_v);

RH = p_v/p_vs;
print ("Relative humidity = %.3f")% (RH)

p_v = 0.01126; 			#bar
R = 287.;
T = 308.; 			#K
V = 150.;

m = (p_t-p_v)*V*10**5/R/T;

print ("(iii) Amount of water vapour added to the air per minute")
amt = m*(W2-W1);
print ("Amount  = %.3f")% (amt), ("kg/min")
(i) Dew point temperature
dew point temperature = 8.692 0C
(ii) Relative humidity of the exit air :
Relative humidity = 0.585
(iii) Amount of water vapour added to the air per minute
Amount  = 0.261 kg/min

Example 10.10 Page no : 471¶

In :
# Variables
p_s = 0.0206; 			#bar
p_t = 1; 			#bar
p_s1 = 0.03782; 			#bar
W_2s = 0.622*p_s/(p_t-p_s);

cp = 1.005;
t_db2 = 18; 			#0C
t_db1 = 28; 			#0C

h_g2 = 2534.4; 			#kJ/kg
h_f2 = 75.6; 			#kJ/kg
h_g1 = 2552.6; 			#kJ/kg

# Calculations
W1 = (cp*(t_db1-t_db2) + W_2s*(h_g2-h_f2))/(h_g1-h_f2);
p_v1 = W1*p_t/(0.622+W1);
RH = p_v1/p_s1; 			#Relative humidity

# Results
print ("Relative humidity %.3f")% (RH)
Relative humidity 0.705

Example 10.11 Page no : 472¶

In :
# Variables
t_db1 = 38.; 			#0C
t_db2 = 18.; 			#0C
phi_1 = 0.75;
phi_2 = 0.85;
p_t = 1.; 			#bar
cp = 1.005;

#At 38 0C
p_vs = 0.0663; 			#bar
h_g1 = 2570.7; 			#kJ/kg

# Calculations
p_v = phi_1*p_vs;
W1 = 0.622*p_v/(p_t-p_v);

#At 18 0C
p_vs = 0.0206; 			#bar
h_g2 = 2534.4; 			#kJ/kg
h_f2 = 75.6; 			#kJ/kg
p_v = phi_2*p_vs;
W2 = 0.622*p_v/(p_t-p_v);

q = (W2*h_g2 - W1*h_g1) + cp*(t_db2-t_db1) + (W1-W2)*h_f2;

# Results
print ("Heat transfer rate = %.3f")% (q), ("kJ/kg of dry air")
Heat transfer rate = -74.052 kJ/kg of dry air

Example 10.12 Page no : 473¶

In :
# Variables
#At 38 0C
p_vs = 0.0663; 			#bar
h_g1 = 2570.7; 			#kJ/kg
phi = 0.25;
p_t = 1.0132;
p_v = phi*p_vs;
cp = 1.005;

#At 18 0C
h_g2 = 2534.4; 			#kJ/kg
p_vs = 0.0206; 			#bar
W1 = 0.622*p_v/(p_t-p_v);

t_db1 = 38.; 			#0C
t_db2 = 18.; 			#0C

# Results
W2 = (cp*(t_db1-t_db2) + W1*h_g1)/h_g2;

#amount of water added  = amt
amt = W2-W1;

# Results
print ("amt = %.5f")% (amt), ("kg/kg of dry air")

p_v2 = amt*p_t/(0.622+amt);
RH = p_v2/p_vs;

print ("Final relative humidity %.3f")% (RH)
amt = 0.00808 kg/kg of dry air
Final relative humidity 0.631

Example 10.13 Page no : 474¶

In :
import math

# Variables
#At 22 0c
p_vs = 0.0264; 			#bar
phi_3 = 0.55;
p_t = 1.0132; 			#bar

p_v3 = phi_3*p_vs;
W3 = 0.622*p_v3/(p_t-p_v3);

#At 3 0C
p_vs1 = 0.0076; 			#bar
p_v1 = p_vs1;

# Calculations and Results
W1 = 0.622*p_v1/(p_t-p_v1);
R = 287.;
T_3 = 295.; 			#K
v = R*T_3/(p_t-p_v3)/10**5;
m = (W3-W1)/v;

print ("(i)Mass of spray water required = %.6f")% (m), ("kg moisture/m**3")

t_dp = 12.5; 			#0C
cp = 1.005;
t_db3 = 22.; 			#0C
h_g3 = 2524.; 			#kJ/kg
h_vapour3 = h_g3 + 1.88*(t_db3 - t_dp);
W2 = 0.0047;
h_g2 = 2524; 			#kJ/kg
h4 = 41.87;

t_db2 = (cp*t_db3 + W3*h_vapour3 -W2*h_g2 + 1.88*W2*t_dp - (W3-W2)*h4)/(cp-W2*1.88);
print ("(ii) Temperature to which the air must be heated"),("t_db2 = %.3f")%(t_db2), ("0C")
(i)Mass of spray water required = 0.005122 kg moisture/m**3
(ii) Temperature to which the air must be heated t_db2 = 33.290 0C

Example 10.14 Page no : 476¶

In :
import math

# Variables
p_vs = 0.0206; 			#bar
phi = 0.6;
p_t = 1.013; 			#bar

# Calculations and Results
p_v1 = phi*p_vs;
p_a1 = p_t-p_v1;
V = 9.; 			#m**3
R = 287.;
T = 291.; 			#K

m_a = p_a1*10**5*V/R/T;

m_v1 = 0.0828; 			#kg/s

#At exit at 26 0C
p_vs = 0.0336; 			#bar
phi = 1;
p_v = p_vs;
W2 = 0.622*p_v/(p_t-p_v);
m_v2 = W2*m_a;
m = m_v2-m_v1;

print ("(i)Make-up water required = %.3f")%(m), ("kg/s")

m_w1 = 5.5; 			#kg/s
m_w2 = m_w1-m;
Wi = 4.75; 			#kJ/s
h_w1 = 184.3; 			#kJ/kg
h_a1 = 18.09; 			#kJ/kg
h_v1 = 2534.74; 			#kJ/kg
h_v2 = 2549; 			#kJ/kg
h_a2 = 26.13; 			#kJ/kg

h_w2 = (Wi + m_w1*h_w1 + m_a*h_a1 + m_v1*h_v1 - m_a*h_a2 - m_v2*h_v2)/m_w2;

#By interpolation, h_w2 corresponds to t
t = 26.7; 			#0C

print ("(ii)final temperature of water = %.3f")%(t), ("0C")
(i)Make-up water required = 0.147 kg/s
(ii)final temperature of water = 26.700 0C

Example 10.15 Page no : 478¶

In :
# Variables
m_water = 60000.; 			#kg/s
c = 4.186;
t1 = 30.; 			#0C
t2 = 35.; 			#0C

# Calculations
Q = m_water*c*(t2-t1);

h1 = 76.5; 			#kJ/kg
W1 = 0.016; 			#kg/kg of air
h2 = 92.5; 			#kJ/kg
W2 = 0.0246; 			#kg/kg of air
m_air = Q/(h2-h1);
A = m_air/10; 			#Quantity of air handled per fan

# Results
print ("Quantity of air handled per fan = %.3f")%(A), ("kg/h")

B = m_air*(W2-W1);

print ("Quantity of make up water = %.3f")% (B),("kg/h")
Quantity of air handled per fan = 7848.750 kg/h
Quantity of make up water = 674.992 kg/h

Example 10.17 Page no : 479¶

In :
# Variables
h1 = 35.4; 			#kJ/kg
h2 = 45.2; 			#kJ/kg
v_s1 = 0.8267; 			#m**3/kg
m_a = 241.9;

# Calculations and Results
RH = 41; 			# From chart
print ("(i) R.H. of heated air  = %.3f")%(RH), ("%")

WBT = 16.1; 			#0C
print ("(ii) WBT of heated air  = "), (WBT), ("ºC")

Q = m_a*(h2-h1);
print ("(iii) Heat added to air per minute  = "), (Q), ("kJ")
(i) R.H. of heated air  = 41.000 %
(ii) WBT of heated air  =  16.1 ºC
(iii) Heat added to air per minute  =  2370.62 kJ

Example 10.18 Page no : 480¶

In :
# Variables
h1 = 29.3; 			#kJ/kg
h2 = 42.3; 			#kJ/kg
h3 = h2;
t_db2 = 24.5; 			#0C
t_db1 = 12; 			#0C
v_s1 = 0.817; 			#m**3/kg
amt = 0.30; 			#Amount of air circulation m**3/min/person
capacity = 60; 			#Seating capacity of office
BF = 0.4; 			#By-pass factor
W3 = 8.6;
W1 = 6.8;

# Calculations and Results
m_a = amt*capacity/v_s1;

Q = m_a*(h2-h1)/60;
print ("(i) Heating capacity of the heating coil  = %.3f")%(Q), ("kW")

t_db4 = (t_db2-BF*t_db1)/(1-BF);
print ("Coil surface temperature  = %.3f")%(t_db4),("ºC")

c = m_a*(W3-W1)/1000*60;
print ("(ii) The capacity of the humidifier  = %.3f")% (c),("kg/h")
(i) Heating capacity of the heating coil  = 4.774 kW
Coil surface temperature  = 32.833 ºC
(ii) The capacity of the humidifier  = 2.379 kg/h

Example 10.19 Page no : 482¶

In :
# Variables
h1 = 82.5; 			#kJ/kg
h2 = 47.5; 			#kJ/kg
h3 = 55.7; 			#kJ/kg
h5 = 36.6; 			#kJ/kg
W1 = 19.6; 			#gm/kg
W3 = 11.8; 			#gm/kg
t_db2 = 17.6; 			#0C
t_db3 = 25.; 			#0C
v_s1 = 0.892; 			#m**3/kg
amt = 250.; 			#m**3/min

# Calculations and Results
m_a = amt/v_s1;
capacity = m_a*(h1-h2)*60/14000.;
print ("(i) The capacity of the cooling coil  = %.3f")%(capacity),("TR")

BF = (h2-h5)/(h1-h5);
print ("by-pass factor of the cooling coil  = %.3f")%(BF)

Q = m_a*(h3-h2)/60.;
print ("(ii) The heating capacity of the heating coil  = %.3f")%(Q),("kW")

BF = 0.3;
t_db6 = (t_db3-BF*t_db2)/(1-BF);
print ("surface temperature of heating coil  = %.3f")% (t_db6), ("C")

m = m_a*(W1-W3)*60/1000.;
print ("(iii) The mass of water vapour removed per hour  = %.3f")% (m), ("kg/h")
(i) The capacity of the cooling coil  = 42.040 TR
by-pass factor of the cooling coil  = 0.237
(ii) The heating capacity of the heating coil  = 38.303 kW
surface temperature of heating coil  = 28.171 C
(iii) The mass of water vapour removed per hour  = 131.166 kg/h