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-rwxr-xr-x2471/CH5/EX5.1/Ex5_1.sce17
-rwxr-xr-x2471/CH5/EX5.10/Ex5_10.sce29
-rwxr-xr-x2471/CH5/EX5.11/Ex5_11.sce18
-rwxr-xr-x2471/CH5/EX5.12/Ex5_12.sce19
-rwxr-xr-x2471/CH5/EX5.13/Ex5_13.sce38
-rwxr-xr-x2471/CH5/EX5.14/Ex5_14.sce45
-rwxr-xr-x2471/CH5/EX5.15/Ex5_15.sce35
-rwxr-xr-x2471/CH5/EX5.2/Ex5_2.sce43
-rwxr-xr-x2471/CH5/EX5.3/Ex5_3.sce19
-rwxr-xr-x2471/CH5/EX5.4/Ex5_4.sce23
-rwxr-xr-x2471/CH5/EX5.5/Ex5_5.sce20
-rwxr-xr-x2471/CH5/EX5.6/Ex5_6.sce35
-rwxr-xr-x2471/CH5/EX5.7/Ex5_7.sce18
-rwxr-xr-x2471/CH5/EX5.8/Ex5_8.sce17
-rwxr-xr-x2471/CH5/EX5.9/Ex5_9.sce27
15 files changed, 403 insertions, 0 deletions
diff --git a/2471/CH5/EX5.1/Ex5_1.sce b/2471/CH5/EX5.1/Ex5_1.sce
new file mode 100755
index 000000000..1322e523f
--- /dev/null
+++ b/2471/CH5/EX5.1/Ex5_1.sce
@@ -0,0 +1,17 @@
+clear ;
+clc;
+// Example 5.1
+printf('Example 5.1\n\n');
+printf('Page No. 110\n\n');
+
+// 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));
+printf('The thickness of scale is %.4f m',L_scale)
diff --git a/2471/CH5/EX5.10/Ex5_10.sce b/2471/CH5/EX5.10/Ex5_10.sce
new file mode 100755
index 000000000..4cb35f3c9
--- /dev/null
+++ b/2471/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,29 @@
+clear ;
+clc;
+// Example 5.10
+printf('Example 5.10\n\n');
+printf('Page No. 123\n\n');
+
+// 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
+printf('the quantity of water requried is %.3f kg/kg\n\n',w)
+
+M = m*w;// in kg/s
+printf('the total mass flow rate of water required is %.3f kg/s \n\n',M)
+
+M_d = M + m;// mass flow rate of desuperheated steam in kg/s
+V = M_d*V_s;// in m^3/s
+printf('the total mass flow rate of desuperheated steam required is %.4f m^3/s \n',V)
+// Deviation in answer due to some approximation in answer in the book
+
diff --git a/2471/CH5/EX5.11/Ex5_11.sce b/2471/CH5/EX5.11/Ex5_11.sce
new file mode 100755
index 000000000..08ded0b32
--- /dev/null
+++ b/2471/CH5/EX5.11/Ex5_11.sce
@@ -0,0 +1,18 @@
+clear ;
+clc;
+// Example 5.11
+printf('Example 5.11\n\n');
+printf('Page No. 130\n\n');
+
+// 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
+printf('The required heat is %.0f W',Q)
diff --git a/2471/CH5/EX5.12/Ex5_12.sce b/2471/CH5/EX5.12/Ex5_12.sce
new file mode 100755
index 000000000..4e58b0d9f
--- /dev/null
+++ b/2471/CH5/EX5.12/Ex5_12.sce
@@ -0,0 +1,19 @@
+clear ;
+clc;
+// Example 5.12
+printf('Example 5.12\n\n');
+printf('Page No. 130\n\n');
+
+// 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
+printf('The mass flow rate of dry saturated eutectic mixture is %.2f kg/s',m)
+
+
diff --git a/2471/CH5/EX5.13/Ex5_13.sce b/2471/CH5/EX5.13/Ex5_13.sce
new file mode 100755
index 000000000..91eecce20
--- /dev/null
+++ b/2471/CH5/EX5.13/Ex5_13.sce
@@ -0,0 +1,38 @@
+clear ;
+clc;
+// Example 5.13
+printf('Example 5.13\n\n');
+printf('Page No. 131\n\n');
+
+// 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 = [2.80 2.65 2.55 2.75]// in W/m^2-k
+Re = [117*10^3 324*10^3 159*10^3 208*10^3]//Reynolds number
+Pr = [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(1)^0.8*Pr(1)^0.3*K_d(1);// //W/m^2-K
+printf('The film heat transfer coefficient using Transcal N is %.0f W/sq.m K \n',h_T)// Deviation in answer due to direct substitution
+
+
+h_D = 0.0232 * Re(2)^0.8*Pr(2)^0.3*K_d(2);// //W/m^2-K
+printf('The film heat transfer coefficient using Dowtherm A is %.0f W/sq.m K \n\n',h_D)// Deviation in answer due to direct substitution
+
+
+h_M = 0.0232 * Re(3)^0.8*Pr(3)^0.3*K_d(3);// //W/m^2-K
+printf('The film heat transfer coefficient using Marlotherm S is %.0f W/sq.m K \n',h_M)// Deviation in answer due to direct substitution
+
+
+h_S = 0.0232 * Re(4)^0.8*Pr(4)^0.3*K_d(4);// //W/m^2-K
+printf('The film heat transfer coefficient using Santotherm 60 is %.0f W/sq.m K \n',h_S)// Deviation in answer due to direct substitution
+
+
+
+
diff --git a/2471/CH5/EX5.14/Ex5_14.sce b/2471/CH5/EX5.14/Ex5_14.sce
new file mode 100755
index 000000000..ab4a15249
--- /dev/null
+++ b/2471/CH5/EX5.14/Ex5_14.sce
@@ -0,0 +1,45 @@
+clear ;
+clc;
+// Example 5.14
+printf('Example 5.14\n\n');
+printf('Page No. 137\n\n');
+
+// 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
+printf('the required humidity is %.3f kg/kg \n',w)
+
+
+//(b) relative humidity
+R_H = 30;// in percentage
+printf('the required relative humidity in percentage is %.0f\n\n',R_H)
+
+//(c) the dew point
+T_w = 20;// in degree celcius
+printf('the required dew-point temperature is %.0f deg C\n',T_w)
+
+//(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
+printf('the humid heat is %.0f J/kg-K\n\n',S )
+
+//(e) the humid volume
+V_G = ((1/29)+(w/18))*22.41*((T2 + 273)/273);//in m^3/kg
+printf('the humid volume is %.3f m^3/kg \n',V_G)
+
+//(f) adiabatic process
+w_A = 0.020;// in kg/kg
+printf('the humidity of the mixture if saturated adiabatically is %.3f kg/kg \n\n',w_A)
+
+// (h) isothermal process
+w_i = 0.049;// in kg/kg
+printf('the humidity of the mixture if saturated isothermally is %.3f kg/kg \n',w_i)
+
+
+
diff --git a/2471/CH5/EX5.15/Ex5_15.sce b/2471/CH5/EX5.15/Ex5_15.sce
new file mode 100755
index 000000000..aeb2df810
--- /dev/null
+++ b/2471/CH5/EX5.15/Ex5_15.sce
@@ -0,0 +1,35 @@
+clear ;
+clc;
+// Example 5.15
+printf('Example 5.15\n\n');
+printf('Page No. 137\n\n');
+
+// 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
+printf('The required heat is %3.2f W \n',Q)// 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
+printf( 'The volumetric flow rate of initial air-water mixture is %3.2f m^3/s',V_f)
+
diff --git a/2471/CH5/EX5.2/Ex5_2.sce b/2471/CH5/EX5.2/Ex5_2.sce
new file mode 100755
index 000000000..2423b1969
--- /dev/null
+++ b/2471/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,43 @@
+clear ;
+clc;
+// Example 5.2
+printf('Example 5.2\n\n');
+printf('Page No. 113\n\n');
+
+// 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 = (%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
+printf(' Heat absorbed by water is %.0f W \n',Q)
+
+//(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
+printf('The film heat transfer coefficient is %.0f W/sq.m K\n',h)// Deviation in answer due to direct substitution and wrongly calculated in the text book
+
+
diff --git a/2471/CH5/EX5.3/Ex5_3.sce b/2471/CH5/EX5.3/Ex5_3.sce
new file mode 100755
index 000000000..b1199e2d4
--- /dev/null
+++ b/2471/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,19 @@
+clear ;
+clc;
+// Example 5.3
+printf('Example 5.3\n\n');
+printf('Page No. 117\n\n');
+
+// 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
+printf('The required heat is %.0f W',Q)
diff --git a/2471/CH5/EX5.4/Ex5_4.sce b/2471/CH5/EX5.4/Ex5_4.sce
new file mode 100755
index 000000000..fd1b65cb4
--- /dev/null
+++ b/2471/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,23 @@
+clear ;
+clc;
+// Example 5.4
+printf('Example 5.4\n\n');
+printf('Page No. 117\n\n');
+
+// 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;
+printf('The oil required is %.3f kg/s',Oil)
diff --git a/2471/CH5/EX5.5/Ex5_5.sce b/2471/CH5/EX5.5/Ex5_5.sce
new file mode 100755
index 000000000..8c30a645c
--- /dev/null
+++ b/2471/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,20 @@
+clear ;
+clc;
+// Example 5.5
+printf('Example 5.5\n\n');
+printf('Page No. 120\n\n');
+
+// 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
+printf('The required heat is %.0f W',Q)// Deviation in answer due to direct substitution and some approximation in answer in book
+
diff --git a/2471/CH5/EX5.6/Ex5_6.sce b/2471/CH5/EX5.6/Ex5_6.sce
new file mode 100755
index 000000000..3aff2922d
--- /dev/null
+++ b/2471/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,35 @@
+clear ;
+clc;
+// Example 5.6
+printf('Example 5.6\n\n');
+printf('Page No. 120\n\n');
+
+// 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));
+printf('the quantity of water to be used at 25 degree celcius is %.2f kg/s \n',y)
+
+
+//(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));
+printf('the quantity of water to be used at 30 degree celcius is %.2f kg/s \n',z)
+
+
+
+
+
diff --git a/2471/CH5/EX5.7/Ex5_7.sce b/2471/CH5/EX5.7/Ex5_7.sce
new file mode 100755
index 000000000..a97fc2aa1
--- /dev/null
+++ b/2471/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,18 @@
+clear ;
+clc;
+// Example 5.7
+printf('Example 5.7\n\n');
+printf('Page No. 121\n\n');
+
+// 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 = sqrt((Q*m)/(v*%pi));
+printf('the required diameter of pipe is %.3f m',d)
diff --git a/2471/CH5/EX5.8/Ex5_8.sce b/2471/CH5/EX5.8/Ex5_8.sce
new file mode 100755
index 000000000..91e7b44c4
--- /dev/null
+++ b/2471/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,17 @@
+clear ;
+clc;
+// Example 5.8
+printf('Example 5.8\n\n');
+printf('Page No. 122\n\n');
+
+// 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
+printf('The required heat is %.0f W',Q)// Deviation in answer due to direct substitution and some approximation in answer in book
diff --git a/2471/CH5/EX5.9/Ex5_9.sce b/2471/CH5/EX5.9/Ex5_9.sce
new file mode 100755
index 000000000..34f05e95f
--- /dev/null
+++ b/2471/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,27 @@
+clear ;
+clc;
+// Example 5.9
+printf('Example 5.9\n\n');
+printf('Page No. 122\n\n');
+
+// 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
+printf('the temperature of the final superheated steam at 1.5 bar is %.0f deg C',T2)
+
+
generated by cgit (git 2.34.1) at 2025-03-24 05:06:48 +0000