initial commit / add all books

6 files changed, 154 insertions, 0 deletions

diff --git a/409/CH27/EX27.1/Example27_1.sce b/409/CH27/EX27.1/Example27_1.sce
new file mode 100755
index 000000000..2f60d9937
--- /dev/null
+++ b/409/CH27/EX27.1/Example27_1.sce
@@ -0,0 +1,19 @@
+clear ;
+clc;
+// Example 27.1
+printf('Example 27.1\n\n');
+//page no. 838
+// Solution E27.1
+
+// Given 
+V_w = 1 ;//  Volume of given water -[L]
+P_atm = 100 ;// Atmospheric pressure - [kPa]
+
+//W = -p*del_V
+V_H2O = 0.001043 ;// Specific volume of water from steam table according to book- [cubic metre] 
+V_vap = 1.694 ;// Specific volume of vapour from steam table according to book- [cubic metre] 
+V1 = 0 ;// Initial volume of H2O in bag-[cubic metre]
+V2 = (V_w*V_vap)/(1000*V_H2O) ;// Final volume of water vapour -[cubic metre] 
+W = -P_atm*(V2 -V1)* 1000 ;// Work done by saturated liquid water -[J]
+ 
+printf(' Work done by saturated liquid water is %.3e J.\n',W) ;
\ No newline at end of file
diff --git a/409/CH27/EX27.2/Example27_2.sce b/409/CH27/EX27.2/Example27_2.sce
new file mode 100755
index 000000000..81e508d7c
--- /dev/null
+++ b/409/CH27/EX27.2/Example27_2.sce
@@ -0,0 +1,27 @@
+clear ;
+clc;
+// Example 27.2
+printf('Example 27.2\n\n');
+//page no. 840
+// Solution E27.2
+
+// Given 
+m_N2 = 1 ;// Moles of N2 taken -[kg mol]
+p = 1000;// Pressure of cylinder-[kPa]
+T = 20 + 273 ;// Temperature of cylinder -[K]
+a_pis = 6 ;// Area of piston - [square centimetre]
+m_pis = 2 ;// Mass of pston - [kg]
+R = 8.31 ;// Ideal gas constant - [(kPa*cubic metre)/(K * kgmol)]
+
+V = (R*T)/p ;// Specific volue of gas at initial stage -[cubic metre/kg mol]
+V1 = V * m_N2 ;// Initial volume of gas - [cubic metre]
+V2 = 2*V1 ;// Final volume of gas according to given condition -[cubic metre]
+
+// Assumed surrounding pressure constant = 1 atm
+p_atm = 101.3 ;// Atmospheric pressure-[kPa]
+del_Vsys = V2 -V1 ;// Change in volume of system -[cubic metre]
+del_Vsurr = - del_Vsys ;// Change in volume of surrounding -[cubic metre]
+W_surr = -p_atm*del_Vsurr ;// Work done by surrounding - [kJ]
+W_sys = -W_surr ;// Work done by system - [kJ]
+
+printf(' Work done by gas(actually gas + piston system) is %.0f kJ.\n',W_sys) ;
\ No newline at end of file
diff --git a/409/CH27/EX27.3/Example27_3.sce b/409/CH27/EX27.3/Example27_3.sce
new file mode 100755
index 000000000..b639cb0be
--- /dev/null
+++ b/409/CH27/EX27.3/Example27_3.sce
@@ -0,0 +1,19 @@
+clear ;
+clc;
+// Example 27.3
+printf('Example 27.3\n\n');
+//page no. 845
+// Solution 
+
+// Given 
+p_plant = 20 ;// Power generated by plant-[MW]
+h = 25 ;// Height of water level - [m]
+V = 100 ;// Flow rate of water -[cubic metre/s]
+d_water = 1000 ;// Density of water - [ 1000 kg / cubic metre]
+g = 9.807 ;// Acceleration due to gravity-[m/square second]
+
+M_flow = V*d_water ;// Mass flow rate of water -[kg/s]
+del_PE = M_flow*g*h ;// Potential energy change of water per second -[W]
+eff = (p_plant*10^6) /(del_PE) ;// Efficiency of plant 
+
+printf(' Efficiency of plant is %.2f .\n',eff) ;
\ No newline at end of file
diff --git a/409/CH27/EX27.4/Example27_4.sce b/409/CH27/EX27.4/Example27_4.sce
new file mode 100755
index 000000000..b61cdfcb9
--- /dev/null
+++ b/409/CH27/EX27.4/Example27_4.sce
@@ -0,0 +1,30 @@
+clear ;
+clc;
+// Example 27.4
+printf('Example 27.4\n\n');
+//page no. 845
+// Solution Fig.E27.4
+
+// Given 
+LHV = 36654 ;// LHV value of fuel - [kJ/ cubic metre]
+Q1 = 16 ;//- [kJ/ cubic metre]
+Q2 = 0 ;//- [kJ/ cubic metre]
+Q3 = 2432 ;//- [kJ/ cubic metre]
+Q4 = 32114 ;//- [kJ/ cubic metre]
+Q41 = 6988 ;//- [kJ/ cubic metre]
+Q8 = 1948 ;//- [kJ/ cubic metre]
+Q9 = 2643 ;//- [kJ/ cubic metre]
+Q81 = 2352 - Q8 ;// - [kJ/ cubic metre]
+Q567 = 9092 ;// Sum of Q5, Q6 and Q7- [kJ/ cubic metre]
+
+//(a)
+G_ef = (LHV+ Q1 +Q2 + Q3 - Q9)/(LHV) ;// Gross efficiency
+printf('(a) Gross efficiency is %.3f .\n',G_ef) ;
+
+//(b)
+T_ef = (Q567+Q8)/(LHV+ Q1 +Q2 + Q3) ;//Thermal efficiency 
+printf(' (b) Thermal efficiency is %.3f .\n',T_ef) ;
+
+//(c)
+C_ef = Q4/(Q4 + Q41) ;// Combustion efficiency
+printf(' (c) Combustion efficiency is %.3f .\n',C_ef) ;
\ No newline at end of file
diff --git a/409/CH27/EX27.5/Example27_5.sce b/409/CH27/EX27.5/Example27_5.sce
new file mode 100755
index 000000000..da9833b0d
--- /dev/null
+++ b/409/CH27/EX27.5/Example27_5.sce
@@ -0,0 +1,29 @@
+clear ;
+clc;
+// Example 27.5
+printf('Example 27.5\n\n');
+//page no. 850
+// Solution 
+
+// Given 
+V1 = 5 ;// Volume of gas initially - [cubic feet]
+P1 = 1 ;// Initial pressure - [atm]
+P2 = 10 ;// Final pressure - [atm]
+T1 = 100 + 460 ;// initial temperature - [degree Rankine]
+R = 0.7302 ;// Ideal gas constant -[(cubic feet*atm)/(lb mol)*(R)]
+//Equation of state pV^1.4 = constant
+
+//(a)
+//Energy balance reduces to del_E = del_U = del_W 
+V2 = V1*(P1/P2)^(1/1.4) ;// Final volume - [cubic feet]    
+W1_rev = integrate('-(P1)*(V1/V)^(1.4)','V',V1,V2) ;// Reversible work done in compresion in a horizontal cylinder with piston -[cubic feet *atm]
+W1 = W1_rev *1.987/.7302 ;// Conversion to Btu -[Btu]
+
+printf('\n (a)Reversible work done in compression in a horizontal cylinder with piston is %.1f Btu .\n ',W1);
+
+//(b)
+n1 = (P1*V1)/(R*T1) ;// Number of moles of gas
+W2_rev = integrate('(V1)*(P1/P)^(1/1.4)','P',P1,P2) ;// Reversible work done in compresion in a rotary compressor -[cubic feet *atm]
+W2 = W2_rev *1.987/.7302 ;// Conversion to Btu -[Btu]
+
+printf('\n (b)Reversible work done in a rotary compressor is %.1f Btu .\n ',W2);
\ No newline at end of file
diff --git a/409/CH27/EX27.6/Example27_6.sce b/409/CH27/EX27.6/Example27_6.sce
new file mode 100755
index 000000000..650f46728
--- /dev/null
+++ b/409/CH27/EX27.6/Example27_6.sce
@@ -0,0 +1,30 @@
+clear ;
+clc;
+// Example 27.6
+printf('Example 27.6\n\n');
+//page no. 853
+// Solution 
+
+// Given 
+m_water = 1 ;// Mass flow rate of water -[lb/min]
+P1 = 100 ;// Initial pressure - [psia]
+P2 = 1000 ;// Final pressure - [psia]
+T1 = 80 + 460 ;// initial temperature - [degree Rankine]
+T2 = 100 + 460 ;// final temperature - [degree Rankine]
+h = 10 ;// Difference in water level between entry and exit of stream-[ft]
+g = 32.2 ;// Accleration due to gravity - [ft/ square second]
+gc = 32.2 ;//[(ft*lbm)/(lbf*square second)]
+
+// The mechanical energy balance reduces to W = PV_work + del_PE ....(A)
+// From steam table , specific volume of liquid water at 80 and 100 degree F is noted , according to book it is as follows-
+v1 = .01607 ;// specific volume of liquid water at 80 degree F -[cubic feet/lbm]
+v2 = .01613 ;// specific volume of liquid water at 100 degree F -[cubic feet/lbm] 
+// But for pratical purposes wwater is taken to be incompressible and specific volume can be taken as v, ith following value
+v= 0.0161 ;// -[cubic feet/lbm]
+
+del_PE = (h*g)/(gc*778) ;// Change in potential energy - [Btu/lbm]
+PV_work = integrate('(v)*(12^2/778)','P',P1,P2) ;// PV work done  -[Btu/lbm]
+//From eqn. (A)
+W = PV_work + del_PE ;// Work per minute required to pump 1 lb water per minute - [Btu/lbm]
+
+printf('\n Work per minute required to pump 1 lb water per minute is %.2f Btu/lbm .\n ',W);
\ No newline at end of file