import math
## Example 5.1
print('Example 5.1');
print('Page No. 110');
## given
Q = 0.30*10**6;## Heat transfer rate in W/sq.m
T1 = 540;## Mean gas temperature in degree celcius
T2 = 207;## Steam temperature in degree celcius
K_tube = 40;## Thermal conductivity of tube in W/m-K
K_scale = 2.5 ;## Thermal conductivity of scale in W/m-K
L_tube = 4*10**-3;## Length of tube in m
## By Fourier equation and neglecting curvature effect, Q/A = [(T1- T2)/((L_tube/K_tube)+(L_scale/K_scale))]
L_scale = K_scale*(((T1-T2)/Q)-(L_tube/K_tube));
print'%s %.2f %s'%('The thickness of scale is ',L_scale,' m',)
import math
## Example 5.2
print('Example 5.2');
print('Page No. 113');
## given
T1 = 10;## in degree celcius
T2 = 70;## in degree celcius
d = 25*10**-3;## Inside diameter in m
v = 1.5;## veocity in m/s
Tm = (T1+T2)/2.;## Arithmetic Mean temperature in degree celcius
## At Tm, All physical properties of water is calculated by using steam table
##(a)Heat absorbed by water
p = 992.;## Density of water in kg/m**3 At Tm
A = (math.pi*d**2)/4.;## Area in m**2
m = p*v*A;## Mass flow rate in kg/s
h_70 = 293*10**3.;## Specific enthalpy of water in J/kg at 70 degree celcius(from steam table)
h_10 = 42*10**3.;## Specific enthalpy of water in J/kg at 10 degree celcius(from steam table)
Q = m*(h_70 - h_10);## in W
print'%s %.2f %s'%(' Heat absorbed by water is ',Q,' W ')
##(b) Film heat transfer
##At Tm, the following properites of water are found by using steam table
u = 650*10**-6.;## viscosity in Ns/m
Cp = 4180.;##Specific heat in J/kg-s
K = 0.632;## Thermal conductivity in W/m-s
Re = (d*v*p)/u;##Reynolds Number ## answer wrongly calculated in the text book
Pr = (Cp*u)/K;## Prandtl Number
Re_d = (Re)**0.8;
Pr_d = (Pr)**0.4;
## By Dittus-Boelter Equation
##Nu = 0.0232 * Re**0.8 Pr**0.4 = (hd)/K
Nu = 0.0232 * Re_d * Pr_d;## Nusselt Number
h = (Nu*K)/d;##W/m**2-K
print'%s %.2f %s'%('The film heat transfer coefficient is',h,'W/sq.m K')## Deviation in answer due to direct substitution and wrongly calculated in the text book
import math
## Example 5.3
print('Example 5.3');
print('Page No. 117');
## given
T1 = 25.;## in degree celcius
T2 = 212.;## in degree celcius
x = 0.96;## dryness fraction
m = 1.25;## Mass flow rate in kg/s
##from steam table
hL_212 = 907.*10**3;## Specific enthalpy at 212 degree celcius in J/kg
hL_25 = 105.*10**3;## Specific enthalpy at 25 degree celcius in J/kg
l_212 = 1890*10**3;## Latent heat of vapourisation at 212 degree celcius in J/kg
Q = m*((hL_212+(x*l_212))-hL_25);## in W
print'%s %.2f %s'%('The required heat is ',Q,' W')
## Example 5.4
print('Example 5.4');
print('Page No. 117');
## given
T = 25;## in degree celcius
x = 0.96;## dryness fraction
m = 3.15;## Mass flow rate in kg/s
CV = 42.6*10**6;## Calorific value in J/kg
P = 15;## Pressure in bar
n = 0.8;## Efficiency
##from steam table
hL_1 = 843*10**3;## Specific enthalpy in J/kg
hL_2 = 293*10**3;## Specific enthalpy in J/kg
l_1 = 1946*10**3;## Latent heat of vapourisation at 70 degree celcius in J/kg
Q = m*((hL_1+(x*l_1))-hL_2);## in W
Q_Ac = Q/n## Actual heat required in Watts
Oil = Q_Ac/CV;
print'%s %.4f %s'%('The oil required is ',Oil,' kg/s')
## Example 5.5
print('Example 5.5');
print('Page No. 120');
## given
T1 = 134;## in degree celcius
T2 = 100;## in degree celcius
x = 0.96;## dryness fraction
m = 0.75;## Mass flow rate in kg/s
##from steam table
hL_134 = 563*10**3;## Specific enthalpy at 134 degree celcius in J/kg
hL_100 = 419*10**3;## Specific enthalpy at 100 degree celcius in J/kg
l_134 = 2162*10**3;## Latent heat of vapourisation at 134 degree celcius in J/kg
Q = m*((hL_134+(x*l_134))-hL_100);## in W
print'%s %.4f %s'%('The required heat is ',Q,' W')## Deviation in answer due to direct substitution and some approximation in answer in book
import math
## Example 5.6
print('Example 5.6');
print('Page No. 120');
## given
x = 0.90;## dryness fraction
m = 0.25;## Mass flow rate in kg/s
P = 0.7;## pressure in bar
T1 = 10;## in degree celcius
##from steam table
h_10= 42*10**3;## Specific enthalpy of water at 10 degree celcius in J/kg
h_25 = 105*10**3;## Specific enthalpy of water at 25 degree celcius in J/kg
h_30 = 126*10**3;## Specific enthalpy of water at 30 degree celcius in J/kg
h_s = 2432*10**3;## Specific enthalpy of steam in J/kg
##(a)T2 = 25;
T2 = 25;## in degree celcius
## By heat balance, heat transfered at 10 degree celcius = heat gained at 25 degree celcius; "(m*h_s)+(h_10*y)= (m*h_25)+(h_25*y)"; where 'y' is the quqntity of water to be used at 25 degree celcius in kg/s
y = (m*(h_s-h_25)/(h_25-h_10));
print'%s %.2f %s'%('the quantity of water to be used at 25 degree celcius is ',y,' kg/s ')
##(b)T2 = 30;
T2 = 30;## in degree celcius
## By heat balance, heat transfered at 10 degree celcius = heat gained at 30 degree celcius; "(m*h_s)+(h_10*y)= (m*h_30)+(h_30*y)"; where 'z' is the quqntity of water to be used at 30 degree celcius in kg/s
z = (m*(h_s-h_30)/(h_30-h_10));
print'%s %.2f %s'%('the quantity of water to be used at 30 degree celcius is ',z,' kg/s ')
import math
## Example 5.7
print('Example 5.7');
print('Page No. 121');
## given
x = 0.97;## dryness fraction
m = 4.0;## Mass flow rate in kg/s
v = 40;## velocity in m/s
P = 10;## pressure in bar
##from steam table
Sp_vol = 0.194;## specific volume at 10 bar dry steam in m**3/kg
Q = Sp_vol*x*m## Volumetric flow rate of steam in m**3/s
d = math.sqrt((Q*m)/(v*math.pi));
print'%s %.2f %s'%('the required diameter of pipe is ',d,' m')
import math
## Example 5.8
print('Example 5.8');
print('Page No. 122');
## given
T1 = 25;## in degree celcius
T2 = 450;## in degree celcius
m = 7.5;## Mass flow rate in kg/s
##from steam table
hL_450 = 3303*10**3;## Specific enthalpy at 450 degree celcius in J/kg
hL_25 = 105*10**3;## Specific enthalpy at 25 degree celcius in J/kg
Q = m*(hL_450 - hL_25);## in W
print'%s %.2f %s'%('The required heat is ',Q,' W')## Deviation in answer due to direct substitution and some approximation in answer in book
import math
## Example 5.9
print('Example 5.9');
print('Page No. 122');
## given
P1 = 15;## Pressure at state 1 in bar
P2 = 1.5;## Pressure at state 2 in bar
T1 = 198;## in degree celcius
## as the process is adiabatic; => Q = 0; => ehthalpy at state1 = enthalpy at state 2
h_1 = 2789*10**3;## specific enthalpy at state 1 in J/kg
h_2 = h_1;##specific enthalpy at state 2 in J/kg
T3 = 150;## in degree celcius
T4 = 200;## in degree celcius
h_3 = 2773*10**3;## specific enthalpy at state 3 in J/kg
h_4 = 2873*10**3;## specific enthalpy at state 4 in J/kg
## Assuming a liner realtionship between temperature and enthalpy for the temperature range 150-200 degree celcius
h = ((h_4 - h_3)/(T4 - T3));## specific enthalpy per degree celcius in J/kg-degC
t = ((h_2 - h_3)/h);## in degree celcius
T2 = T3 + t;## in degree celcius
print'%s %.2f %s'%('the temperature of the final superheated steam at 1.5 bar is',T2,' deg C')
import math
## Example 5.10
print('Example 5.10');
print('Page No. 123');
## given
m = 0.45;## Mass flow rate in kg/s
P = 2.;## pressure in bar
T1 = 60.;## in degree celcius
T2 = 250.;## in degree celcius
h_s = 2971.*10**3;## Specific enthalpy of superheated steam in J/kg
h_d = 2706.*10**3;## Specific enthalpy of dry saturated steam in J/kg
h_e = h_s - h_d;##excess Specific enthalpy in J/kg
h = 251*10**3.;## in J/kg
V_s = 0.885;## specific volume of dry saturated steam at 2bar in m**3/kg
h_r = h_d- h;## heat required to convert water at 60 deg C into dry saturated steam at 2 bar
w = (h_e/h_r);## in kg/kg
print'%s %.3f %s'%('the quantity of water requried is ',w,' kg/kg')
M = m*w;## in kg/s
print'%s %.3f %s'%('the total mass flow rate of water required is ',M,' kg/s ')
M_d = M + m;## mass flow rate of desuperheated steam in kg/s
V = M_d*V_s;## in m**3/s
print'%s %.2f %s'%('the total mass flow rate of desuperheated steam required is',V,' m**3/s ')
## Deviation in answer due to some approximation in answer in the book
import math
## Example 5.11
print('Example 5.11');
print('Page No. 130');
## given
T1 = 180.;## in degree celcius
T2 = 350.;## in degree celcius
m = 0.5;## Mass flow rate in kg/s
##from steam table
hL_180 = 302*10**3.;## Specific enthalpy at 180 degree celcius in J/kg
hL_350 = 690*10**3.;## Specific enthalpy at 350 degree celcius in J/kg
Q = m*(hL_350 - hL_180);## in W
print'%s %.2f %s'%('The required heat is ',Q,' W')
import math
## Example 5.12
print('Example 5.12');
print('Page No. 130');
## given
T1 = 200.;## in degree celcius
T2 = 300.;## in degree celcius
m_l = 0.55;## Mass flow rate of liquid in kg/s
P = 3; ##pressure in bar
Cp = 2.34*10**3;## Mean haet capacity in J/kg-K
h = 272*10**3;## Latent heat of eutectic mixture at 3 bar
Q = m_l*Cp*(T2 -T1);## in Watts
m = Q/h;## in kg/s
print'%s %.2f %s'%('The mass flow rate of dry saturated eutectic mixture is',m,' kg/s')
import math
## Example 5.13
print('Example 5.13');
print('Page No. 131');
import numpy
## given
T = 300.;## in degree celcius
v = 2.;## velocity in m/s
d = 40.*10**-3;## diameter in m
## From the table 5.3 and 5.4 given in the book
K_d = numpy.array([2.80, 2.65, 2.55, 2.75])## in W/m**2-k
Re = numpy.array([117*10**3 ,324*10**3, 159*10**3 ,208*10**3])##Reynolds number
Pr = numpy.array([12, 4.50, 10.0, 7.3])##Prandtl Number
## By Dittus-Boelter Equation
##Nu = 0.0232 * Re**0.8*Pr**0.3 = (hd)/K
##h = 0.0232 * Re**0.8*Pr**0.3 *(K/d)
h_T = 0.0232 * Re[0]**0.8*Pr[0]**0.3*K_d[0];## ##W/m**2-K
print('The film heat transfer coefficient using Transcal N is',h_T,' W/sq.m K ')## Deviation in answer due to direct substitution
h_D = 0.0232 * Re[1]**0.8*Pr[1]**0.3*K_d[1];## ##W/m**2-K
print'%s %.2f %s'%('The film heat transfer coefficient using Dowtherm A is ',h_D,' W/sq.m K ')## Deviation in answer due to direct substitution
h_M = 0.0232 * Re[2]**0.8*Pr[2]**0.3*K_d[2];## ##W/m**2-K
print'%s %.2f %s'%('The film heat transfer coefficient using Marlotherm S is ',h_M,' W/sq.m K \n')## Deviation in answer due to direct substitution
h_S = 0.0232 * Re[3]**0.8*Pr[3]**0.3*K_d[3];## ##W/m**2-K
print'%s %.2f %s'%('The film heat transfer coefficient using Santotherm 60 is ',h_S,' W/sq.m K \n')## Deviation in answer due to direct substitution
import math
## Example 5.14
print('Example 5.14');
print('Page No. 137');
## given
T1 = 25;## Wet-bulb temperature in degree celcius
T2 = 40;##Dry-bulb temperature in degree celcius
##By using the humidity chart and steam tables for air-water mixtures at the given temperatures, the all following data can be obtained
##(a) humidity
w = 0.014;## in kg/kg
print'%s %.2f %s'%('the required humidity is ',w,' kg/kg ')
##(b) relative humidity
R_H = 30;## in percentage
print'%s %.2f %s'%('the required relative humidity in percentage is ',R_H,'')
##(c) the dew point
T_w = 20;## in degree celcius
print'%s %.2f %s'%('the required dew-point temperature is ',T_w,' deg C')
##(d) the humid heat
Cpa = 1.006*10**3;## Heat Capacity of bone dry air in J/kg-K
Cpwv = 1.89*10**3;## Heat Capacity of water vapour in J/kg-K
S = Cpa + (w*Cpwv);##in J/kg-K
print'%s %.2f %s'%('the humid heat is',S,' J/kg-K' )
##(e) the humid volume
V_G = ((1/29.)+(w/18.))*22.41*((T2 + 273.)/273.);##in m**3/kg
print'%s %.2f %s'%('the humid volume is ',V_G,' m**3/kg ')
##(f) adiabatic process
w_A = 0.020;## in kg/kg
print'%s %.2f %s'%('the humidity of the mixture if saturated adiabatically is ',w_A,' kg/kg ')
## (h) isothermal process
w_i = 0.049;## in kg/kg
print'%s %.2f %s'%('the humidity of the mixture if saturated isothermally is ',w_i,' kg/kg ')
import math
## Example 5.15
print('Example 5.15');
print('Page No. 137');
## given
T = 25.;## Wet-bulb temperature in degree celcius
T1 = 30.;##Dry-bulb temperature in degree celcius
V = 5.;## Volumetric flow rate of initial air-water mixture in m**3/s
T2 = 70.;## Final Dry-bulb temperature in degree celcius
##By using the humidity chart and steam tables for air-water mixtures at the given temperatures, the all following data can be obtained
w = 0.018;## humidity at 25/30 degree celcius in kg/kg
Cpa_1 = 1.00*10**3;## Heat Capacity of bone dry air at 30 degree celcius in J/kg-K
Cpwv_1 = 1.88*10**3;## Heat Capacity of water vapour at 30 degree celcius in J/kg-K
Cpa_2 = 1.008*10**3;## Heat Capacity of bone dry air at 70 degree celcius in J/kg-K
Cpwv_2 = 1.93*10**3;## Heat Capacity of water vapour at 70 degree celcius in J/kg-K
lo = 2.50*10**6;## Specifc Latent heat of vapourisation of water at 0 degree celcius in J/kg
S_1 = Cpa_1 + (w*Cpwv_1);## the humid heat at 30 degree celcius in J/kg-K
S_2 = Cpa_2 + (w*Cpwv_2);##the humid heat at 70 degree celcius in J/kg-K
hG_1 = ((S_1*T1) + (w*lo));##the specific enthalpy at 30 degree celcius in J/kg
hG_2 = ((S_2*T2) + (w*lo));##the specific enthalpy at 70 degree celcius in J/kg
VG_1 = ((1/29.)+(w/18.))*22.41*((T1 + 273)/273.);## Humid volume at 30 degree celcius in m**3/kg
m = V/VG_1;## Mass flow rate in kg/s
Q = m*(hG_2 - hG_1);## in Watts
print'%s %.2f %s'%('The required heat is ',Q,' W')## Deviation in answer is due to some approximation in calculation in the book
w_2 = w;## given in the question
VG_2 = ((1/29.)+(w_2/18.))*22.41*((T2 + 273)/273.);## Humid volume at 70 degree celcius in m**3/kg
V_f = m*VG_2;## in m**3/s
print'%s %.2f %s'%( 'The volumetric flow rate of initial air-water mixture is ',V_f,' m^3/s')