Chapter5-Heat transfer media

Ex1-pg110

In [1]:
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',)
Example 5.1
Page No. 110
The thickness of scale is  0.00  m

Ex2-pg113

In [2]:
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
Example 5.2
Page No. 113
 Heat absorbed by water is  183335.49  W 
The film heat transfer coefficient is 6725.39 W/sq.m K

Ex3-pg117

In [3]:
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 5.3
Page No. 117
The required heat is  3270500.00  W

Ex4-pg117

In [4]:
## 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 5.4
Page No. 117
The oil required is  0.2235  kg/s

Ex5-pg120

In [5]:
## 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
Example 5.5
Page No. 120
The required heat is  1664640.0000  W

Ex6-pg120

In [6]:
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 ')
Example 5.6
Page No. 120
the quantity of water to be used at 25 degree celcius is  9.23  kg/s 
the quantity of water to be used at 30 degree celcius is  6.86  kg/s 

Ex7-pg121

In [7]:
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')
Example 5.7
Page No. 121
the required diameter of pipe is  0.15  m

Ex8-pg122

In [8]:
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
Example 5.8
Page No. 122
The required heat is  23985000.00  W

Ex9-pg122

In [9]:
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')
Example 5.9
Page No. 122
the temperature of the final superheated steam at 1.5 bar is 158.00  deg C

Ex10-pg123

In [10]:
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
Example 5.10
Page No. 123
the quantity of water requried is  0.108  kg/kg
the total mass flow rate of water required is  0.049  kg/s 
the total mass flow rate of desuperheated steam required is 0.44  m**3/s 

Ex11-pg130

In [11]:
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')
Example 5.11
Page No. 130
The required heat is  194000.00  W

Ex12-pg130

In [12]:
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')
Example 5.12
Page No. 130
The mass flow rate of dry saturated eutectic mixture is 0.47  kg/s

Ex13-pg131

In [16]:
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 
Example 5.13
Page No. 131
('The film heat transfer coefficient using Transcal N is', 1552.2050049396955, ' W/sq.m K ')
The film heat transfer coefficient using Dowtherm A is  2472.48  W/sq.m K 
The film heat transfer coefficient using Marlotherm S is  1710.59  W/sq.m K 

The film heat transfer coefficient using Santotherm 60 is  2080.98  W/sq.m K 

Ex14-pg137

In [17]:
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 ')


 
Example 5.14
Page No. 137
the required humidity is  0.01  kg/kg 
the required relative humidity in percentage is  30.00 
the required dew-point temperature is  20.00  deg C
the humid heat is 1032.46  J/kg-K
the humid volume is  0.91  m**3/kg 
the humidity of the mixture if saturated adiabatically is  0.02  kg/kg 
the humidity of the mixture if saturated isothermally is  0.05  kg/kg 

Ex15-pg137

In [18]:
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')
Example 5.15
Page No. 137
The required heat is  237814.35  W
The volumetric flow rate of initial air-water mixture is  5.66  m^3/s