diff options
Diffstat (limited to '2345/CH15')
35 files changed, 492 insertions, 0 deletions
diff --git a/2345/CH15/EX15.1/Ex15_1.sce b/2345/CH15/EX15.1/Ex15_1.sce new file mode 100755 index 000000000..169315e96 --- /dev/null +++ b/2345/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,18 @@ +//consumer voltage and energy loss
+//Example 15.1(pg 392)
+clc
+clear
+R=0.2//total resistance of cable in ohms
+I=200//current in A
+t=100//time in hours
+V=240//voltage in volts
+c=0.8//cost of electrical energy in Rs per unit
+V1=I*R//voltage drop in the cable
+//(i)consumer voltage
+Vc=V-V1
+//(ii)Power loss in the cable
+P=I*I*R//in watts
+E=P*t/1000//energy loss in kWh
+C=E*c//cost of energy loss in Rs.
+printf('(i)Consumer voltage is %3.1f Volts \n',Vc)
+printf('(ii)cost of energy loss is Rs %3.2f ',C)
diff --git a/2345/CH15/EX15.10/Ex15_10.sce b/2345/CH15/EX15.10/Ex15_10.sce new file mode 100755 index 000000000..82187aa3c --- /dev/null +++ b/2345/CH15/EX15.10/Ex15_10.sce @@ -0,0 +1,13 @@ +//Finding resistance
+//Example 15.10(pg 398)
+clc
+clear
+rho=1.7*(10^-6)//resistivity of copper in ohm-cm
+l=5//length in metres
+t=0.005//thickness in m
+D=0.08//external diameter in m
+d=D-(2*t)//internal diameter in m
+a=%pi*(D^2-d^2)/4//cross section area in cm^2
+R=rho*l/a//resistance of copper tube in ohm
+R1=R/(10^-4)//resistance in micro-ohm
+printf('Thus the resistance of copper tube is %3.2f micro-ohm',R1)
diff --git a/2345/CH15/EX15.11/Ex15_11.sce b/2345/CH15/EX15.11/Ex15_11.sce new file mode 100755 index 000000000..81b247e91 --- /dev/null +++ b/2345/CH15/EX15.11/Ex15_11.sce @@ -0,0 +1,11 @@ +//Conductivity and conductance
+//Example 15.11(pg 399)
+clc
+clear
+rho=1.7*(10^-8)//resistivity in ohm-m
+K=1/rho//conductivity in mho/m
+a=0.125*(10^-4)//cross sectional area of cable in m^2
+l=2000//length of cable in meters
+G=K*a/l//conductance
+printf('Thus conductivity of cable is %e mho/metres \n',K)
+printf('and conductance of cable is %3f mho',G)
diff --git a/2345/CH15/EX15.12/Ex15_12.sce b/2345/CH15/EX15.12/Ex15_12.sce new file mode 100755 index 000000000..4d69a26ba --- /dev/null +++ b/2345/CH15/EX15.12/Ex15_12.sce @@ -0,0 +1,9 @@ +//Finding resistivity
+//Example 15.12(pg 399)
+clc
+clear
+V=0.05//volume in m^3
+l=300//length in m
+R=0.0306//resistance of conductor in ohm
+rho=R*V/(l^2)//resistivity of conducting material
+printf('Thus resistivity of conducting material is %e ohm-m',rho)
diff --git a/2345/CH15/EX15.13/Ex15_13.sce b/2345/CH15/EX15.13/Ex15_13.sce new file mode 100755 index 000000000..c5f5513e7 --- /dev/null +++ b/2345/CH15/EX15.13/Ex15_13.sce @@ -0,0 +1,11 @@ +//Finding resistivity
+//Example 15.12(pg 399)
+clc
+clear
+rho=0.67*(10^-6)//resistivity in ohm-inch
+m=39.4//1meter = 39.4inch
+m2=1525//1 meter2=1525 square inch
+rhoc=rho*m/m2//resistivity of copper in ohm/m^3
+rho1=rhoc/(10^-6)
+printf('Thus resistivity of copper is %e ohm/m^3',rhoc)
+printf('/n which is equal to %2.4f micro-ohm/m^3',rho1)
diff --git a/2345/CH15/EX15.14/Ex15_14.sce b/2345/CH15/EX15.14/Ex15_14.sce new file mode 100755 index 000000000..69b4a7fbe --- /dev/null +++ b/2345/CH15/EX15.14/Ex15_14.sce @@ -0,0 +1,13 @@ +//Finding resistance
+//Example 15.14(pg. 400)
+clc
+clear
+R1=0.12//old conductor resistance in ohm
+d1=15//diameter of old conductor in cm
+d2=0.4*d1//diameter of new conductor in cm
+a1=%pi*(d1^2)/4//area of cross section of old conductor
+a2=%pi*(d2^2)/4//area of cross section of new conductor
+//R=rho*l/a=rho*V/a^2
+//Henec R is proportional to 1/a^2
+R2=R1*((a1/a2)^2)//resistance of new conductor
+printf('Thus resistance of new conductor is %2.4f ohm',R2)
diff --git a/2345/CH15/EX15.15/Ex15_15.sce b/2345/CH15/EX15.15/Ex15_15.sce new file mode 100755 index 000000000..ae0c135a8 --- /dev/null +++ b/2345/CH15/EX15.15/Ex15_15.sce @@ -0,0 +1,11 @@ +//Finding resistance
+//Example 15.15(pg. 401)
+clc
+clear
+lab=10//la=10*lb ratio of length of A to length of B.
+Aab=1/2//Aa=1/2*Ab ratio of area of A to area of B
+RHOab=1/2//RHOa=2*RHOb ratio of resistivity of A to resistivity of B
+Ra=2//resistance of A in ohm
+Rb=(Ra*Aab)/(lab*RHOab)//resistance of B in ohm
+//Since Ra=RHOa*la/Aa and Rb=RHOb*lb/Ab so from ratio of two we get Rb
+printf('Thus resistance of resistor B is %2.2f ohm',Rb)
diff --git a/2345/CH15/EX15.16/Ex15_16.sce b/2345/CH15/EX15.16/Ex15_16.sce new file mode 100755 index 000000000..a4f415448 --- /dev/null +++ b/2345/CH15/EX15.16/Ex15_16.sce @@ -0,0 +1,14 @@ +//Finding resistance
+//Example 15.16(pg. 402)
+clc
+clear
+RHOo=10.3*(10^-6)//resistivity of platinum wire at 0 degree in ohm-cm
+d=0.0074//diameter of platinum wire
+a=%pi*(d^2)/4//area of cross section of platinum wire in sq cm
+Ro=4//resistance of wire in ohm
+l=Ro*a/RHOo//length of wire in cm
+alphao=0.0038
+t=100//temp in degree C
+R100=Ro*(1+(alphao*t))
+printf('Thus length of wire required is %3.2f cms\n',l)
+printf('and Resistance of wire at 100 degreeC is %2.2f ohms',R100)
diff --git a/2345/CH15/EX15.17/Ex15_17.sce b/2345/CH15/EX15.17/Ex15_17.sce new file mode 100755 index 000000000..95d4f69d8 --- /dev/null +++ b/2345/CH15/EX15.17/Ex15_17.sce @@ -0,0 +1,11 @@ +//Finding resistance
+//Example 15.17(pg. 403)
+clc
+clear
+Ra=1//resistance of A in ohm
+lab=20//ratio of length of A to length of B
+Aab=1/3//ratio of area of A to area of B
+//resistivity is same for both wires
+Rb=Ra*(Aab/lab)//resistance of wire B in ohm
+//since Ra=rho*la/Aa and Rb=rho*lb/Ab so from ratio of both we get Rb
+printf('Thus resistance of wire B is %2.4f omhs',Rb)
diff --git a/2345/CH15/EX15.19/Ex15_19.sce b/2345/CH15/EX15.19/Ex15_19.sce new file mode 100755 index 000000000..d05c3c386 --- /dev/null +++ b/2345/CH15/EX15.19/Ex15_19.sce @@ -0,0 +1,11 @@ +//Finding potential difference
+//Example 15.19(pg. 405)
+clc
+clear
+I1=2/(2+3)//current across 2V battery in circuit EBD in A
+Vbe=3*I1//voltage dropp across BE in V
+I2=4/(5+3)//current across 4V battery in circuit AFC in A
+Vaf=3*I2//voltage dropp across AF in V
+V=Vbe+4-Vaf//sum of potential drops starting from E and ending at F
+//V is the P.D. between E and F
+printf('Thus the P.D. between E and F is %2.1f Volts',V)
diff --git a/2345/CH15/EX15.2/Ex15_2.sce b/2345/CH15/EX15.2/Ex15_2.sce new file mode 100755 index 000000000..03a645516 --- /dev/null +++ b/2345/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,17 @@ +//Resistance and BOT units
+//Example 15.2(pg 393)
+clc
+clear
+Vi=220//voltage in volts supplied by dynamo
+Vo=200//voltage in volts required for lighting
+I=40//current in Amperes
+Pi=Vi*I//power output of dynamo
+Po=Vo*I//power consumed for lighting
+L=Pi-Po//line losses
+R=L/(I^2)//resistance of lines since line losses=I^2*R
+t=10//time in hrs
+N=(Po*t)/1000//no of units of consumed in B.O.T units
+Nw=(L*t)/1000//No of units wasted in B.O.T units
+printf('(i)Resistance of lines is %3.1f Ohms \n',R)
+printf('(ii)No. of B.O.T units consumed in 10hrs is %3.2f B.O.T units\n',N)
+printf('(iii)No. of B.O.T units wasted in 10hrs is %3.2f B.O.T units\n',Nw)
diff --git a/2345/CH15/EX15.20/Ex15_20.sce b/2345/CH15/EX15.20/Ex15_20.sce new file mode 100755 index 000000000..10e1de2af --- /dev/null +++ b/2345/CH15/EX15.20/Ex15_20.sce @@ -0,0 +1,9 @@ +//Finding current
+//Example 15.20(pg. 405)
+clc
+clear
+//Let current in XA=I, in XY=I1, in AY=I-40, in YB=I-40+I1-60, in BX=I+I1-150.
+//By Kirchhoff's second law, in circuit XAYA I-I1=20
+// and in circuit XAYBX 25I+15I1=1950
+I1=(1950-500)/(15+25)//in Amperes
+printf('Thus the current in branch XY is I1=%2.2f Amps',I1)
diff --git a/2345/CH15/EX15.21/Ex15_21.sce b/2345/CH15/EX15.21/Ex15_21.sce new file mode 100755 index 000000000..6342c129a --- /dev/null +++ b/2345/CH15/EX15.21/Ex15_21.sce @@ -0,0 +1,12 @@ +//Finding loss
+//Example 15.21(pg. 407)
+clc
+clear
+A=30//area of hysteresis material in cm^2
+s1=0.4//scale is 1cm=0.4Wb/m^2
+s2=400// and 1cm=400AT/m
+V=1.2*(10^-3)
+f=50//frequency in Hz
+H=A*s1*s2//hysteresis loss/m^3/cycle in joules
+Hp=H*V*f//hysteresis power loss in Watts
+printf('Thus hysteresis power loss is %3.2f Watts',Hp)
diff --git a/2345/CH15/EX15.22/Ex15_22.sce b/2345/CH15/EX15.22/Ex15_22.sce new file mode 100755 index 000000000..c175f9c27 --- /dev/null +++ b/2345/CH15/EX15.22/Ex15_22.sce @@ -0,0 +1,12 @@ +//Finding energy loss
+//Example 15.22(pg. 407)
+clc
+clear
+d=7500//density of iron in kg/m^3
+w=12//weight of iron in kgm
+V=w/d//volume of iron in m^3
+f=25//frequency in Hz
+N=3600*f//number of cycle per hour
+A=300//area in joules/m^3
+E=A*V*N//Total energy loss per hour in joules
+printf('Thus total energy loss per hour is %5.2f Joules',E)
diff --git a/2345/CH15/EX15.23/Ex15_23.sce b/2345/CH15/EX15.23/Ex15_23.sce new file mode 100755 index 000000000..46a3099a1 --- /dev/null +++ b/2345/CH15/EX15.23/Ex15_23.sce @@ -0,0 +1,15 @@ +//Finding inductance and energy
+//Example 15.23(pg. 407)
+clc
+clear
+l=0.5//length of coil in meters
+d=0.1//diameter of coil
+N=1500//no of turns of coil
+a=%pi*(d^2)/4//cross sectional area of coil in m^2
+Ur=1//relative permeability
+Uo=4*%pi*(10^-7)//permeability
+I=8//current in A
+L=((N^2)*a*Uo*Ur)/l//self inductance of coil in H
+E=(1/2)*L*(I^2)//Energy stored in Joules
+printf('Thus Self Inductance of coil is %2.3f H\n',L)
+printf('and Energy stored is %1.2f Joules',E)
diff --git a/2345/CH15/EX15.24/Ex15_24.sce b/2345/CH15/EX15.24/Ex15_24.sce new file mode 100755 index 000000000..10c99412c --- /dev/null +++ b/2345/CH15/EX15.24/Ex15_24.sce @@ -0,0 +1,15 @@ +//Finding flux and field strength
+//Example 15.24(pg. 408)
+clc
+clear
+N=600//number of turns on the coil
+I=2//current passing through solenoid in A
+l=0.6//length of solenoid in meter
+H=N*I/l//magnetic field at the centre in AT/m
+Ur=1//relative permeability
+Uo=4*%pi*(10^-7)//permeability
+d=0.025//diameter in meters
+a=%pi*(d^2)/4//cross sectional area of coil in m^2
+phi=Uo*Ur*H*a//flux in Wb
+printf('Thus Magenetic field at centre is %3.2f AT/m',H)
+printf('\n and Flux is %e Wb',phi)
diff --git a/2345/CH15/EX15.25/Ex15_25.sce b/2345/CH15/EX15.25/Ex15_25.sce new file mode 100755 index 000000000..92aeac097 --- /dev/null +++ b/2345/CH15/EX15.25/Ex15_25.sce @@ -0,0 +1,11 @@ +//Finding Ampere-Turns
+//Example 15.25(pg. 408)
+clc
+clear
+Ur=1//relative permeability
+B=1.257//flux density in Wb/m^2
+Uo=4*%pi*(10^-7)//permeability
+H=B/(Uo*Ur)//magnetising force in AT/m
+l=0.004//length of air gap in meter
+AT=H*l//AT required for the air gap
+printf('Thus AT required for the air gap is %3.1f ',AT)
diff --git a/2345/CH15/EX15.26/Ex15_26.sce b/2345/CH15/EX15.26/Ex15_26.sce new file mode 100755 index 000000000..8e9c29e99 --- /dev/null +++ b/2345/CH15/EX15.26/Ex15_26.sce @@ -0,0 +1,14 @@ +//Finding flux
+//Example 15.26(pg. 409)
+clc
+clear
+D=0.3//diameter of anchor ring in m
+l=%pi*D//length of iron ring in m
+N=400//number of turns on the iron ring
+a=0.0012//area of cross section of iron path in m^2
+Ur=1000//relative permeability
+Uo=4*%pi*(10^-7)//permeability
+I=2//current in A
+phi=(N*I)/(l/(Uo*Ur*a))//flux through iron path in WB
+phi1=phi/(10^-3)//flux in mWb
+printf('Thus flux through iron path is %2.2f mWb',phi1)
diff --git a/2345/CH15/EX15.27/Ex15_27.sce b/2345/CH15/EX15.27/Ex15_27.sce new file mode 100755 index 000000000..eb528372d --- /dev/null +++ b/2345/CH15/EX15.27/Ex15_27.sce @@ -0,0 +1,22 @@ +//Finding magnetising current
+//Example 15.27(pg. 409)
+clc
+clear
+a=0.01//crosssectional area of ring in m^2
+Uo=4*(%pi)*(10^-7)//absolute permeability
+lf=1.25//leakage factor
+Ur=400//permeability
+N=175//no of turns
+phig=0.8*(10^-3)//flux through air gap in Wb
+Bg=phig/a//Flux density in air gap in Wb/m^2
+Hg=Bg/Uo//magnetising force in air gap in AT/m
+Lg=0.004//length of air gap in m
+ATg=Hg*Lg//AT required for air gap in AT
+phii=phig*lf//flux through iron path in Wb
+Bi=phii/a//Flux density in iron path in Wb/m^2
+Hi=Bi/(Uo*Ur)//magnetising force in iron path in AT/m
+Li=1.5//length of iron path in m
+ATi=Hi*Li//At required for iron path in AT
+AT=ATi+ATg//total AT required
+I=ATg/N//Magnetising current required in A
+printf('Thus the magnetising current required is %2.2f Amps',I)
diff --git a/2345/CH15/EX15.28/Ex15_28.sce b/2345/CH15/EX15.28/Ex15_28.sce new file mode 100755 index 000000000..576f3fdfa --- /dev/null +++ b/2345/CH15/EX15.28/Ex15_28.sce @@ -0,0 +1,12 @@ +//Finding charge and capacity
+//Example 15.28(pg. 411)
+clc
+clear
+SI=0.2//steady current in A
+t=0.2//time in sec
+Q=SI*t//charge given to condenser in Coulomb
+V=220//PD across condenser in Volts
+C=Q/V//Capacitance of condenser in F
+C1=C*(10^6)//Capacitance in mircoF
+printf('Thus the Charge of condenser is %2.2f Coulomb\n',Q)
+printf('And the Capacitance of condenser is %3.2f microF',C1)
diff --git a/2345/CH15/EX15.29/Ex15_29.sce b/2345/CH15/EX15.29/Ex15_29.sce new file mode 100755 index 000000000..09702de7e --- /dev/null +++ b/2345/CH15/EX15.29/Ex15_29.sce @@ -0,0 +1,9 @@ +//Finding heat
+//Example 15.29(pg. 411)
+clc
+clear
+C=2*(10^-6)//capacitance of condenser in F
+V=10000//PD across condenser in Volts
+E=(1/2)*C*(V^2)//energy stored in condenser in Joules
+H=E/4.2//heat produced in the wire in calories
+printf('Thus heat produced in the wire is %2.2f calories',H)
diff --git a/2345/CH15/EX15.3/Ex15_3.sce b/2345/CH15/EX15.3/Ex15_3.sce new file mode 100755 index 000000000..7fca59d5e --- /dev/null +++ b/2345/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,15 @@ +//Finding current
+//Example 15.3(pg 393)
+clc
+clear
+M=250000//weight of water lifted per hr in kg
+h=50//height in metres
+g=9.81//gravitational const.
+WD=M*h*g//work done by pump per hr in watt-sec
+P=WD/3600//Power output of pump per sec in watts
+V=500//supply voltage in volts
+Ep=0.8//efficiency of pump
+Em=0.9//efficiency of motor
+E=Em*Ep//overall efficiency
+I=P/(V*E)//current in amperes
+printf('Current drawn by the motor is %3.2f Amperes',I)
diff --git a/2345/CH15/EX15.30/Ex15_30.sce b/2345/CH15/EX15.30/Ex15_30.sce new file mode 100755 index 000000000..c33ebde03 --- /dev/null +++ b/2345/CH15/EX15.30/Ex15_30.sce @@ -0,0 +1,14 @@ +//Finding K and flux density
+//Example 15.30(pg. 411)
+clc
+clear
+V=15*(10^3)//potential difference applied in V
+A=0.02//surface area of plate in m^2
+d=0.001//distance between plates in m
+C=4.5*(10^-10)//Capacitance of capacitor in F
+Ko=8.854*(10^-12)//constant
+K=(C*d)/(Ko*A)//dielectric constant
+q=C*V//charge on condenser in C
+D=q/A//Electric flux density in C/m^2
+printf('Thus the Charge of condenser is %e Coulomb\n',q)
+printf('And the electric flux density of condenser is %e microF',D)
diff --git a/2345/CH15/EX15.31/Ex15_31.sce b/2345/CH15/EX15.31/Ex15_31.sce new file mode 100755 index 000000000..c96145cbb --- /dev/null +++ b/2345/CH15/EX15.31/Ex15_31.sce @@ -0,0 +1,17 @@ +//Finding resistance
+//Example 15.31(pg. 412)
+clc
+clear
+m=0.6//mass of water in kgm
+S=4200//specific heat of water
+T1=100//temperature in degreeC
+T2=10//temperature in degreeC
+t=5*60//time in sec
+V=230//Supply voltage in Volts
+H=m*S*(T1-T2)//Heat required to raise the temp of water from 0 to 100 degree. in J
+e=0.78//efficiency of kettle
+Ei=H/e//Energy input in Joules
+Ei1=Ei/(100*3600)//Energy input in kWh
+W=Ei/t//Rating of kettle in watts
+R=(V*V)/W//Resistance of heating element in ohms
+printf('Thus Resistance of heating element is %2.1f ohms',R)
diff --git a/2345/CH15/EX15.32/Ex15_32.sce b/2345/CH15/EX15.32/Ex15_32.sce new file mode 100755 index 000000000..79e15dfcd --- /dev/null +++ b/2345/CH15/EX15.32/Ex15_32.sce @@ -0,0 +1,18 @@ +//Finding time
+//Example 15.32(pg. 413)
+clc
+clear
+m1=120//mass of water to be heated in kg
+m2=20//mass of copper tank in kg
+S1=1//specific heat of water
+S2=0.095//specific heat of copper
+T1=10//temp in degreeC
+T2=60//temp in degreeC
+H=(m1*S1*(T2-T1))+(m2*S2*(T2-T1))//heat required to raise the temp of water and tank in kcal
+H1=H*4200//heat required in Joules
+e=0.8//thermal efficiency
+E=H1/e//Energy input in joules
+E1=E/(1000*3600)//energy input in kWh
+r=3//rating of heater in kW
+t=E1/r//time taken in hours
+printf('Thus the time taken to raise the temp is %2.3f hours',t)
diff --git a/2345/CH15/EX15.33/Ex15_33.sce b/2345/CH15/EX15.33/Ex15_33.sce new file mode 100755 index 000000000..624932d3d --- /dev/null +++ b/2345/CH15/EX15.33/Ex15_33.sce @@ -0,0 +1,10 @@ +//Finding frequency
+//Example 15.33(pg. 414)
+clc
+clear
+rho=5*(10^-5)//specific resistance for steel in ohm-cm
+U=1//relative permeability
+d=0.15//depth of penetration in cm
+f=(rho*(10^9))/(U*d*d*4*(%pi^2))//frequency required in cycles per sec
+f1=f/1000//frquency in k.cycles/sec
+printf('Thus the frequency required is %3.3f k.cycles/sec',f1)
diff --git a/2345/CH15/EX15.34/Ex15_34.sce b/2345/CH15/EX15.34/Ex15_34.sce new file mode 100755 index 000000000..dd4721118 --- /dev/null +++ b/2345/CH15/EX15.34/Ex15_34.sce @@ -0,0 +1,30 @@ +//Finding current,voltage and power
+//Example 15.34(pg. 414)
+clc
+clear
+v=50*20*2//Volume of board to be heated in cm^3
+Mw=0.56//weight of wood in gm/cm^3
+m=Mw*v/1000//mass of wood in kgm
+S=0.35//specific heat of wood
+t=15/60//time in hrs
+f=30*(10^6)//frequency in cycles/sec
+t2=150,t1=30//temp in degreeC
+H=m*S*(t2-t1)//heat required to raise the temp in kcal
+Hw=H*1000/860//heat required in kW
+P=Hw/t//power required in Watts
+e=0.5//efficiency of dielectric heating process
+Pi=P/e//power input required in Watts
+Ko=8.854*(10^-12)//absolute permittivity
+K=5//relative permittivity
+A=0.5*0.2//area in m
+i=0.02
+C=Ko*K*A/i//capacitance of parallel plate capacitor in F
+Xc=1/(2*%pi*f*C)//capacitive reactance in ohms
+cosx=0.05
+tanx=19.97
+R=Xc*tanx//resistance
+V=sqrt(Pi*R)//voltage in volts
+Ic=V/Xc//current through the board in Amps
+printf('Thus the power required is %2.1f Watts\n',Pi)
+printf('And Voltage across the board is %3.2f volts\n',V)
+printf('And the current through the board is %2.3f Amps',Ic)
diff --git a/2345/CH15/EX15.35/Ex15_35.sce b/2345/CH15/EX15.35/Ex15_35.sce new file mode 100755 index 000000000..ee2dd5bc5 --- /dev/null +++ b/2345/CH15/EX15.35/Ex15_35.sce @@ -0,0 +1,14 @@ +//Finding efficiency
+//Example 15.35(pg. 416)
+clc
+clear
+m=2//quantity of aluminium to be melted in kg
+t1=15,t2=660//temp in degreeC
+S=0.212//specific heat of aluminium
+L=78.8//latent heat of aluminium in kcal/kg
+H=(m*S*(t2-t1))+(m*L)//total heat required to melt Al in kcal
+i=5//input to furnace in kW
+E=i*(1000*10*60)//Energy input to furnace in watt-sec
+E1=E/4180//energy input in kcal
+e=H*100/E1//efficiency of furnace
+printf('Thus the efficiency of furnace is %2.3f percent',e)
diff --git a/2345/CH15/EX15.36/Ex15_36.sce b/2345/CH15/EX15.36/Ex15_36.sce new file mode 100755 index 000000000..782d77e04 --- /dev/null +++ b/2345/CH15/EX15.36/Ex15_36.sce @@ -0,0 +1,12 @@ +//Finding cost
+//Example 15.36(pg. 417)
+clc
+clear
+O=5*735.5//output of motor in W
+e=0.85//efficiency of motor
+c=2//cost of energy per unit in Rs
+I=O/e//input of motor in Watts
+t=4//time in hrs
+E=I*t/1000//energy consumed in kWh
+C=c*E//cost of using the motor in Rs
+printf('Thus the cost of using the motor is %2.3f Rs',C)
diff --git a/2345/CH15/EX15.37/Ex15_37.sce b/2345/CH15/EX15.37/Ex15_37.sce new file mode 100755 index 000000000..6d6faf91f --- /dev/null +++ b/2345/CH15/EX15.37/Ex15_37.sce @@ -0,0 +1,10 @@ +//Finding no of electrons
+//Example 15.37(pg. 417)
+clc
+clear
+I=2.5*(10^-3)//current in Amp
+t=30*(10^-3)//time in sec
+Q=I*t//charge passing through the person in Coulumbs
+e=1.602*(10^-19)//charge of 1 electron in C
+N=Q/e//no of electrons passing through the person
+printf('Thus the no of electrons passing through the person is %e electrons',N)
diff --git a/2345/CH15/EX15.38/Ex15_38.sce b/2345/CH15/EX15.38/Ex15_38.sce new file mode 100755 index 000000000..b28bf1fb8 --- /dev/null +++ b/2345/CH15/EX15.38/Ex15_38.sce @@ -0,0 +1,15 @@ +//Finding resistance
+//Example 15.38(pg. 417)
+clc
+clear
+//(a)Finding resistance between 2 ends
+l=1//length in m
+a=2.5*(10^-2)*0.05*(10^-2)//area of cross section in m^2
+rho=1.724*(10^-8)//specific resistance of copper in ohm-m
+R=rho*l/a//resistance of the strip in ohm
+//(b) Finding resistance between 2 faces
+l1=0.05*(10^-2)//length in m
+a1=2.5*(10^-2)*1//area of cross section in m^2
+R1=rho*l1/a1//resistance in ohm
+printf('Thus the resistance of the strip is %e ohms\n ',R)
+printf('And the resistance between the faces is %e ohms',R1)
diff --git a/2345/CH15/EX15.39/Ex15_39.sce b/2345/CH15/EX15.39/Ex15_39.sce new file mode 100755 index 000000000..8a9fb6d82 --- /dev/null +++ b/2345/CH15/EX15.39/Ex15_39.sce @@ -0,0 +1,20 @@ +//Finding resistance and cost
+//Example 15.39(pg. 418)
+clc
+clear
+m=2//weight of water to be heated in kg
+t2=98,t1=15//temp in degreeC
+s=1//specific heat of water
+V=200//voltage in volts
+H=m*s*(t2-t1)//energy required to raise the temp of water in kcal
+H1=H*4200//energy in Watt-sec or Joules
+e=0.85//efficiency of kettle
+E=H1/e//energy input required in watt-sec
+E1=E/(1000*3600)//energy input in kWh
+c=35//cost per unit in paise
+C=c*E1//ocst of energy used in paise
+t=10/60//time in hrs
+W=E1*1000/t//wattage of kettle in watts
+R=V*V/W//resistance of heating element in ohms
+printf('Thus the resistance of heating element is %2.0f ohms\n',R)
+printf('And the cost of energy used is %2.0f paisa',C)
diff --git a/2345/CH15/EX15.4/Ex15_4.sce b/2345/CH15/EX15.4/Ex15_4.sce new file mode 100755 index 000000000..6001db4b3 --- /dev/null +++ b/2345/CH15/EX15.4/Ex15_4.sce @@ -0,0 +1,15 @@ +//Finding torque
+//Example 15.4(pg 394)
+clc
+clear
+P=10//Power developed by motor in H.P
+N=600//Speed of motor in rpm
+//1HP=735.5Nw-m/sec=75kgm/sec
+a=75
+b=735.5
+//Torque in kg-m
+Tkgm=(P*a*60)/(2*%pi*N)//since P=2*pi*NT/60
+//Torque in Nw-m
+TNwm=(P*b*60)/(2*%pi*N)//since P=2*pi*NT/60
+printf('(i)Torque in kg.meter is %3.2f kg-m \n',Tkgm)
+printf('(ii)Torque in Newton.meter is %3.2f Nw-m',TNwm)
diff --git a/2345/CH15/EX15.40/Ex15_40.sce b/2345/CH15/EX15.40/Ex15_40.sce new file mode 100755 index 000000000..cf87200c5 --- /dev/null +++ b/2345/CH15/EX15.40/Ex15_40.sce @@ -0,0 +1,18 @@ +//Finding current
+//Example 15.40(pg. 418)
+clc
+clear
+phi=70000/(10^8)//flux to be set up in Wb since 10^8lines =1Wb
+d=0.03//diameter in m
+a=%pi*d*d/4//area of cross section in m^2
+B=phi/a//flux density in Wb/m^2
+Lg=0.002//length of air gap in m
+Ls=(%pi*0.2)-Lg//length of steel path
+Uo=4*%pi*(10^-7)//absolute permitivity
+Ur=800//relative permitivity of steel
+Hg=B/Uo
+Hs=B/(Uo*Ur)
+AT=(Hg*Lg)+(Hs*Ls)//total ampere turns required
+N=500// no of turns
+I=AT/N//exciting current in amps
+printf('Thus the value of exciting current is %2.3f A',I)
diff --git a/2345/CH15/EX15.5/Ex15_5.sce b/2345/CH15/EX15.5/Ex15_5.sce new file mode 100755 index 000000000..38560dd52 --- /dev/null +++ b/2345/CH15/EX15.5/Ex15_5.sce @@ -0,0 +1,14 @@ +//finding mass and energy
+//Example 15.5(pg 395)
+clc
+clear
+P=25//Output of diesel engine in kW
+s=12500//calorific value of fuel oil in k-cal/kgm
+e=0.35//overall efficiency of diesel set
+P1=P/e//input energy required in 1 hour in kWh
+P2=P1*860//input energy in kcal
+m=P2/s//mass of oil needed per hr in kgm
+w=1000//weight of 1 ton of oil in kgm
+Eg=(P*w)/m//Energy generated by 1ton of oil in kWh
+printf('(i)Mass of oil required per hr is %3.3f kgm \n',m)
+printf('(ii)Eletrical energy generated per ton of fuel is %4.1f Kwh',Eg)
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