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diff --git a/3731/CH10/EX10.1/Ex10_1.sce b/3731/CH10/EX10.1/Ex10_1.sce
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+//Chapter 10:Traction Drives

+//Example 1

+clc;

+

+//Variable Initialization

+Ma=480    //mass of each motor armature in kg  0.48tonne=480kg

+Da=0.5    //average diameter of each motor in m

+m=450     //mass of each wheel in kg

+R=0.54    //radius of each wheel tread in m

+M=40      //combine wight of one motor and one trailer coach in ton

+alpha=5   //accelaration in metres per second

+N=4       //number of DC motors 

+a=0.4     //gear ratio

+r=20      //train resistance in ohms

+

+//Solution

+Jw=1/2*m*R**2   //inertia of the each wheel in kg metre square

+nw=2*(N*2)      //total number of wheels

+J1=nw*Jw        //total inertia of all the wheels in kg metre square

+

+Jm=N*(1/2*Ma*(Da/2)**2)  //approximate inertia of all the motors in kg metre square

+J2=Jm/a**2               //approximate innertia of the motor referred to the wheels in kg metre square

+

+Fa2=(J1+J2)*alpha*1000/3600/R    //Tractive efforts for driving rorating parts

+Fa1=277.8*M*alpha                //tractive efforts to accelerate the train mass horizontally

+Fr=r*M                           //Tractive efforts required to overcome train resistance

+Ft=Fa1+Fa2+Fr                    //Tractive efforts required to move the train

+Tm=a*R*Ft/N                      //torque per motor

+

+//Result

+mprintf("\nTorque per motor: %.1f N-m",Tm)

diff --git a/3731/CH10/EX10.2/Ex10_2.sce b/3731/CH10/EX10.2/Ex10_2.sce
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index 000000000..4686669c2
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+//Chapter 10:Traction Drives

+//Example 2

+clc;

+

+//Variable Initialization

+M=100      //mass of each motor armature in tonne

+Me=100

+Tm=5000    //torque of each motor in N-m

+Da=0.5     //average diameter of each motor in m

+m=450      //mass of each wheel in kg

+R=0.54     //radius of each wheel tread in m

+N=4        //number of DC motors 

+r=25       //train resistance N/tonne

+a=0.25     //gear ratio 

+nt=0.98    //gear transmission efficiency

+G=50       //up gradient

+Vm=100     //speed in kmph

+

+//Solution

+Ft=nt*Tm*N/a/R     //Tractive efforts required to move the train

+alpha=(Ft-(9.81*M*G+M*r))/(277.8*1.1*Me)  //accelaration in metre per second

+t=Vm/alpha         //time taken to attain speed of Vm in sec

+

+//Result

+mprintf("\n Time taken to reach a speed of 100kmph is : t=%.1f sec",t)

diff --git a/3731/CH10/EX10.3/Ex10_3.sce b/3731/CH10/EX10.3/Ex10_3.sce
new file mode 100644
index 000000000..c31064d73
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+++ b/3731/CH10/EX10.3/Ex10_3.sce
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+//Chapter 10:Traction Drives

+//Example 3

+clc;

+

+//Variable Initialization

+G=8       //up gradient

+r=25      //train resistance N/tonne

+M=500     //mass of the electric train in tonne

+n=0.8     //combine effiency of transmission and motor

+

+//Speed-Time curve characteristics

+t1=60       //characteristic for uniform accelaration at v1 in sec

+alpha=2.5   //given accelaration in km/hr/sec at t1

+t2=5*60     //characteristic for constant speed in sec 

+t3=3*60     //characteristic for coasting in sec

+B=3         //dynamic braking deceleration in km/hr/sec

+

+//Solution

+Vm=alpha*t1  //peak voltage in V

+Fg=9.81*M*G  //tractive effort required to overcome the force of gravity

+Fr=M*r       //tractive effort required to overcome the train resistance

+F_bc=Fg+Fr   //retarding force during coasting in N

+

+Me=1.1*M

+B_c=F_bc/(277.8*Me)   //deceleration during coasting in metre per second square

+V=Vm-B_c*t3           //speed after coasting in m/s

+t4=V/B                //characteristic for a dynamic braking of 3km/hr/sec

+

+d1=1/2*Vm*t1/3600     //distance covered during accelaration 

+d2=Vm*t2/3600         //distance covered during constant speed

+d3=1/2*(Vm+V)*t3/3600 //distance covered coasting

+d4=1/2*V*t4/3600      //distance covered during braking

+D=d1+d2+d3+d4         //distance during stops

+D1=d1+d2

+x=D1/D

+y=1-x

+E=(0.01072*Vm**2/D)*(Me/M)+2.725*G*x+0.2778*r*x   //specific energy output in Whptpkm

+Eo=E/n    //specific energy consumption in Whptpkm

+

+//Result

+mprintf("\n Specific energy consumption is : Eo= %.1f Whptpkm",Eo)

diff --git a/3731/CH10/EX10.4/Ex10_4.sce b/3731/CH10/EX10.4/Ex10_4.sce
new file mode 100644
index 000000000..c92b4c8de
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+++ b/3731/CH10/EX10.4/Ex10_4.sce
@@ -0,0 +1,50 @@
+//Chapter 10:Traction Drives

+//Example 4

+clc;

+

+//Variable Initialization

+G=20              //up gradient

+r=25              //train resistance N/tonne

+M=500             //mass of the electric train in tonne

+n=0.8             //combine effiency of transmission and motor

+

+//Speed-Time curve characteristics

+t1=50       //characteristic for uniform accelaration at v1 in sec

+alpha=3     //given accelaration in km/hr/sec at t1

+t2=10*60    //characteristic for constant speed in sec 

+B=2.5       //uniform braking deceleration in kmphs to rest

+

+//Solution 

+Vm=alpha*t1  //peak voltage in V

+t3=Vm/B      //characteristic for a uniform braking of B=2.5 kmphs

+

+//(i)during accelaration total tractive effort 

+Me=1.1*M

+Fta=277.8*Me*alpha-9.81*M*G+M*r   //total tractive effort during accelaration

+Da=1/2*Vm*t1/3600     //distance covered during accelaration ,and t1 is in seconds

+Ea=Fta*Da*1000/3600   //energy output during accleration in Wh

+

+//(ii)during uniform speed net tractive effort

+Ftu=-9.81*M*G+M*r     //total tractive effort during uniform speed

+//Ftu in the book is -105220 N  which is wrong, hence the other answers are incorrect

+

+Du=Vm*t2/3600         //distance traveled,and t2 is in seconds

+Eu=Ftu*Du*1000/3600   //energy output in Wh

+

+//(iii)during braking net tractive effort

+Ftb=-277.8*Me*B-9.81*M*G+M*r  //total tractive effort for the net braking

+Db=1/2*Vm*t3/3600     //distance covered during braking

+Eb=Ftb*Db*1000/3600   //energy output during braking in Wh

+

+E=Ea/n+n*(Eu+Eb)      //net energy consumption in Wh

+D=Da+Du+Db            //total distance traveled in km

+Eo=E/(M*D)            //specific energy consumption in Wh

+

+//Results 

+mprintf("(i)Energy consumption during accelaration is :Ea : %.1f Wh",Ea)

+mprintf("\n(ii)Energy consumption during uniform speed is :Eu : %d Wh",Eu) 

+mprintf("\n(iii)Energy consumption during braking is :Eb : %.1f Wh",Eb)    

+mprintf("\n Net Energy consumption  is E : %.1f Wh",E)    

+mprintf("\n Total Distance traveled is D : %.4f km",D)

+mprintf("\n Specific Energy consumption  is Eo : %.2f Whptpkm",Eo),

+//Answers provided in the textbook are incorrect

diff --git a/3731/CH10/EX10.5/Ex10_5.sce b/3731/CH10/EX10.5/Ex10_5.sce
new file mode 100644
index 000000000..f910c99a2
--- /dev/null
+++ b/3731/CH10/EX10.5/Ex10_5.sce
@@ -0,0 +1,31 @@
+//Chapter 10:Traction Drives

+//Example 5

+clc;

+

+//Variable Initialization

+Mm=40       //weight of the motor coach in tonne

+Mt=35       //weight of the trailer in tonne

+u=0.2       //co-efficient of adhesion

+r=30        //train resistance N/tonne

+

+//Solution

+//(a)when the motor to trailer ratio is 1:2

+M=Mm+2*Mt     //weight of one unit

+Me=1.1*M

+Md=40         //weight on the driving wheels

+Fm=9810*u*Md  //total tractive effort without the wheel

+Ft=Fm         //at maximum accelaration    

+alpha=(Ft-M*r)/(277.8*Me) //required accelaration since Ft=277.8*u*alpha*M*r

+

+//(b)when the motor to trailer ratio is 1:1

+M=Mm+Mt       //weight of one unit

+Me=1.1*M

+Md=40         //weight on the driving wheels

+Fm=9810*u*Md  //total tractive effort wihout the wheel

+Ft=Fm         //at maximum accelaration    

+alpha1=(Ft-M*r)/(277.8*Me) //required accelaration since Ft=277.8*u*alpha*M*r

+

+

+//Results

+mprintf("(a)Maximum accelaration on a level track  is alpha : %.4f kmphps",alpha)

+mprintf("\n(b)Maximum accelaration when motor to trailer coaches ratio is 1:1 is alpha : %.3f kmphps",alpha1)

diff --git a/3731/CH10/EX10.6/Ex10_6.sce b/3731/CH10/EX10.6/Ex10_6.sce
new file mode 100644
index 000000000..01d73e110
--- /dev/null
+++ b/3731/CH10/EX10.6/Ex10_6.sce
@@ -0,0 +1,33 @@
+//Chapter 10:Traction Drives

+//Example 6

+clc;

+

+//Variable Initialization

+G=10        //up gradient of the locomotive

+Ml=110      //weight of the locomotive coach in tonne

+Mt=500      //weight of the train in tonne

+r=35        //train resistance N/tonne

+n=0.8       //80% of the locomotive weight is carried by the driving wheels

+alpha=1     //acelaration in kmphps

+

+//Solution

+//when only the 110 tonne locomotive is present

+Md=Ml*n     //weight of the motor

+M=Mt+Ml     //total mass of the train

+Me=1.1*M

+Ft=277.8*Me*alpha+9.81*M*G+M*r  //total tractive effort required to move the train

+Fm=Ft

+u=Fm/(9810*Md)   //co-efficient of adhesion ,since Fm=9810*u*Md

+

+//when another locomotive of 70 is added together

+Md=Ml*n+70      // mass of the motor

+M_=Mt+Ml+70     // mass of the train

+Fm=9810*u*Md

+Ft=Fm

+M=Ft/(277.8*1.1*alpha+9.81*G+r)  //total mass of the train, since Ft=277.8*Me*alpha+9.81*M*G+M*r

+W=M-M_          //weight of additional bogies to be attached

+

+

+//Results

+mprintf("\n Given co-efficient of adhesion is: %.2f",u)

+mprintf("\n Weight of additional bogies to be attached is:%.1f T",W)

diff --git a/3731/CH10/EX10.7/Ex10_7.sce b/3731/CH10/EX10.7/Ex10_7.sce
new file mode 100644
index 000000000..cde5382d0
--- /dev/null
+++ b/3731/CH10/EX10.7/Ex10_7.sce
@@ -0,0 +1,38 @@
+//Chapter 10:Traction Drives

+//Example 6

+clc;

+

+//Variable Initialization

+Ml=1000      //weight of the empty train in tonne

+Mt=5000      //weight of the fully loaded train in tonne

+G=15         //gradient of the track

+V=30         //maximum speed of the train 

+r=40         //train resistance in N/tonne

+u=0.25       //co-efficient of adhesion

+alpha=0.3    //acelaration in kmphps

+

+W=100            //weight of each locomotive

+

+//Solution

+Md=W//Md=W*n

+Fm=9810*u*Md

+//By expanding and clubbing similar terms we get

+//(G*9.81*Mt)+(9.81*W*n*G)-((r*Mt)+(r*W*n))

+//(G*9.81*Mt)-(r*Mt)+(9.81*W*n*G)-(r*W*n)

+Fb1=(9.81*Mt*G)-(r*Mt)          //By expanding we get 

+Fb2=(9.81*W*G)-(r*W)//By expanding we get Mt*r+W*n*r

+mprintf("\nFm=%d*n",Fm)

+mprintf("\nFb=%d*n+%d",Fb2,Fb1)

+mprintf("\nEquating Fb and Fm we get")

+n=535750/(245250-10715)

+if (n>2) then 

+    n=3

+end

+mprintf("\nThe number of locomotives is n:%d",n)

+Md=W*n

+M=Ml+W*n

+Ft=277.8*1.1*M*alpha+9.81*M*G+M*r  

+Fm=9810*0.3*Md

+if (Fm>Ft) then

+    mprintf("\nThe train can be accelarated with %d locomotives",n)

+end