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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3733/CH22 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
| download | Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.gz Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.bz2 Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.zip | |
initial commit / add all books
Diffstat (limited to '3733/CH22')
30 files changed, 1081 insertions, 0 deletions
diff --git a/3733/CH22/EX22.1/Ex22_1.sce b/3733/CH22/EX22.1/Ex22_1.sce new file mode 100644 index 000000000..63c5abcf8 --- /dev/null +++ b/3733/CH22/EX22.1/Ex22_1.sce @@ -0,0 +1,35 @@ +// Example 22_1
+clc;funcprot(0);
+//Given data
+p_1=30;// The boiler pressure in bar
+p_2=1;// The condenser pressure in bar
+
+//Calculation
+//(a)
+// From steam tables, at pressure P_b=30 bar
+h_1=2796;// kJ/kg
+//For finding the dryness-fraction of steam at the point 'c',we can equate the entropies.
+// At pressure 30 bar=At pressure 1 bar
+// From steam tables, at pressure P_1=30 bar and P_2=1 bar
+T_s1=232.8;//°C
+T_s2=99.1;//°C
+h_f2=414.6;// kJ/kg
+h_fg1=1797;// kJ/kg
+h_fg2=2253;// kJ/kg
+v_f2=0.001043;// m^3/kg
+// Assume x_2=y(1)
+function[X]=drynessfraction(y)
+ X(1)=((2.3026*log10((T_s2+273)/273))+((y(1)*h_fg2)/(T_s2+273)))-((2.3026*log10((T_s1+273)/273))+(h_fg1/(T_s1+273)));
+endfunction
+y=[0.1];
+z=fsolve(y,drynessfraction);
+x_2=z(1);
+//x_2=z(1);// Dryness fraction
+h_2=h_f2+(x_2*h_fg2);// kJ/kg
+n_r1=((h_1-h_2)/(h_1-h_f2))*100;// The thermal efficiency of the cycle without feed pump work in %
+//(b)
+W_p=(v_f2*(p_1-p_2)*10^5)/1000;// kJ
+n_r2=(((h_1-h_2)-W_p)/(h_1-(h_f2+W_p)))*100;// The thermal efficiency of the plant feed pump work in %
+printf('\nThe thermal efficiency of the cycle without feed pump work=%0.2f percentage \nThe thermal efficiency of the cycle with feed pump work=%0.2f percentage',n_r1,n_r2);
+// The answer vary due to round off error
+
diff --git a/3733/CH22/EX22.10/Ex22_10.sce b/3733/CH22/EX22.10/Ex22_10.sce new file mode 100644 index 000000000..274e89dbb --- /dev/null +++ b/3733/CH22/EX22.10/Ex22_10.sce @@ -0,0 +1,37 @@ +// Example 22_10
+clc;funcprot(0);
+//Given data
+p_2=100;// bar
+T_1=500;// °C
+p_3=11.5;// bar
+p_5=0.05// bar
+n_i1=85;// Isentropic efficiency of each stage expansion in %
+n_i2=80;//Isentropic efficiency of one stage expansion with no reheat in %
+P=100;// The capacity of the plant in MW
+
+//Calculation
+//(a)
+// In this case the processes are shown in Fig.Prob.22.10(b)
+// From h-s chart:
+h_2=3370;// kJ/kg
+h_3aa=2860;// kJ/kg
+h_4=3500;// kJ/kg
+h_5aa=2530;// kJ/kg
+// From steam tables, at 0.06 bar
+h_f6=137.6;// kJ/kg
+W_t1=h_2-h_3aa;// (H.P turbine) kJ/kg
+W_t2=h_4-h_5aa;// (L.P turbine) kJ/kg
+Q_b=h_2-h_f6;// Heat supplied in the boiler in kJ/kg
+Q_r=h_4-h_3aa;// Heat supplied in the reheater in kJ/kg
+n_a=((W_t1+W_t2)/(Q_b+Q_r))*100;// Efficiency of the cycle iin %
+m_s=(P*10^3)/(W_t1+W_t2);// The mass flow of steam per scond in kJ/sec
+m_sa=(m_s*3600)/1000;// tons/hr
+
+//(b)
+// In this case the processes are shown in Fig.Prob.22.10(c)
+h_2=3370;// kJ/kg
+h_3a=2300;// kJ/kg
+h_f4=137.8;// kJ/kg
+n_b=((h_2-h_3a)/(h_2-h_f4))*100;//Efficiency of the cycle in %
+m_sb=(((P*10^3*3600))/((h_2-h_3a)*1000));// tons/hr
+printf('\n(a)Efficiency of the plant with reheating=%0.1f percentage \n The steam required per hour=%0.2f kJ/sec \n(b)Efficiency of the plant with no reheating=%0.1f percentage \n The steam consumption per hour=%0.2f kJ/sec',n_a,m_sa,n_b,m_sb);
diff --git a/3733/CH22/EX22.11/Ex22_11.sce b/3733/CH22/EX22.11/Ex22_11.sce new file mode 100644 index 000000000..bbeee15b0 --- /dev/null +++ b/3733/CH22/EX22.11/Ex22_11.sce @@ -0,0 +1,41 @@ +// Example 22_11
+clc;funcprot(0);
+//Given data
+P=27000;// kW
+p_1=60;// bar
+T_1=450;// °C
+p_v=707.5;// The condenser vaccum in mm of Hg
+p_2=3;//bar
+n_t=87;// The turbine efficiency
+n_b=90;// The boiler efficiency in %
+n_a=95;//The alternator efficiency in %
+n_m=98;//The mechanical efficiency in %
+p_b=760;// cm of Hg
+
+//Calculation
+p_3=((p_b-p_v)/p_b)*1.013;//The condenser pressure bar
+// From h-s chart:
+h_1=3296;// kJ/kg
+h_2a=2606;// kJ/kg
+h_3a=2163;// kJ/kg
+h_2=h_1-((n_t/100)*(h_1-h_2a));// kJ/kg
+h_3=h_2-((n_t/100)*(h_2-h_3a));// kJ/kg
+//From steam tables
+h_f4=162;// kJ/kg (at 0.07 bar)
+h_f5=558;// kJ/kg (at 3 bar)
+//Assume m=y(1)
+function[X]=bled(y)
+ X(1)=((1-y(1))*(h_f5-h_f4))-(y(1)*(h_2-h_f5));
+endfunction
+y=[0.1]
+z=fsolve(y,bled);
+m=z(1);// kg/kg of steam generated
+W=(h_1-h_2)+((1-m)*(h_2-h_3));//Work developed per kg of steam in kJ/kg
+W_act=(P/((n_a/100)*(n_m/100)));//Actual work developed by the turbine kW
+m_s=(W_act/W)*(3600/1000);// Steam generated per second in tons/hr
+P_p=P*(10/100);// Pump power in kW
+P_net=P*(1-(10/100));// Net power available in kW
+Q_s=((m_s*1000*(h_1-h_f5))/((n_b/100)*3600));// Heat supplied in the boiler in kW
+n_o=(P_net/Q_s)*100;// The overall efficiency of the plant in %
+printf('\n(a)The steam bled per kg of steam supplied to the turbine=%0.3f kg/kg of steam generated \n(b)Steam generated per hour=%0.1f tons/hr \n(c)The overall efficiency of the plant=%0.1f percentage',m,m_s,n_o);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.12/Ex22_12.sce b/3733/CH22/EX22.12/Ex22_12.sce new file mode 100644 index 000000000..29894dd0f --- /dev/null +++ b/3733/CH22/EX22.12/Ex22_12.sce @@ -0,0 +1,27 @@ +// Example 22_12
+clc;funcprot(0);
+//Given data
+T_1=300;// °C
+p_1=35;// bar
+p_2=25;// bar
+p_2a=1.5;// bar
+p_3=0.1;// bar
+n_t=80/100;// The isentropic efficiency for both sections of the turbine
+gradT=10;// °C
+m_w=1;// kg
+C_p=4.2;// kJ/kg.°C
+
+// Calculation
+// From h-s chart:
+h_1=2970;// kJ/kg
+h_2=2504;// kJ/kg
+h_3=2197;// kJ/kg
+h_f2=264;// kJ/kg (at 1.5 bar)
+h_f2a=h_f2-(m_w*C_p*gradT);// kJ/kg
+h_f3=190;// kJ/kg (at 0.1 bar)
+m=(h_f2a-h_f3)/(h_2-h_f3);// kg/kg of steam
+W=(h_1-h_2)+((1-m)*(h_2-h_3));// kJ/kg
+Q_s=h_1-h_f2a;// kJ/kg
+n_th=(W/Q_s)*100;// Thermal efficiency of the plant
+printf('\n(a)Bleed steam per kg of steam supplied to the steam turbine=%0.3f kg/kg of steam \n(b)The thermal efficiency of the plant=%0.1f percentage',m,n_th);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH22/EX22.13/Ex22_13.sce b/3733/CH22/EX22.13/Ex22_13.sce new file mode 100644 index 000000000..eaa72df93 --- /dev/null +++ b/3733/CH22/EX22.13/Ex22_13.sce @@ -0,0 +1,40 @@ +// Example 22_13
+clc;funcprot(0);
+//Given data
+T_1=300;// °C
+p_1=40;// bar
+p_2=14;// bar
+p_3=3.4;// bar
+p_4=0.07;// bar
+n_t=80/100;// The turbine efficiency of each portion of the expansion
+
+//Calculation
+//(a)
+// From h-s chart:
+h_1=2953;// kJ/kg
+h_2a=2738;// kJ/kg
+h_2=h_1-((n_t)*(h_1-h_2a));// kJ/kg
+// From h-s chart:
+h_3a=2529;// kJ/kg
+h_3=h_2-((n_t)*(h_2-h_3a));// kJ/kg
+// From h-s chart:
+h_4a=2040;// kJ/kg
+h_4=h_3-((n_t)*(h_3-h_4a));// kJ/kg
+// From steam tables
+h_f5=162;// kJ/kg
+h_f7=575;// kJ/kg
+h_f8=825;// kJ/kg
+m_1=(h_f8-h_f7)/(h_2-h_f8);// kJ/kg of steam
+//Assume m_2=y(1);h_f6=y(2)
+function[X]=mass(y)
+ X(1)=(y(1)*(h_3-h_f7))-(1*(h_f7-y(2)));
+ X(2)=(((m_1+y(1))*h_f7)+((1-m_1-y(1))*h_f5))-(1*y(2));
+endfunction
+y=[0.1 100];
+z=fsolve(y,mass)
+m_2=z(1);// kJ/kg of steam supplied to turbine
+h_f6=z(2);// kJ/kg
+//(b)
+n=(((h_1-h_2)+((1-m_1)*(h_2-h_3))+((1-m_1-m_2)*(h_3-h_4)))/(h_1-h_f8))*100;//The efficiency of the cycle in %
+printf('\n(a)The optimum mass of bled steam=%0.2f kJ/kg \n(b)The cycle efficiency=%0.1f percentage',m_2,n);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.14/Ex22_14.sce b/3733/CH22/EX22.14/Ex22_14.sce new file mode 100644 index 000000000..5e1807b7e --- /dev/null +++ b/3733/CH22/EX22.14/Ex22_14.sce @@ -0,0 +1,40 @@ +// Example 22_14
+clc;funcprot(0);
+//Given data
+T_1=350;// °C
+p_1=30;// bar
+p_2=6;// bar
+p_3=1;// bar
+p_4=0.07;// bar
+P=10;// Power developed by the turbine in MW
+n_t=80/100;// Isentropic efficiency of each stage
+
+// Calculation
+// From h-s chart:
+h_1=3106;// kJ/kg
+h_2=2811;// kJ/kg
+h_3=2560;// kJ/kg
+h_4=2259;// kJ/kg
+// From steam tables
+h_f2=777;// kJ/kg (at 6 bar)
+h_f3=415;// kJ/kg (at 1 bar)
+h_f5=162;//kJ/kg (at 0.07 bar)
+h_f8=h_f2;// kJ/kg
+h_f6=h_f3;// kJ/kg
+//Assume m_1=y(1);m_2=y(2)
+function[X]=mass(y)
+ X(1)=(y(1)*(h_2-h_f2))-(1*(h_f8-h_f6));
+ X(2)=((y(2)*(h_3-h_f3))+(y(1)*(h_f2-h_f3)))-((1-y(1)-y(2))*(h_f6-h_f5));
+ X(3)=(((1-y(1)-y(2)))*h_f6)+((y(1)+y(2))*h_f3)-(y(3));
+endfunction
+y=[0.1 0.01 100];
+z=fsolve(y,mass);
+m_1=z(1);// kg/kg of steam generated
+m_2=z(2);// kg/kg of steam generated
+W_t=(h_1-h_2)+((1-m_1)*(h_2-h_3))+((1-m_1-m_2)*(h_3-h_4));// kJ/kg
+m_s=((P*10^3)/W_t)*60;// kg/sec
+m_s6=(m_s*m_1);// Quantity of steam extracted per minute at 6 bar pressure in kg/min
+m_s1=(m_s*m_2);// Quantity of steam extracted per minute at 1 bar pressure in kg/min
+C=m_s6+m_s1;// Capacity of feed pump extraction pump in kg/min
+printf('\nQuantity of steam extracted per minute at 6 bar pressure=%0.1f kg/min \nQuantity of steam extracted per minute at 1 bar pressure=%0.1f kg/min \nCapacity of feed pump extraction pump=%0.1f kg/min',m_s6,m_s1,C);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH22/EX22.15/Ex22_15.sce b/3733/CH22/EX22.15/Ex22_15.sce new file mode 100644 index 000000000..29e02dfc1 --- /dev/null +++ b/3733/CH22/EX22.15/Ex22_15.sce @@ -0,0 +1,45 @@ +// Example 22_15
+clc;funcprot(0);
+//Given data
+T_1=400;// °C
+p_1=40;// bar
+p_2=2;// bar
+p_3=0.5;// bar
+p_4=0.05;// bar
+n_t1=75/100;// The isentropic efficiency of the first stage of the turbine
+n_t2=80/100;// The isentropic efficiency of the second stage of the turbine
+n_t3=85/100;// The isentropic efficiency of the third stage of the turbine
+m_s=10;// The steam flow in kg/sec
+
+// Calculation
+// From h-s chart:
+h_1=3210;// kJ/kg
+h_2a=2562;// kJ/kg
+h_2=h_1-((n_t1)*(h_1-h_2a));// kJ/kg
+h_3a=2508;// kJ/kg
+h_3=h_2-((n_t2)*(h_2-h_3a));// kJ/kg
+h_4a=2232;// kJ/kg
+h_4=h_3-((n_t3)*(h_3-h_4a));// kJ/kg
+// From steam tables
+h_f8=502;// kJ/kg(2 bar)
+h_f10=h_f8;// kJ/kg
+h_f6=339;// kJ/kg(0.5 bar)
+h_f7=h_f6;// kJ/kg
+h_f9=h_f6;// kJ/kg
+h_f5=136;// kJ/kg(0.05 bar)
+//Assume m_1=y(1);m_2=y(2)
+function[X]=mass(y)
+ X(1)=(y(1)*(h_2-h_f10))-((1-y(1))*(h_f8-h_f7));
+ X(2)=(y(2)*(h_3-h_f9))-((1-y(1)-y(2))*(h_f6-h_f5));
+endfunction
+y=[0.01 0.01];
+z=fsolve(y,mass);
+m_1=z(1);// kJ/kg
+m_2=z(2);// kJ/kg
+W=(h_1-h_2)+((1-m_1)*(h_2-h_3))+((1-m_1-m_2)*(h_3-h_4));// kJ/kg
+P=W*m_s;// Power developed by the turbine in kW
+Q_s=h_1-h_f10;// Heat supplied per kg of steam in kJ/kg
+n_th=(W/Q_s)*100;// Thermal efficiency of the cycle in %
+printf('\n(a)Steam bled for regenerative heaters per kg of steam to turbine,m_1=%0.4f kJ/kg & m_2=%0.4f kJ/kg \n(b)Power developed by the turbine=%0.0f kW \n(c)Thermal efficiency of the cycle=%0.2f percentage',m_1,m_2,P,n_th);
+// The answer provided in the textbook is wrong
+
diff --git a/3733/CH22/EX22.16/Ex22_16.sce b/3733/CH22/EX22.16/Ex22_16.sce new file mode 100644 index 000000000..a2d368439 --- /dev/null +++ b/3733/CH22/EX22.16/Ex22_16.sce @@ -0,0 +1,43 @@ +// Example 22_16
+clc;funcprot(0);
+//Given data
+T_1=459;// °C
+T_3=420;// °C
+p_1=70;// bar
+p_2=25;// bar
+p_3=10;// bar
+p_4=0.07;// bar
+n_t1=78.5/100;// The isentropic efficiency of the H.P turbine
+n_t2=83/100;// The isentropic efficiency of the L.P turbine 1
+n_t3=83/100;// The isentropic efficiency of the L.P turbine 2
+T_7=179;// °C
+P=20;// MW
+n_m=85/100;// Mechanical efficiency of the turbine
+n_t=95/100;// Transmission efficiency
+n_g=95/100;// Generation efficiency
+
+// Calculation
+// From h-s chart:
+h_1=3280;// kJ/kg
+h_2a=2997;// kJ/kg
+h_2=h_1-((n_t1)*(h_1-h_2a));// kJ/kg
+h_3=3277;// kJ/kg
+h_4a=3020;// kJ/kg
+h_4=h_3-((n_t2)*(h_3-h_4a));// kJ/kg
+h_5a=2220;// kJ/kg
+h_5=h_4-((n_t3)*(h_4-h_5a));// kJ/kg
+// From steam tables
+h_f6=162;// kJ/kg(at 0.07 bar)
+h_f7=758;// kJ/kg(at 10 bar)
+function[X]=mass(y)
+ X(1)=(y(1)*(h_4-h_f7))-((1-y(1))*(h_f7-h_f6));
+endfunction
+y=[0.1];// kg
+z=fsolve(y,mass);
+m=z(1);// kg
+W=(h_1-h_2)+(h_3-h_4)+((1-m)*(h_4-h_5));// kJ/kg
+E_g=W*n_m*n_t*n_g;// Energy coverted for generating the electrical energy in kJ
+m_s=((P*10^3)/E_g)*60;// Steam generated in kg/min
+Q_s=(h_1-h_f7)+(h_3-h_2);// Heat supplied per kg of steam in kJ/kg
+n_th=(W/Q_s)*100;// Thermal efficiency of the cycle in %
+printf('\n(a)Thermal efficiency of the cycle=%0.1f percentage \n(b)Quantity of steam supplied per minute=%0.0f kg/min',n_th,m_s);
diff --git a/3733/CH22/EX22.17/Ex22_17.sce b/3733/CH22/EX22.17/Ex22_17.sce new file mode 100644 index 000000000..f72fd0dcb --- /dev/null +++ b/3733/CH22/EX22.17/Ex22_17.sce @@ -0,0 +1,38 @@ +// Example 22_17
+clc;funcprot(0);
+//Given data
+P=500;// Plant capacity in kW
+T_1=300;// °C
+p_4=30;// bar
+p_5=7;// bar
+p_6=0.04;// bar
+dT=5;// The rise in cooling water temperature in °C
+C_pw=4.2;// kJ/kg.°C
+
+// Calculation
+// From h-s chart:
+h_4=3000;// kJ/kg
+h_5=2700;// kJ/kg
+h_6=1970;// kJ/kg
+// From steam tables
+h_f1=121.4;// kJ/kg(at 0.04 bar)
+h_f2=697;// kJ/kg(at 7 bar)
+function[X]=mass(y)
+ X(1)=((y(1)*h_5)+((1-y(1))*h_f1))-(1*h_f2);
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+W=(1*(h_4-h_5))+((1-m)*(h_5-h_6));// kJ/kg
+Q_s=h_4-h_f2;// Heat supplied in kJ/kg
+n_s=(W/Q_s)*100;// Efficiency in %
+m_s=(P/W)*3600;//Steam generated per second in kg/hr
+m_w=((h_6-h_f1)*(m_s/3600)*(1-m))/(C_pw*dT);// kg/sec
+// If there ie no feed water,then
+W_1=h_4-h_6;// kJ/kg
+Q_s=h_4-h_f1;// kJ/kg
+n=(W_1/Q_s)*100;// Efficiency in %
+m_s1=(P/W_1)*3600;//Steam generated per second in kg/hr
+m_w1=((m_s/3600)*(h_6-h_f1))/(C_pw*dT);// The amount of cooling water in kg/sec
+printf('\n(a)The rankine efficiency=%0.1f percentage \n Steam generation rate of boiler=%0.1f kg/hr \n The amount of cooling water=%0.2f kg/sec \n(b)The rankine efficiency=%0.1f percentage \n Steam generation rate of boiler=%0.1f kg/hr \n The amount of cooling water=%0.2f kg/sec',n_s,m_s,m_w,n,m_s1,m_w1);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.18/Ex22_18.sce b/3733/CH22/EX22.18/Ex22_18.sce new file mode 100644 index 000000000..5a1577b72 --- /dev/null +++ b/3733/CH22/EX22.18/Ex22_18.sce @@ -0,0 +1,37 @@ +// Example 22_18
+clc;funcprot(0);
+//Given data
+T_1=500;// °C
+p_1=40;// bar
+p_2=10;// bar
+p_3=0.04;// bar
+m_b=50;// The boiler generation rate in tons/hour
+n_m=85/100;// Mechanical efficiency
+n_g=95/100;// Electrical generation efficiency
+
+// Calculation
+// From h-s chart:
+h_1=3400;// kJ/kg
+h_2=3050;// kJ/kg
+h_3=2150;// kJ/kg
+// From steam tables
+h_f4=121.4;// kJ/kg(at 0.04 bar)
+h_f5=762.6;// kJ/kg(at 10 bar)
+h_f6=h_f5;// kJ/kg
+//Assume m_1=y(1);h_fm=y(2)
+function[X]=mass(y)
+ X(1)=((y(1)*h_f6)+((1-y(1))*h_f4))-(y(1)*y(2));
+ X(2)=(y(1)*(h_2-h_f5))-(1*(h_f5-y(2)));
+endfunction
+y=[0.1 100];
+z=fsolve(y,mass);
+m=z(1);// kg/kg of steam generated
+h_fm=z(2);// kJ/kg
+W=(h_1-h_2)+((1-m)*(h_2-h_3));// kJ/kg
+m_b=m*100;// Bled steam in %
+Q_s=h_1-h_f5;// Heat supplied per kg of steam in kJ/kg
+n=(W/Q_s)*100;// Efficiency in %
+P=(((m_b*10^3)*W*n_m*n_g)/3600)/1000;// Power developed in MW
+printf('\nThe percentage of bled steam=%0.0f percentage \nThe thermal efficiency of the plant=%0.1f percentage \nThe generating capacity of the plant=%0.1f MW',m_b,n,P);
+// The answer provided in the textbook is wrong
+
diff --git a/3733/CH22/EX22.19/Ex22_19.sce b/3733/CH22/EX22.19/Ex22_19.sce new file mode 100644 index 000000000..8c800e64a --- /dev/null +++ b/3733/CH22/EX22.19/Ex22_19.sce @@ -0,0 +1,32 @@ +// Example 22_19
+clc;funcprot(0);
+//Given data
+P=100;// MW
+p_2=80;// bar
+p_3=7;// bar
+p_5=0.05;// bar
+T_4=350;// °C
+
+// Calculation
+// From h-s chart:
+h_2=2990;// kJ/kg
+h_3=2350;// kJ/kg
+h_4=3170;// kJ/kg
+h_5=2180;// kJ/kg
+// From steam tables
+h_f6=138;// kJ/kg
+h_f7=697;// kJ/kg
+function[X]=mass(y)
+ X(1)= (y(1)*(h_3-h_f7))-((1-y(1))*(h_f7-h_f6));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);
+m_p=m*100;// Percentage of bled steam in %
+W=(h_2-h_3)+((1-m)*(h_4-h_5));// kJ/kg
+Q_s=(h_2-h_f7)+((1-m)*(h_4-h_3));// kJ/kg
+n=(W/Q_s)*100;// The efficiency of the power plant in %
+m_b=((P*10^3)/((h_2-h_3)+((1-m)*(h_4-h_5))));// tons/hr
+printf('\nThe percentage of bled steam=%0.1f percentage \nThe thermal efficiency of the cycle=%0.0f percentage \nBoiler generating rate=%0.0f tons/hr',m_p,n,m_b);
+// The answer provided in the textbook is wrong
+
diff --git a/3733/CH22/EX22.2/Ex22_2.sce b/3733/CH22/EX22.2/Ex22_2.sce new file mode 100644 index 000000000..f3b7097f6 --- /dev/null +++ b/3733/CH22/EX22.2/Ex22_2.sce @@ -0,0 +1,34 @@ +// Example 22_2
+clc;funcprot(0);
+//Given data
+p_a=10;// bar
+p_b=0.08;// bar
+T_1=450;// °C
+p_1=30;// bar
+p_3=25;// bar
+T_4=33;// °C
+p_4=0.04;// bar
+
+
+//Calculation
+//From tables of mercury the following enthalpy values and entropy values are taken
+h_1=359.11;// kJ/kg
+h_f2=33.21;// kJ/kg
+h_3=h_f2;// kJ/kg
+s_1=0.5089;// kJ/kg.K
+s_f2=0.087;// kJ/kg.K
+Q_fg2=0.5721;// kJ/kg
+h_g2=294.7;// kJ/kg
+x_2=(s_1-s_f2)/(Q_fg2);
+h_2=h_f2+(x_2*h_g2);// kJ/kg
+// From steam tables and chart(For steam cycle)
+h_4=3348.6;// kJ/kg
+h_5=2183;// kJ/kg
+h_6=138;// kJ/kg
+h_7=972;// kJ/kg
+h_8=2803;// kJ/kg
+// Assume m_r=m_hg/m_H2O
+m_r=(h_8-h_7)/(h_2-h_3);
+// For each kg of steam generated,8.42 kg of mercury is to be used
+n=((m_r*(h_1-h_2))+(h_4-h_5))/((m_r*(h_1-h_f2))+(h_7-h_6)+(h_4-h_8));// The cycle efficiency
+printf('\nCycle efficiency=%0.3f',n);
diff --git a/3733/CH22/EX22.20/Ex22_20.sce b/3733/CH22/EX22.20/Ex22_20.sce new file mode 100644 index 000000000..240a3a64c --- /dev/null +++ b/3733/CH22/EX22.20/Ex22_20.sce @@ -0,0 +1,59 @@ +// Example 22_20
+clc;funcprot(0);
+//Given data
+T_1=556;// °C
+T_2=222;// °C
+m_s=20;// kg/sec
+n_m=80/100;// Mechanical efficiency
+n_t=95/100;// Transmission efficiency
+n_g=85/100;// Generator efficiency
+W_act=50/100;
+h_f1=76;// kJ/kg
+h_f2=29;// kJ/kg
+h_fg1=290;// kJ/kg
+h_fg2=302;// kJ/kg
+h_g1=366;// kJ/kg
+h_g2=331;// kJ/kg
+s_f1=0.152;// kJ/kg-K
+s_f2=0.08;// kJ/kg-K
+s_fg1=0.359;// kJ/kg-K
+s_fg2=0.626;// kJ/kg-K
+s_g1=0.511;// kJ/kg-K
+s_g2=0.706;// kJ/kg-K
+p_a=17;// bar
+p_b=0.035;// bar
+h_fa=874;// kJ/kg
+h_fb=111;// kJ/kg
+h_fga=1932;// kJ/kg
+h_fgb=2453;// kJ/kg
+h_ga=2806;// kJ/kg
+h_gb=2564;// kJ/kg
+s_fa=2.37;// kJ/kg-K
+s_fb=0.388;// kJ/kg-K
+s_ga=6.42;// kJ/kg-K
+s_gb=0.388;// kJ/kg-K
+
+// Calculation
+//(a)
+x_2=(s_g1-s_f2)/s_fg2;// The condition of the mercury vapour at point2
+m_hg=h_fga/(x_2*h_fg2);// kg
+//(b)
+W=h_g1-(h_f2+(x_2*h_fg2));// kJ/kg
+W_m=W*m_hg;//Work done per kg of Hg vapour in kJ
+//(c)
+// From steam tables,
+T_sup=380+273;// K
+T_sa=203.4+273;// K
+T_b=26.5+273;// K
+x_b=(((s_ga+(2*2.303*log(T_sup/T_sa)))*(T_b))-s_fb)/(T_b/h_fga);
+T_sup=383+273;// K
+x_2=0.72;
+W_s=(h_ga+(2*(T_sup-T_sa)))-(h_fb+(x_2*h_fgb));// Work done per kg of steam in kJ/kg
+//(d)
+W=W_m+W_s;// Total work done in kJ
+Q_s=(m_hg*(h_g1-h_f2))+(1*(h_fa-h_fb))+(2*(T_sup-T_sa));// Heat supplied in kJ
+n_o=(W/Q_s)*100;// Overall efficiency of the cycle in %
+E=((m_s*W*W_act)*n_m*n_t*n_g)/1000;// Total energy generated per sec in MW
+printf('\n(a)Mass of Hg required per kg of steam used=%0.1f kg \n(b)Work done per kg of Hg vapour=%0.1f kJ/kg \n(c)Work done per kg of steam=%0.0f kJ/kg \n(d)Overall efficiency of the cycle=%0.1f percentage \n(e)Total energy generated per sec=%0.3f MW',m_hg,W_m,W_s,n_o,E);
+// The answer provided in the textbook is wrong
+
diff --git a/3733/CH22/EX22.21/Ex22_21.sce b/3733/CH22/EX22.21/Ex22_21.sce new file mode 100644 index 000000000..e4a169780 --- /dev/null +++ b/3733/CH22/EX22.21/Ex22_21.sce @@ -0,0 +1,28 @@ +// Example 22_21
+clc;funcprot(0);
+//Given data
+p_1=30;// bar
+T_1=550;// °C
+p_2=2.6;// bar
+p_3=0.2;// bar
+m_s=30;// kg/sec
+
+//Calculation
+//From h-s chart:
+h_1=3580;// kJ/kg
+h_2=2870;// kJ/kg
+h_3=2440;// kJ/kg
+// From steam tables
+h_f2=541;// kJ/kg
+h_f3=251.5;// kJ/kg
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*(h_f2-h_f3));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+P=m_s*((h_1-h_2)+((1-m)*(h_2-h_3)))/1000;// MW
+n_r=((h_1-h_3)/(h_1-h_f3))*100;// The efficiency of the rankine cycle in %
+n_b=(((h_1-h_2)+((1-m)*(h_2-h_3)))/(h_1-h_f2))*100;
+printf('\nThe power generating capacityof the plant=%0.2f MW \nThe efficiency of the rankine cycle=%0.0f percentage \nThe efficiency of the cycle with bled heating=%0.0f percentage',P,n_r,n_b);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.22/Ex22_22.sce b/3733/CH22/EX22.22/Ex22_22.sce new file mode 100644 index 000000000..567947ec7 --- /dev/null +++ b/3733/CH22/EX22.22/Ex22_22.sce @@ -0,0 +1,31 @@ +// Example 22_22
+clc;funcprot(0);
+//Given data
+P=30;// MW
+p_1=0.04;// bar
+p_2=7;// bar
+p_3=60;// bar
+T_1=550;// °C
+p_c=730;// mm of Hg
+p_v=760;// mm of Hg
+n_t=90/100;// The isentropic efficiency of the turbine
+
+//Calculation
+p_1=((p_v-p_c)*133.3)/10^5;// bar
+//From h-s chart:
+h_1=3420;// kJ/kg
+h_2a=2860;// kJ/kg
+h_2=2900;// kJ/kg
+h_3=2410;// kJ/kg
+h_3a=2190;// kJ/kg
+// From steam tables
+h_f3=121.5;// kJ/kg(liquid heat at 0.04 bar)
+h_f2=697;// kJ/kg(liquid heat at 7 bar)
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*(h_f2-h_f3));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+m_s=(P*10^3)/((h_1-h_2)+((1-m)*(h_2-h_3a)));// kg/sec
+printf('\n(a)Fraction of steam bled for feed heating=%0.3f kg \n(b)Boiler generating capacity=%0.1f kg/sec',m,m_s);
diff --git a/3733/CH22/EX22.23/Ex22_23.sce b/3733/CH22/EX22.23/Ex22_23.sce new file mode 100644 index 000000000..ec6eac066 --- /dev/null +++ b/3733/CH22/EX22.23/Ex22_23.sce @@ -0,0 +1,32 @@ +// Example 22_23
+clc;funcprot(0);
+//Given data
+P=120;// MW
+p_1=86;// bar
+p_2=7;// bar
+p_3=0.35;// bar
+T_1=350;// °C
+
+//Calculation
+//From h-s chart:
+h_1=2980;// kJ/kg
+h_2=2520;// kJ/kg
+h_3=3170;// kJ/kg
+h_4=2550;// kJ/kg
+// From steam tables
+h_f1=304.3;// kJ/kg(liquid heat at 0.35 bar)
+T_s1=72.7;// °C
+h_f2=697;// kJ/kg(liquid heat at 7 bar)
+T_s2=165;// °C
+h_f4=h_f1;// kJ/kg
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*(h_f2-h_f4));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// tons/hr
+S=(1/m);// The ratio of steam bled to steam generated
+m_s=((P*10^3)/((h_1-h_2)+((1-m)*(h_3-h_4))))*(3600/1000);// kg/sec
+n_th=(((h_1-h_2)+((1-m)*(h_3-h_4)))/((h_1-h_f1)+((1-m)*(h_3-h_2))))*100;
+printf('\n(a)The ratio of steam bled to steam generated=%0.2f \n(b)The boiler generating capacity=%0.1f tons/hr \n(c)The thermal efficiency of the cycle=%0.1f percentage',S,m_s,n_th);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.24/Ex22_24.sce b/3733/CH22/EX22.24/Ex22_24.sce new file mode 100644 index 000000000..048e941bf --- /dev/null +++ b/3733/CH22/EX22.24/Ex22_24.sce @@ -0,0 +1,40 @@ +// Example 22_24
+clc;funcprot(0);
+//Given data
+T_1=300;// °C
+p_1=30;// bar
+p_2=10;// bar
+p_4=5;// bar
+T_4=270;// °C
+p_6=0.07;// bar
+m_s=20;// tons/hr
+C_pw=4.2;// kJ/kg.°C
+T_9=180;// °C
+T_8=38;// °C
+
+//Calculation
+//From h-s chart:
+h_1=3000;// kJ/kg
+h_2=2780;// kJ/kg
+h_3=2640;// kJ/kg
+// From steam tables
+h_f2=762.5;// kJ/kg
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*C_pw*(T_9-T_8));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+//From h-s chart:
+h_4=3000;// kJ/kg
+h_5=(((1/3)*h_4)+(((2/3)-m)*h_3))/(1-m);// kJ/kg
+//From h-s chart:
+h_6=2150;// kJ/kg
+// From steam tables
+h_f7=h_f2;// kJ/kg
+W=(((2/3)*(h_1-h_2))+(((2/3)-m)*(h_2-h_3))+((1-m)*(h_5-h_6)));// kJ/kg
+n=((((2/3)*(h_1-h_2)+((2/3)-m)*(h_2-h_3))+((1-m)*(h_5-h_6)))/(((2/3)*h_1)+((1/3)*h_4)-h_f7))*100;// Efficiency of the cycle in %
+m_s=(m_s*1000)/3600;// Steam generated per second in kg/sec
+P=m_s*W;// Power generating capacity of the plant in kW
+printf('\nFraction of steam bled=%0.4f \nEfficiency of the plant=%0.1f percentage \nPower generating capacity of the plant=%0.0f kW',m,n,P);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH22/EX22.25/Ex22_25.sce b/3733/CH22/EX22.25/Ex22_25.sce new file mode 100644 index 000000000..2cce37478 --- /dev/null +++ b/3733/CH22/EX22.25/Ex22_25.sce @@ -0,0 +1,29 @@ +// Example 22_25
+clc;funcprot(0);
+//Given data
+P=30;// MW
+p_1=60;// bar
+p_2=3;// bar
+T_1=500;// °C
+p_v=73;// mm of Hg
+p_b=76;// mm of Hg
+n_t=90/100;// The isentropic efficiency of the turbine
+
+//Calculation
+p_3=(((p_b-p_v)/p_b)*1.013);// bar
+//From h-s chart:
+h_1=3410;// kJ/kg
+h_2=2720;// kJ/kg
+h_3=2220;// kJ/kg
+// From steam tables
+h_f2=361.4;// kJ/kg(liquid heat at 0.04 bar)
+h_f3=121.4;// kJ/kg(liquid heat at 7 bar)
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*(h_f2-h_f3));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg/kg of steam
+m_s=(P*10^3)/((h_1-h_2)+((1-m)*(h_2-h_3)));// kg/sec
+m_s=m_s*(3600/1000);// tons/hr
+printf('\n(a)Fraction of steam bled for feed heating=%0.3f kg/kg of steam \n(b)Steam supplied by the boiler=%0.1f tons/hr',m,m_s);
diff --git a/3733/CH22/EX22.26/Ex22_26.sce b/3733/CH22/EX22.26/Ex22_26.sce new file mode 100644 index 000000000..ec1b9e026 --- /dev/null +++ b/3733/CH22/EX22.26/Ex22_26.sce @@ -0,0 +1,37 @@ +// Example 22_26
+clc;funcprot(0);
+//Given data
+p_1=80;// bar
+T_1=470;// °C
+p_2=7;// bar
+T_1=350;// °C
+p_3=0.35;// bar
+m_s=50;// kg/sec
+
+//Calculation
+//From h-s chart:
+h_1=3310;// kJ/kg
+h_2=2780;// kJ/kg
+h_3=3170;// kJ/kg
+h_4=2220;// kJ/kg
+// From steam tables
+h_f2=697;// kJ/kg
+h_6=h_f2;// kJ/kg
+h_5=111.85;// kJ/kg
+h_f4=h_5;// kJ/kg
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*(h_f2-h_f4));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+m_b=m*100;// Amount of steam bled off in %
+m_l=(100-m_b);// Amount of steam supplied to L.P turbine in %
+Q_b=h_1-h_6;// kJ/kg
+Q_r=(1-m)*(h_3-h_2);// kJ
+Q_s=Q_b+Q_r;// Total amount of heat supplied by the boiler and reheater in kJ/kg
+W=(h_1-h_2)+((1-m)*(h_3-h_4));// kJ/kg
+n=(W/Q_s)*100;
+P=(m_s*W)/1000;// Power developed by the steam in MW
+printf('\n(a)Amount of steam bled off for feed heating=%0.0f percentage \n(b)Amount of steam in LP turbine=%0.0f percentage \n(c)Heat supplied in the boiler and reheater=%0.1f kJ/kg \n(d)Cycle efficiency=%0.1f percentage \n(e)Power developed by the steam=%0.1f MW',m_b,m_l,Q_s,n,P);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH22/EX22.27/Ex22_27.sce b/3733/CH22/EX22.27/Ex22_27.sce new file mode 100644 index 000000000..fdad23996 --- /dev/null +++ b/3733/CH22/EX22.27/Ex22_27.sce @@ -0,0 +1,45 @@ +// Example 22_27
+clc;funcprot(0);
+//Given data
+T_1=600;// °C
+p_1=150;// bar
+T_3=600;// °C
+p_3=40;// bar
+p_4=5;// bar
+p_5=0.1;// bar
+
+//Calculation
+//From h-s chart:
+h_1=3570;// kJ/kg
+h_2=3280;// kJ/kg
+h_3=3650;// kJ/kg
+h_4=2920;// kJ/kg
+h_5=2280;// kJ/kg
+// From steam tables
+h_f1=1610;// kJ/kg(at 150 bar)
+h_f2=1087;// kJ/kg(at 40 bar)
+h_f4=640;// kJ/kg(at 5 bar)
+h_f5=192;// kJ/kg(at 0.1 bar)
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*(h_f2-h_f4));
+ X(2)=(y(2)*(h_4-h_f4))-((1-y(1)-y(2))*(h_f4-h_f5));
+endfunction
+y=[0.1 0.1];
+z=fsolve(y,mass);
+m_1=z(1);// kg/kg of steam supplied by the boiler
+m_2=z(2);// kg/kg of steam supplied by the boiler
+W_t=(h_1-h_2)+((1-m_1)*(h_3-h_4))+((1-m_1-m_2)*(h_4-h_5));// Total workdone per kg of steam supplied by the boiler in kJ/kg
+v_w1=1/1000;// m^3/kg
+v_w2=v_w1;// m^3/kg
+v_w3=v_w1;// m^3/kg
+W_p1=(v_w1*(1-m_1-m_2)*(p_4-p_5*10^5))/1000;// kJ/kg
+W_p2=(v_w2*(1-m_1)*(p_1-p_4)*10^5)/1000;// kJ/kg
+W_p3=(v_w3*(m_1)*(p_1-p_3)*10^5)/1000;// kJ/kg
+W_pt=W_p1+W_p2+W_p3;// kJ/kg
+W_n=W_t-W_pt;// Net work done by the turbine per kg of steam supplied by the boiler in kJ
+Q_f=((1-m_1)*h_f1)+(m_1*h_f1);// Heat of feed water entering the boiler in kJ
+Q_s1=h_1-Q_f;// Heat supplied by the boiler per kg of steam in kJ
+Q_s2=(1-m_1)*(h_3-h_2);// Heat supplied in the reheater in kJ/kg
+Q_st=Q_s1+Q_s2;// Total heat supplied in kJ/kg
+n=(W_n/Q_st)*100;// Thermal efficiency in %
+printf('\nm_1=%0.2f kg/kg of steam \nm_2=%0.3f kg/kg of steam \nThermal efficiency=%0.1f percentage',m_1,m_2,n);
diff --git a/3733/CH22/EX22.28/Ex22_28.sce b/3733/CH22/EX22.28/Ex22_28.sce new file mode 100644 index 000000000..fe2edcc83 --- /dev/null +++ b/3733/CH22/EX22.28/Ex22_28.sce @@ -0,0 +1,32 @@ +// Example 22_28
+clc;funcprot(0);
+//Given data
+p_1=80;// bar
+T_1=470;// °C
+p_2=7;// bar
+T_1=350;// °C
+p_3=0.035;// bar
+m_s=100;// kg/sec
+
+//Calculation
+//From h-s chart:
+h_1=3350;// kJ/kg
+h_2=2770;// kJ/kg
+h_3=3170;// kJ/kg
+h_4=2220;// kJ/kg
+// From steam tables
+h_f5=112;// kJ/kg
+h_f6=697;// kJ/kg
+function[X]=mass(y)
+ X(1)= (y(1)*(h_2-h_f6))-((1-y(1))*(h_f6-h_f5));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+m_b=m*100;// The fraction of steam bled for reheating in %
+Q_s=(h_1-h_f6)+((1-m)*(h_3-h_2));// Heat supplied in the boiler and reheater in kJ/kg
+W=(h_1-h_2)+((1-m)*(h_3-h_4));// Power output in kJ/kg
+P=(W*100)/1000;// Capacity of the plant in MW
+n=(W/Q_s)*100;// The efficiency of the plant in %
+printf('\n(a)Fraction of steam bled for feed heating=%0.0f percentage \n(b)Heat supplied per kg of steam in boiler and turbine=%0.0f kJ/kg \n(c)Power output of the plant=%0.0f MW \n(d)Thermal efficiency of the plant=%0.1f percentage',m_b,Q_s,P,n);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.29/Ex22_29.sce b/3733/CH22/EX22.29/Ex22_29.sce new file mode 100644 index 000000000..ef1dd006d --- /dev/null +++ b/3733/CH22/EX22.29/Ex22_29.sce @@ -0,0 +1,46 @@ +// Example 22_29
+clc;funcprot(0);
+//Given data
+p_a=4.5;// bar
+p_b=0.04;// bar
+p_1=15;// bar
+p_2=0.04;// bar
+m_s=48000;// kg/hr
+T_a=450;// °C
+T_b=217;// °C
+h_fa=62.9// kJ/kg
+h_fb=30.0;// kJ/kg
+h_ga=356;// kJ/kg
+h_gb=330;// kJ/kg
+s_fa=0.135;// kJ/kg-K
+s_fb=0.081;// kJ/kg-K
+s_ga=0.539;// kJ/kg-K
+s_gb=0.693;// kJ/kg-K
+v_sfa=80*10^-6;// m^3/kg
+v_sfb=76.5*10^-6;// m^3/kg
+v_sga=0.068;// m^3/kg
+v_sgb=5.178;// m^3/kg
+
+//Calculation
+m_h2o=(m_s/3600);// kg/sec
+// s_a=s_b
+x_b=(s_ga-s_fb)/(s_gb-s_fb);
+h_b=h_fb+(x_b*(h_gb-h_fb));// kJ/kg
+h_c=30;// kJ/kg
+h_fc=h_c;// kJ/kg
+//From h-s chart:
+h_1=2800;// kJ/kg
+h_2=1920;// kJ/kg
+// From steam tables
+h_f3=121.4;// kJ/kg
+h_f4=844.6;// kJ/kg
+m_hg=(m_h2o*(h_1-h_f3))/(h_b-h_fc);// kg/sec
+m=m_hg/m_h2o;
+W_Hg=m_hg*(h_ga-h_b);// kW
+W_H2o=m_h2o*(h_1-h_2);// kW
+W_t=(W_Hg+W_H2o)/1000;//Total work done per second in MW
+Q_s=m_hg*(h_ga-h_fc);// The total heat supplied in kJ/sec
+n_o=((W_t*1000)/Q_s)*100;// Overall efficiency in %
+printf('\nThe overall efficiency of the cycle=%0.1f percentage \nThe flow of mercury through mercury turbine=%0.1f kg/sec \nTotal work done per second=%0.1f MW',n_o,m_hg,W_t);
+// The answer vary due to round off error
+
diff --git a/3733/CH22/EX22.3/Ex22_3.sce b/3733/CH22/EX22.3/Ex22_3.sce new file mode 100644 index 000000000..f3f01c34d --- /dev/null +++ b/3733/CH22/EX22.3/Ex22_3.sce @@ -0,0 +1,35 @@ +// Example 22_3
+clc;funcprot(0);
+//Given data
+p_1=30;// bar
+p_3=0.04;// bar
+x_1=0.841;// Dryness fraction
+
+//Calculation
+//From h-s chart:
+h_1=2803;// kJ/kg
+h_2=2370;// kJ/kg
+h_3=2717;// kJ/kg
+h_4=2124;// kJ/kg
+x_2=0.824;// kJ/kg
+p_7=2.5// bar
+p_2=p_7;// bar
+//From steam tables at p=2.5bar & p=0.04 bar
+v_s1=0.00106;// kJ/kg
+v_s2=0.00104;// kJ/kg
+h_f5=121;// kJ/kg
+h_f2=535;// kJ/kg
+
+W_ph=(p_1-p_2)*10^2*v_s1;// Pump work for higher pressure stage in kJ/kg
+W_pl=(p_1-p_2)*10^2*v_s2;// Pump work for lower pressure side in kJ/kg
+m_s=x_1;// mass flow inkg
+m_f=0.159;// Mass flow through first feed pump in kg
+n_ws=(((h_1-h_2)+(m_s*(h_3-h_4))-(m_s*W_ph)-(m_f*W_pl))/((m_s*(h_1-h_f5))+(m_f*(h_1-h_f2))))*100;// Efficiency of the cycle
+W_p=(p_1-p_2)*10^2*v_s2;// Pump work in kJ/kg
+n_wos=(((h_1-h_4)-W_p)/(h_1-h_f5))*100;// Efficiency of the cycle without seperation
+//From steam table,at p=0.04 bar
+h_fg4=2433.1;// kJ/kg
+h_f4=121.4;// kJ/kg
+x_4=(h_4-h_f4)/(h_fg4);// Dryness at exit
+printf('\n Efficiency of the cycle with seperation=%0.1f percentage \n Efficiency of the cycle without seperation=%0.1f percentage \n Dryness at exit,x_4=%0.3f',n_ws,n_wos,x_4 );
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.30/Ex22_30.sce b/3733/CH22/EX22.30/Ex22_30.sce new file mode 100644 index 000000000..6befc7cd7 --- /dev/null +++ b/3733/CH22/EX22.30/Ex22_30.sce @@ -0,0 +1,45 @@ +// Example 22_30
+clc;funcprot(0);
+//Given data
+p_a=10;// bar
+p_b=0.2;// bar
+p_1=40;// bar
+T_1=400;// °C
+T_2=40;// °C
+m_s=500;// kg/sec
+T_sa=515.5;// °C
+T_sb=277.3;// °C
+h_fa=72.33// kJ/kg
+h_fb=38.35;// kJ/kg
+h_ga=363.0;// kJ/kg
+h_gb=336.55;// kJ/kg
+s_fa=0.1478;// kJ/kg-K
+s_fb=0.0967;// kJ/kg-K
+s_ga=0.5167;// kJ/kg-K
+s_gb=0.6385;// kJ/kg-K
+v_fa=80.9*10^-6;// m^3/kg
+v_fb=77.4*10^-6;// m^3/kg
+v_ga=0.0333;// m^3/kg
+v_gb=1.163;// m^3/kg
+
+//Calculation
+//From h-s chart:
+h_1=3230;// kJ/kg
+h_2=2120;// kJ/kg
+// From steam tables
+h_3=167.5;// kJ/kg
+h_4=h_3;// kJ/kg
+// s_a=s_b
+x_b=(s_ga-s_fb)/(s_gb-s_fb);
+h_b=h_fb+(x_b*(h_gb-h_fb));// kJ/kg
+h_c=38.35;// kJ/kg
+h_d=h_c;// kJ/kg
+//(a)
+h_a=h_ga;// kJ/kg
+m_Hg=(h_1-h_4)/(h_b-h_c);// kg/kg of steam
+n_Hg=((h_a-h_b)/(h_a-h_d))*100;// The efficiency of the mercury cycle in %
+n_H2o=((h_1-h_2)/(h_1-h_3))*100;// The efficiency of the steam cycle in %
+n_o=(((m_Hg*(h_a-h_b))+(1*(h_1-h_2)))/(m_Hg*(h_a-h_c)))*100;// The over all efficiency of the plant in %
+P=((m_s/60)*((m_Hg*(h_a-h_b))+(1*(h_1-h_2))))/1000;// Total power generated in the system in MW
+printf('\nMass of mercury required to generate one kg of steam=%0.2f kg/kg of steam \nThe efficiency of the mercury cycle=%0.1f percentage \nThe efficiency of the steam cycle=%0.2f percentage \nThe over all efficiency of the plant=%0.1f percentage \nThe power generating capacity of the plant=%0.2f MW',m_Hg,n_Hg,n_H2o,n_o,P);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.4/Ex22_4.sce b/3733/CH22/EX22.4/Ex22_4.sce new file mode 100644 index 000000000..7d6b9b48f --- /dev/null +++ b/3733/CH22/EX22.4/Ex22_4.sce @@ -0,0 +1,26 @@ +// Example 22_4
+clc;funcprot(0);
+//Given data
+p_1=90;// bar
+T_1=480;// °C
+p_2=12;// bar
+p_3=0.07;// bar
+m=1;// Steam flow rate in kg/sec
+
+//Calculation
+//From h-s chart:
+h_1=3333.5;// kJ/kg
+h_2=2815;// kJ/kg
+h_3=3425.5;// kJ/kg
+h_4=2364;// kJ/kg
+//From steam tables at p=0.07 bar
+h_f5=161.8;// kJ/kg
+v_sw1=0.001013;// m^3/kg
+h_6=h_f5+((v_sw1*(p_1-p_3)*10^5)/(1000*m));// kJ/kg
+W_p=(h_6-h_f5);// Pump work in kJ/kg
+W_net=(h_1-h_2)+((h_3-h_4))-W_p;// Net Work done in kJ/kg
+P=W_net*m;// Power generating capacity of the plant in kW
+H_s=(h_1-h_6)+(h_3-h_2);// Heat supplied per kg of steam in kJ/kg
+n=(W_net/H_s)*100;// Efficiency of the cycle
+printf('\nEfficiency of the cycle=%0.1f percentage \nNet work done per kg steam=%0.1f kJ/kg',n,W_net);
+// The answer vary due to round off error
diff --git a/3733/CH22/EX22.5/Ex22_5.sce b/3733/CH22/EX22.5/Ex22_5.sce new file mode 100644 index 000000000..38824e28c --- /dev/null +++ b/3733/CH22/EX22.5/Ex22_5.sce @@ -0,0 +1,33 @@ +// Example 22_5
+clc;funcprot(0);
+//Given data
+p_1=100;// bar
+T_1=500;// °C
+p_2=8.5;// bar
+p_3=p_2-0.5;// bar
+p_4=0.05;// bar
+n_t=80;// The isentropic efficiency of the turbine in %
+n_lt=85;// The isentropic efficiency of lower stage of the turbine in %
+
+//Calculation
+//From h-s chart:
+h_1=3377;// kJ/kg
+h_2a=2750;// kJ/kg
+h_3=3478;// kJ/kg
+h_4a=2738;// kJ/kg
+// The isentropic efficiency of the expansion 1-2 is 80% as given in problem
+h_2=h_1-((n_t/100)*(h_1-h_2a));// kJ/kg
+// The isentropic efficiency of the expansion 3-4 is 85% as given in problem
+h_4=h_3-((n_lt/100)*(h_3-h_4a));// kJ/kg
+//From steam tables ,
+h_f5=137;// kJ/kg
+n_th1=(((h_1-h_2)+(h_3-h_4))/((h_1-h_f5)+(h_3-h_2)))*100;// The efficiency of the cycle in %
+// From h-s diagram
+h_6a=2305;// kJ/kg
+// The isentropic efficiency of the expansion 2-6 is 75% as given in problem
+n_lt=75;//The isentropic efficiency of the turbine in %
+h_6=h_2-((n_lt/100)*(h_2-h_6a));// kJ/kg
+n_th2=(((h_1-h_2)+(h_2-h_6))/(h_1-h_f5))*100;// The thermal efficiency of the cycle without reheating in %
+printf('\nThe thermal efficiency of the cycle with reheating=%0.1f percentage \nThe thermal efficiency of the cycle without reheating=%0.1f percentage',n_th1,n_th2);
+// The answer is bit different due to calculation error in the book
+
diff --git a/3733/CH22/EX22.6/Ex22_6.sce b/3733/CH22/EX22.6/Ex22_6.sce new file mode 100644 index 000000000..792a27b47 --- /dev/null +++ b/3733/CH22/EX22.6/Ex22_6.sce @@ -0,0 +1,35 @@ +// Example 22_6
+clc;funcprot(0);
+//Given data
+p_1=215;// bar
+T_1=500;// °C
+p_2=40;// bar
+T_2=280;// °C
+p_3=p_2-1;// bar
+p_4=8;// bar
+T_4=270;//°C
+p_5=p_4-0.5;// bar
+p_6=0.07;// bar
+m=10;// The flow of steam in kg/sec
+
+//Calculation
+// From h-s diagram
+h_1=3234;// kJ/kg
+h_2a=2822;// kJ/kg
+h_2=2910;// kJ/kg
+h_3=3435;// kJ/kg
+h_4a=2977;// kJ/kg
+h_4=2998;// kJ/kg
+h_5=3473;// kJ/kg
+h_6a=2444;// kJ/kg
+h_6=2578;// kJ/kg
+//From steam tables,
+h_f7=162;// kJ/kg
+W=(h_1-h_2)+(h_3-h_4)+(h_5-h_6);// Work done per kg of steam kJ/kg
+Q=(h_1-h_f7)+(h_3-h_2)+(h_5-h_4);// Heat supplied per kg of steam kJ/kg
+n_th=(W/Q)*100;//The thermal efficiency of the cycle in %
+P=(W*m);// Power developed by the plant in kW
+n_i1=((h_1-h_2)/(h_1-h_2a))*100;//Isentropic efficiency of the first stage in %
+n_i2=((h_3-h_4)/(h_3-h_4a))*100;//Isentropic efficiency of the second stage in %
+n_i3=((h_5-h_6)/(h_5-h_6a))*100;//Isentropic efficiency of the third stage in %
+printf('\n(a)The thermal efficiency of the cycle=%0.1f percentage \n Power developed by the plant=%0.0f kW \n(b)Isentropic efficiency of the first stage=%0.1f percentage \n Isentropic efficiency of the second stage=%0.1f percentage \n Isentropic efficiency of the third stage=%0.0f percentage',n_th,P,n_i1,n_i2,n_i3);
diff --git a/3733/CH22/EX22.7/Ex22_7.sce b/3733/CH22/EX22.7/Ex22_7.sce new file mode 100644 index 000000000..6bb3057d8 --- /dev/null +++ b/3733/CH22/EX22.7/Ex22_7.sce @@ -0,0 +1,27 @@ +// Example 22_7
+clc;funcprot(0);
+//Given data
+p_1=100;// bar
+T_1=400;// °C
+p_2=20;// bar
+p_l=1;// bar
+p_3=p_2-p_l;// bar
+T_3=380;//°C
+n_i=80;// Isentropic efficiency of both the expansions in %
+n_t=98;//The transmission efficiency in %
+n_g=95;// The generator efficiency in %
+P=60;// The generator output in MW
+
+//Calculation
+// From h-s diagram
+h_1=3093;// kJ/kg
+h_2a=2734;// kJ/kg
+h_3=3203;// kJ/kg
+h_4a=2157;// kJ/kg
+// The isentropic efficiency of the expansion 1-2 and 3-4 is 80% as given in problem
+h_2=h_1-((n_i/100)*(h_1-h_2a));// kJ/kg
+h_4=h_3-((n_i/100)*(h_3-h_4a));// kJ/kg
+W=(h_1-h_2)+(h_3-h_4);//Work done per kg of steam kJ/kg
+m_s=(P*1000)/(W*(n_t/100)*(n_g/100));// Mass of steam passing through the turbine in kg/sec
+m_s=(m_s*3600)/1000;// tons/hr
+printf('\nThe quantity of steam circulated per hour=%0.1f tons/hr',m_s)
diff --git a/3733/CH22/EX22.8/Ex22_8.sce b/3733/CH22/EX22.8/Ex22_8.sce new file mode 100644 index 000000000..0063e8861 --- /dev/null +++ b/3733/CH22/EX22.8/Ex22_8.sce @@ -0,0 +1,33 @@ +// Example 22_8
+clc;funcprot(0);
+//Given data
+p_1=50;// bar
+T_1=400;// °C
+x=0.96;// Dryness
+p_2=5;// bar
+p_3=0.03;// bar
+T_3=250;// °C
+n_i=80;// Isentropic efficiency of both the expansions in %
+n_m=99;//The mechanical efficiency in %
+n_g=96;// The generator efficiency in %
+
+//Calculation
+//From h-s chart:
+h_1=3198;// kJ/kg
+h_2a=2675;// kJ/kg
+h_3=2955;// kJ/kg
+h_4a=2153;// kJ/kg
+dh_1=((n_i/100)*(h_1-h_2a));//(h_1-h_2) kJ/kg
+h_2=h_1-((n_i/100)*(h_1-h_2a));// kJ/kg
+dh_2=((n_i/100)*(h_3-h_4a));//(h_3-h_4) in kJ/kg
+W=dh_1+dh_2;//Work done per kg of steam kJ/kg
+W_e=W*(n_m/100)*(n_g/100);// The work used out of 1060 kJ for the generation of electricity in kJ/kg
+m_g=(1000/W_e)*3.6;// The steam generated in the boiler per 1 kW power generation in kg/kW-hr
+//From steam tables,
+h_fg=1643.5;// kJ/kg
+Lh=x*h_fg;// The latent heat of steam lost per kg
+m_s=(m_g*(h_3-h_2))/Lh;// The steam used in the reheater in kg
+m=m_g+m_s;// Steam generated by the boiler per kW-hr output from the generator in kg
+printf('\nThe mass of steam generated by the boiler per kW-hr=%0.3f kg',m);
+// The answer vary due to round off error
+
diff --git a/3733/CH22/EX22.9/Ex22_9.sce b/3733/CH22/EX22.9/Ex22_9.sce new file mode 100644 index 000000000..4d5d036b2 --- /dev/null +++ b/3733/CH22/EX22.9/Ex22_9.sce @@ -0,0 +1,19 @@ +// Example 22_9
+clc;funcprot(0);
+//Given data
+p_1=100;// bar
+p_4=0.035;// bar
+T_1=500;// bar
+
+//Calculation
+// From the (Mollier) chart:
+h_1=3373;// kJ/kg
+h_2=2778;// kJ/kg
+h_3=3478;// kJ/kg
+h_4=2322;// kJ/kg
+x_4=0.907;
+// From steam tables
+h_f5=112;// kJ/kg
+W_p=10;// Pump work as calculated in kJ/kg
+n=(((h_1-h_2)+(h_3-h_4)-W_p)/((h_1-h_f5)+(h_3-h_2)));// Efficiency of the cycle
+printf('\n The efficiency of the cycle=%0.2f(or)%0.0f percentage',n,n*100);
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