initial commit / add all books

21 files changed, 297 insertions, 0 deletions

diff --git a/323/CH5/EX5.1/ex5_1.sci b/323/CH5/EX5.1/ex5_1.sci
new file mode 100755
index 000000000..f72889599
--- /dev/null
+++ b/323/CH5/EX5.1/ex5_1.sci
@@ -0,0 +1,9 @@
+//Chapter5,Ex5.1,Pg5.4

+clc;

+//(i)

+V2=110*110/220 // V2/V1 = E2/E1

+printf("\n V2=%.0f V \n",V2)

+//(ii)

+printf("\n V2=%.0f V \n",0)

+

+

diff --git a/323/CH5/EX5.11/ex5_11.sci b/323/CH5/EX5.11/ex5_11.sci
new file mode 100755
index 000000000..4730dfb60
--- /dev/null
+++ b/323/CH5/EX5.11/ex5_11.sci
@@ -0,0 +1,22 @@
+//Chapter 5,Ex5.10,Pg5.13

+clc;

+E1=6600 //Primary voltage

+E2=400 //Secondary voltage

+R1=2.5 //Primary resistance

+X1=3.9 //Primary reactance

+X2=0.025 //Secondary reactance

+R2=0.01 //Secondary resistance

+K=E2/E1 

+//Equivalent resistance referred to primary

+R01=R1+(R2/(K^2))

+printf("\n Equivalent resistance referred to primary=%.2f ohms \n",R01)

+//Equivalent reactance referred to primary

+X01=X1+(X2/(K^2))

+printf("\n Equivalent reactance referred to secondary =%.2f ohms \n",X01)

+//Equivalent resistance referred to secondary

+R02=R2+((K^2)*R1)

+printf("\n Equivalent resistance referred to secondary=%.2f ohms \n",R02)

+//Equivalent resistance referred to secondary

+X02=X2+((K^2)*X2)

+printf("\n Equivalent resistance referred to secondary=%.2f ohms \n",X02)

+

diff --git a/323/CH5/EX5.12/ex5_12.sci b/323/CH5/EX5.12/ex5_12.sci
new file mode 100755
index 000000000..6d7466165
--- /dev/null
+++ b/323/CH5/EX5.12/ex5_12.sci
@@ -0,0 +1,30 @@
+//Chapter 5,Ex5.12,Pg5.14

+clc;

+E1=4400 //Primary EMF

+R1=3.45 //Primary resistance in ohms

+X1=5.2 //Primary reactance in ohms

+E2=220 //Secondary EMF

+R2=0.009 //Secondary resistance in ohms

+X2=0.015 //Secondary reactance in ohms

+K=E2/E1

+I1=50*1000/E1 //Using the formula I1=kVA rating*1000/E

+printf("\n Full load Primary current I1=%.2f A \n",I1)

+I2=50*1000/220

+printf("\n Full load secondary current I2=%.2f A \n",I2)

+R01=R1+(R2/(K*K))

+printf("\n Equivalent resistance referred to primary=%.2f ohms \n",R01)

+X01=X1+(X2/(K*K))

+printf("\n Equivalent reactance referred to primary =%.1f ohms \n",X01)

+Z01=sqrt((R01^2)+(X01*X01))

+printf("\n Equivalent impedance referred to primary=%.2f ohms \n",Z01)

+R02=(K^2)*R01

+printf("\n Equivalent resistance referred to secondary=%.2f ohms \n",R02)

+X02=(K^2)*X01

+printf("\n Equivalent reactance referred to secondary=%.3f ohms \n",X02)

+Z02=(K^2)*Z01

+printf("\n Equivalent impedance referred to secondary=%.2f ohms \n",Z02)

+CL1=(I1^2)*R1 + (I2^2)*R2

+printf("\n Copper losses with individual resistances=%.2f W \n",CL1)

+CL2=(I1^2)*R01

+printf("\n Copper loss with equivalent resistances=%.1f W \n",CL2)

+

diff --git a/323/CH5/EX5.13/ex5_13.sci b/323/CH5/EX5.13/ex5_13.sci
new file mode 100755
index 000000000..73ff83543
--- /dev/null
+++ b/323/CH5/EX5.13/ex5_13.sci
@@ -0,0 +1,7 @@
+//Chapter 5,Ex5.13,Pg5.17

+clc;

+E2=440 //Secondary voltage

+V2=400 //voltage at full load

+//Given that power factor=0.8(lagging)

+percentreg=((E2-V2)/E2)*100

+printf("\n Percentage regulation=%.2f percent\n",percentreg)

diff --git a/323/CH5/EX5.15/ex5_15.sci b/323/CH5/EX5.15/ex5_15.sci
new file mode 100755
index 000000000..795fdb806
--- /dev/null
+++ b/323/CH5/EX5.15/ex5_15.sci
@@ -0,0 +1,20 @@
+//Chapter 5,Ex5.16,Pg5.18

+clc;

+//Let x=cos(phi) and y=sin(phi)

+x=0.8

+y=0.6

+vr=1

+vx=5

+//For 0.8 lagging power factor

+percentreg=vr*x+vx*y

+printf("\n Percentage regulation=%.1f percent \n",percentreg)

+//For unity power factor

+x=1

+y=0

+percentreg=vr*x+vx*y

+printf("\n Percent regulation=%.0f percent \n",percentreg)

+//For 0.8 leading pf

+x=0.8

+y=0.6

+percentreg=vr*x-vx*y

+printf("\n Percent regulation=%.1f percent \n",percentreg)

diff --git a/323/CH5/EX5.17/ex5_17.sci b/323/CH5/EX5.17/ex5_17.sci
new file mode 100755
index 000000000..6d322834a
--- /dev/null
+++ b/323/CH5/EX5.17/ex5_17.sci
@@ -0,0 +1,18 @@
+//Chapter5,Ex5.17,Pg5.19

+clc;

+E1=230 //EMF in primary winding

+E2=460 //EMF in secondary winding

+R1=0.2 //Primary resistance

+R2=0.75 //Secondary resistance

+X1=0.5 //Reactance in ohms

+X2=1.8 //Secondary reactance in ohms

+I2=10 //secondary current in amperes

+pf=0.8 //cos(phi)=0.8

+K=E2/E1 

+printf("K=%.0f \n",K)

+R02=R2+(K^2)*R1 //Effective secondary resistance

+X02=X2+(K^2)*X1 //Effective secondary reactance

+y=sqrt(1-(pf^2)) //sin(phi)=y

+V2=E2-I2*((R02*pf)+(X02*y))

+printf("\n Secondary terminal voltage=%.1f V \n",V2)

+

diff --git a/323/CH5/EX5.18/ex5_18.sci b/323/CH5/EX5.18/ex5_18.sci
new file mode 100755
index 000000000..d07afa1e3
--- /dev/null
+++ b/323/CH5/EX5.18/ex5_18.sci
@@ -0,0 +1,15 @@
+//Example 5.18,Pg5.20

+clc;

+//Given Full load kVA=100kVA

+//Wi=600W(iron loss)

+//WCu=1.5kW(copper loss)

+//(i)

+x=1 //Full load

+pf=0.8

+n=(x*100*pf/((x*100*pf)+0.6+((x^2)*1.5)))*100

+printf("\n Efficiency=%.2f percent \n",n)

+//(ii)

+x=0.5

+pf=1

+n=(x*100*pf/((x*100*pf)+0.6+((x^2)*1.5)))*100

+printf("\n Efficiency=%.2f percent \n",n)

diff --git a/323/CH5/EX5.19/EX5_19.sci b/323/CH5/EX5.19/EX5_19.sci
new file mode 100755
index 000000000..142384d82
--- /dev/null
+++ b/323/CH5/EX5.19/EX5_19.sci
@@ -0,0 +1,26 @@
+//Ex5.19,Pg5.21

+clc;

+flkva=25 //Full load kVA

+R1=1.8 //Primary resistance in ohms

+R2=0.02 //Secondary resistance in ohms

+E1=2200 //Primary EMF in volts

+E2=220 //Secondary EMF in volts

+Wi=1000 //Iron loss in watts

+I2=flkva*1000/220

+printf("\n I2=%.2f A \n",I2)

+K=E2/E1

+printf("\n K=%.1f \n",K)

+R02=R2+(K^2)*R1

+printf("\n Effective secondary resistance=%.3f ohms \n",R02)

+Wcu=(I2^2)*R02

+printf("\n Copper loss=%.2f W \n",Wcu)

+//(i)

+x=1 //Full load

+pf=1

+n=(x*flkva*pf/((x*flkva*pf)+Wi/1000+((x^2)*Wcu/1000)))*100

+printf("\n Efficiency=%.2f percent \n",n)

+//(ii)

+x=0.5

+pf=0.8

+n=(x*flkva*pf/((x*flkva*pf)+Wi/1000+((x^2)*Wcu/1000)))*100

+printf("\n Efficiency=%.2f percent \n",n)

diff --git a/323/CH5/EX5.2/ex5_2.sci b/323/CH5/EX5.2/ex5_2.sci
new file mode 100755
index 000000000..a29042856
--- /dev/null
+++ b/323/CH5/EX5.2/ex5_2.sci
@@ -0,0 +1,5 @@
+//Chapter5,Ex5.2,Pg5.5

+clc;

+//Given data: Flux required=4.13mWb,V1=110V f=50,

+N1=110/(4.44*50*0.001*4.13) //No. of turns= Voltage of operation/(Flux required*4.44*frequency of input signal)

+printf("\n No. of turns=%.0f turns \n",N1)

diff --git a/323/CH5/EX5.20/ex5_20.sci b/323/CH5/EX5.20/ex5_20.sci
new file mode 100755
index 000000000..5a65e2bc7
--- /dev/null
+++ b/323/CH5/EX5.20/ex5_20.sci
@@ -0,0 +1,13 @@
+//Ex5.20,Pg5.22

+clc;

+n1=98.135 //Given efficiency

+n2=97.751 //Given efficiency

+x=1 //Full load

+pf=0.8 //Power factor

+//Using the above data we have to solve 2 simultaneous equations by substituting the values in the formula for calculating the efficiency

+A=[1 1;1 0.25]

+B=[3.8; 2.3]

+W=A\B

+printf("\n Full load copper loss =%.0f kW \n",W(2))

+printf("\n Iron loss =%.1f kW \n",W(1))

+

diff --git a/323/CH5/EX5.21/ex5_21.sci b/323/CH5/EX5.21/ex5_21.sci
new file mode 100755
index 000000000..09946ff84
--- /dev/null
+++ b/323/CH5/EX5.21/ex5_21.sci
@@ -0,0 +1,13 @@
+//Ex5.21,Pg5.23

+clc;

+//Given x=1 and pf=1 we obtain the first equation

+//With x=0.5 and pf=1 we obtain the second equation

+A=[1 1;1 0.25]

+B=[52.2;26.1]

+W=A\B

+printf("\n Copper loss=%.1f kW \n",W(2))

+printf("\n Iron loss=%.1f kW \n",W(1))

+//Now if x=0.6 and pf=1

+n= (0.6*600*1/((0.6*600*1)+W(1)+((0.6^2)*W(2))))*100

+printf("\n Efficiency=%.2f percent \n",n)

+ 
\ No newline at end of file
diff --git a/323/CH5/EX5.22/ex5_22.sci b/323/CH5/EX5.22/ex5_22.sci
new file mode 100755
index 000000000..8cb2bd414
--- /dev/null
+++ b/323/CH5/EX5.22/ex5_22.sci
@@ -0,0 +1,15 @@
+//Ex5.22,Pg5.23

+clc;

+flkva=150 //Given

+Wi=1.4 //Iron loss in kW

+Wcu=1.6 //Copper loss in kW

+//(a)

+lkva=flkva*sqrt(Wi/Wcu)

+printf("\n Load kVA=%.2f kVA \n",lkva)

+//For maximum efficiency Wi=Wcu=1.4kW and pf=0.8

+n= (lkva*0.8/((lkva*0.8)+Wi+Wcu))*100

+printf("\n Efficiency=%.2f percent \n",n)

+//(b)

+n= (0.5*flkva*0.8/((0.5*flkva*0.8)+Wi+(0.5^2)*Wcu))*100

+printf("\n Efficiency=%.2f percent \n",n)

+ 
\ No newline at end of file
diff --git a/323/CH5/EX5.23/ex5_23.sce b/323/CH5/EX5.23/ex5_23.sce
new file mode 100755
index 000000000..272f8c25b
--- /dev/null
+++ b/323/CH5/EX5.23/ex5_23.sce
@@ -0,0 +1,11 @@
+//Example5.23,Pg5.24

+clc;

+flkva=100 //Full load kVA given

+x=1 //Full load

+pf=0.8 //Power factor lagging

+A=[1 1;1 -0.64]

+B=[2.474;0]

+W=A\B

+n= (x*flkva*pf/((x*flkva*pf) + W(1) +(x^2)*W(2)))*100

+printf("\n Efficiency=%.2f percent \n",n)

+

diff --git a/323/CH5/EX5.28/ex5_28.sci b/323/CH5/EX5.28/ex5_28.sci
new file mode 100755
index 000000000..c45ba5893
--- /dev/null
+++ b/323/CH5/EX5.28/ex5_28.sci
@@ -0,0 +1,9 @@
+//Example5.28,Pg5.28

+clc;

+Wi=1 //Iron loss in kW

+op=50*0.8*10+25*0.6*10+0*4

+Wcu=1.2 //Copper loss in kW

+Culoss=1*1.2*10+(25/50)*1.2*10+0 //Copper loss in the entire day considering the load cycle for a day as given in the question

+Iloss=1*24

+nallday= (op/(op+Iloss+Culoss))*100

+printf("\n All day efficiency=%.2f percent \n",nallday)

diff --git a/323/CH5/EX5.3/ex5_3.sci b/323/CH5/EX5.3/ex5_3.sci
new file mode 100755
index 000000000..3a9fa4ae8
--- /dev/null
+++ b/323/CH5/EX5.3/ex5_3.sci
@@ -0,0 +1,12 @@
+//Chapter 5,Ex5.3,Pg5.5

+clc;

+//Given f=50Hz V1=240V N1=80 N2=280 A=200sq cm

+//V1 is approximately equal to E1 for a transformer

+//(i)

+B=240/(4.44*50*200*0.0001*80) //E1=4.44fBmAN1

+printf("\n Maximum flux density Bm=%.2f Wb/m2 \n",B)

+//(ii)

+E2=(280/80)*240 //Induced Emf E2=N2/N1*E1

+printf("\n Induced EMF E2=%.0f V \n",E2)

+

+

diff --git a/323/CH5/EX5.4/ex5_4.sci b/323/CH5/EX5.4/ex5_4.sci
new file mode 100755
index 000000000..ceeccd68e
--- /dev/null
+++ b/323/CH5/EX5.4/ex5_4.sci
@@ -0,0 +1,14 @@
+//Chapter 5,Ex5.4,Pg5.5

+clc;

+//Given E1=3200V E2=400V f=50Hz N2=111

+//Part(i)

+N1=(3200/400)*111 //E2/E1=N2/N1

+printf("\n No of turns in primary=%.0f turns \n",N1)

+//Part(ii)

+I2=80*1000/400 //I2=KVA Rating*1000/V2 where I2=secondary current

+printf("\n Secondary current I2=%.0f A \n",I2)

+//Part(iii)

+A=400/(4.44*50*1.2*111) //Using the formula E=4.44BmAfN2

+printf("\n Cross sectional area=%.4f sq m \n",A)

+

+

diff --git a/323/CH5/EX5.5/ex5_5.sci b/323/CH5/EX5.5/ex5_5.sci
new file mode 100755
index 000000000..70b8c94f3
--- /dev/null
+++ b/323/CH5/EX5.5/ex5_5.sci
@@ -0,0 +1,15 @@
+//Chapter5,Ex5.5,Pg5.6

+clc;

+//Given: kVA rating=5kVA E1=240V E2=2400V f=50Hz Bm=1.2Tesla

+N1=240/8 //Since it is given that EMF per turn is 8

+//(i)

+printf("\n No. of turns in primary=%.0f \n",N1)

+//(ii)

+N2=(2400/240)*N1 //E2/E1=N2/N1

+A=2400/(4.44*50*1.2*300) //using the formula E=4.44BmAfN2

+printf("\n Cross sectional area=%.2f sq m \n",A)

+//(iii)

+I1=5*1000/240 //Using the formula I=kVA rating*1000/V1

+printf("\n I1=%.2f A \n",I1)

+I2=5*1000/2400

+printf("\n I2=%.2f A \n",I2)

diff --git a/323/CH5/EX5.6/ex5_6.sci b/323/CH5/EX5.6/ex5_6.sci
new file mode 100755
index 000000000..2985a63d1
--- /dev/null
+++ b/323/CH5/EX5.6/ex5_6.sci
@@ -0,0 +1,11 @@
+//Chapter5,Ex5.6,Pg5.6

+clc;

+//Given: kVA rating=250kVA f=50Hz N2/N1=0.1 E2=240V

+//(i)

+E1=240/0.1 //E2/E1=N2/N1

+printf("\n Primary EMF E1=%.0f V \n",E1)

+//(ii)

+I1=250*1000/2400 //Using the formula I=kVA rating*1000/V

+printf("\n I1=%.2f A \n",I1)

+I2=250*1000/240 

+printf("\n I2=%.2f A \n",I2)

diff --git a/323/CH5/EX5.7/ex5_7.sci b/323/CH5/EX5.7/ex5_7.sci
new file mode 100755
index 000000000..98a7d14d5
--- /dev/null
+++ b/323/CH5/EX5.7/ex5_7.sci
@@ -0,0 +1,13 @@
+//Chapter 5,Ex5.7,Pg 5.8

+clc;

+W0=200 //Power

+R1=3.5 //Primary resistance

+V1=2300 //Primary voltage

+I0=0.3 //no load current

+cl=(I0^2)*R1

+printf("\n Copper loss=%.3f W \n",cl)

+coreloss=W0-cl //Core loss=Input power-copper loss

+printf("\n Core loss=%.3f W \n",coreloss)

+pf=W0/(V1*I0)

+printf("\n Power factor = %.2f (lagging) \n",pf)

+

diff --git a/323/CH5/EX5.8/ex5_8.sci b/323/CH5/EX5.8/ex5_8.sci
new file mode 100755
index 000000000..0be613208
--- /dev/null
+++ b/323/CH5/EX5.8/ex5_8.sci
@@ -0,0 +1,13 @@
+//Chapter 5,Ex5.8,Pg 5.8

+clc;

+//Primary voltage=230V no-load primary current=5A pf=0.25 N1=200 f=50Hz

+//(i)

+fluxm=230/(4.44*50*200) //Using E=4.44fN1*fluxm

+printf("\n Max flux in the core=%.5f Wb \n",fluxm)

+//(ii)

+W=230*5*0.25 //Using the formula W=V1*I0*powerfactor

+printf("\n Core loss =%.1f W \n",W)

+//(iii)

+x=sqrt(1-(0.25^2)) //x=sin(phi)

+Iu=5*x

+printf("\n Magnetising current=%.2f A \n",Iu)

diff --git a/323/CH5/EX5.9/ex5_9.sci b/323/CH5/EX5.9/ex5_9.sci
new file mode 100755
index 000000000..93dc092d8
--- /dev/null
+++ b/323/CH5/EX5.9/ex5_9.sci
@@ -0,0 +1,6 @@
+//Chapter 5,Ex5.9,Pg 5.13

+clc;

+tr=4 //tr=N1/N2=4 which is given in the question

+K=1/tr

+Rp=50/(K^2) // Using the formula Equivalent resistance referred to primary=R'=R/(K^2)

+printf("\n Rp=%.0f ohms \n",Rp)
\ No newline at end of file