diff options
Diffstat (limited to '608/CH21')
| -rwxr-xr-x | 608/CH21/EX21.18/21_18.sce | 33 | ||||
| -rwxr-xr-x | 608/CH21/EX21.21/21_21.sce | 38 | ||||
| -rwxr-xr-x | 608/CH21/EX21.22/21_22.sce | 50 | ||||
| -rwxr-xr-x | 608/CH21/EX21.23/21_23.sce | 46 | ||||
| -rwxr-xr-x | 608/CH21/EX21.28/21_28.sce | 52 |
5 files changed, 109 insertions, 110 deletions
diff --git a/608/CH21/EX21.18/21_18.sce b/608/CH21/EX21.18/21_18.sce index d88fa62c6..cb5c14118 100755 --- a/608/CH21/EX21.18/21_18.sce +++ b/608/CH21/EX21.18/21_18.sce @@ -1,18 +1,17 @@ -//Problem 21.18: The shaft torque of a diesel motor driving a 100 V d.c. shunt-wound generator is 25 Nm. The armature current of the generator is 16 A at this value of torque. If the shunt field regulator is adjusted so that the flux is reduced by 15%, the torque increases to 35 Nm. Determine the armature current at this new value of torque.
-
-//initializing the variables:
-T1 = 25; // in Nm
-T2 = 35; // in Nm
-Ia1 = 16; // in Amperes
-V = 100; // in Volts
-x = 0.15;
-
-//calculation:
-//the shaft torque T of a generator is proportional to (phi*Ia), where Phi is the flux and Ia is the armature current. Thus, T = k*Phi*Ia, where k is a constant.
-//The torque at flux phi1 and armature current Ia1 is T1 = k*Phi1*Ia1.
-//similarly T2 = k*Phi2*Ia2
-Phi2 = (1 - x)*Phi1
-Ia2 = T2*Ia1*Phi1/(Phi2*T1)
-
-printf("\n\n Result \n\n")
+//Problem 21.18: The shaft torque of a diesel motor driving a 100 V d.c. shunt-wound generator is 25 Nm. The armature current of the generator is 16 A at this value of torque. If the shunt field regulator is adjusted so that the flux is reduced by 15%, the torque increases to 35 Nm. Determine the armature current at this new value of torque. + +//initializing the variables: +T1 = 25; // in Nm +T2 = 35; // in Nm +Ia1 = 16; // in Amperes +V = 100; // in Volts +x = 0.15; + +//calculation: +//the shaft torque T of a generator is proportional to (phi*Ia), where Phi is the flux and Ia is the armature current. Thus, T = k*Phi*Ia, where k is a constant. +//The torque at flux phi1 and armature current Ia1 is T1 = k*Phi1*Ia1. +//similarly T2 = k*Phi2*Ia2 +Ia2 = T2*Ia1/(0.85*T1) + +printf("\n\n Result \n\n") printf("\n armature current at the new value of torque is %.2f A ",Ia2)
\ No newline at end of file diff --git a/608/CH21/EX21.21/21_21.sce b/608/CH21/EX21.21/21_21.sce index 2cc2c0368..2311a2062 100755 --- a/608/CH21/EX21.21/21_21.sce +++ b/608/CH21/EX21.21/21_21.sce @@ -1,20 +1,20 @@ -//Problem 21.21: A 200 V, d.c. shunt-wound motor has an armature resistance of 0.4 ohm and at a certain load has an armature current of 30 A and runs at 1350 rev/min. If the load on the shaft of the motor is increased so that the armature current increases to 45 A, determine the speed of the motor, assuming the flux remains constant.
-
-//initializing the variables:
-Ia1 = 30; // in Amperes
-Ia2 = 45; // in Amperes
-Ra = 0.4; // in ohm
-n1 = 1350/60; // in Rev/sec
-V = 200; // in Volts
-
-//calculation:
-//The relationship E proportional to (Phi*n) applies to both generators and motors. For a motor,
-//E = V - (Ia*Ra)
-E1 = V - (Ia1*Ra)
-E2 = V - (Ia2*Ra)
-//The relationship, E1/E2 = Phi1*n1/Phi2*n2, applies to both generators and motors. Since the flux is constant, Phi1 = Phi2
-Phi2 = Phi1
-n2 = E2*Phi1*n1/(Phi2*E1)
-
-printf("\n\n Result \n\n")
+//Problem 21.21: A 200 V, d.c. shunt-wound motor has an armature resistance of 0.4 ohm and at a certain load has an armature current of 30 A and runs at 1350 rev/min. If the load on the shaft of the motor is increased so that the armature current increases to 45 A, determine the speed of the motor, assuming the flux remains constant. + +//initializing the variables: +Ia1 = 30; // in Amperes +Ia2 = 45; // in Amperes +Ra = 0.4; // in ohm +n1 = 1350/60; // in Rev/sec +V = 200; // in Volts + +//calculation: +//The relationship E proportional to (Phi*n) applies to both generators and motors. For a motor, +//E = V - (Ia*Ra) +E1 = V - (Ia1*Ra) +E2 = V - (Ia2*Ra) +//The relationship, E1/E2 = Phi1*n1/Phi2*n2, applies to both generators and motors. Since the flux is constant, Phi1 = Phi2 +//Phi2 = Phi1 +n2 = E2*n1/(E1) + +printf("\n\n Result \n\n") printf("\n the speed of the motor is %.2f rev/sec ",n2)
\ No newline at end of file diff --git a/608/CH21/EX21.22/21_22.sce b/608/CH21/EX21.22/21_22.sce index e6fc14636..45c50cd08 100755 --- a/608/CH21/EX21.22/21_22.sce +++ b/608/CH21/EX21.22/21_22.sce @@ -1,26 +1,26 @@ -//Problem 21.22: A 220 V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 ohm. Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant.
-
-//initializing the variables:
-Ia1 = 30; // in Amperes
-Ra = 0.4; // in ohm
-n = 800/60; // in Rev/sec
-V = 220; // in Volts
-x= 0.1;
-
-//calculation:
-//For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,
-E1 = V - (Ia1*Ra)
-//The generated e.m.f. is also such that E proportional to (Phi*n) so at the instant the flux is reduced, the speed has not had time to change, and
-E = E1*(1-x)
-//Hence, the voltage drop due to the armature resistance is
-Vd = V - E
-//The instantaneous value of the current is
-Ia = Vd/Ra
-//T proportional to (Phi*Ia), since the torque is constant,
-//Phi1*Ia1 = Phi2*Ia2, The flux 8 is reduced by 10%, hence
-Phi2 = (1-x)*Phi1
-Ia2 = Phi1*Ia1/Phi2
-
-printf("\n\n Result \n\n")
-printf("\n (a)instantaneous value of the current %.0f A ",Ia)
+//Problem 21.22: A 220 V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 ohm. Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant. + +//initializing the variables: +Ia1 = 30; // in Amperes +Ra = 0.4; // in ohm +n = 800/60; // in Rev/sec +V = 220; // in Volts +x= 0.1; + +//calculation: +//For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f., +E1 = V - (Ia1*Ra) +//The generated e.m.f. is also such that E proportional to (Phi*n) so at the instant the flux is reduced, the speed has not had time to change, and +E = E1*(1-x) +//Hence, the voltage drop due to the armature resistance is +Vd = V - E +//The instantaneous value of the current is +Ia = Vd/Ra +//T proportional to (Phi*Ia), since the torque is constant, +//Phi1*Ia1 = Phi2*Ia2, The flux 8 is reduced by 10%, hence +//Phi2 = (1-x)*Phi1 +Ia2 = Ia1/0.9 + +printf("\n\n Result \n\n") +printf("\n (a)instantaneous value of the current %.0f A ",Ia) printf("\n (b)steady state value of armature current, %.2f A ",Ia2)
\ No newline at end of file diff --git a/608/CH21/EX21.23/21_23.sce b/608/CH21/EX21.23/21_23.sce index ebf977d2b..0b8538445 100755 --- a/608/CH21/EX21.23/21_23.sce +++ b/608/CH21/EX21.23/21_23.sce @@ -1,24 +1,24 @@ -//Problem 21.23: A series motor has an armature resistance of 0.2 ohm and a series field resistance of 0.3 ohm. It is connected to a 240 V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply. (a) Determine the generated e.m.f. at this load. (b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux.
-
-//initializing the variables:
-Ia1 = 15; // in Amperes
-Ia2 = 30; // in Amperes
-Rf = 0.3; // in ohms
-Ra = 0.2; // in ohm
-n1 = 24; // in Rev/sec
-V = 240; // in Volts
-x= 2;
-
-//calculation:
-//generated e.m.f., E, at initial load, is given by
-E1 = V - Ia1*(Ra + Rf)
-//When the current is increased to 30 A, the generated e.m.f. is given by:
-E2 = V - Ia2*(Ra + Rf)
-//E proportional to (Phi*n)
-//E1/E2 = Phi1*n1/Phi2*n2
-Phi2 = x*Phi1
-n2 = E2*Phi1*n1/(Phi2*E1)
-
-printf("\n\n Result \n\n")
-printf("\n (a)generated e.m.f., E is %.1f V ",E1)
+//Problem 21.23: A series motor has an armature resistance of 0.2 ohm and a series field resistance of 0.3 ohm. It is connected to a 240 V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply. (a) Determine the generated e.m.f. at this load. (b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux. + +//initializing the variables: +Ia1 = 15; // in Amperes +Ia2 = 30; // in Amperes +Rf = 0.3; // in ohms +Ra = 0.2; // in ohm +n1 = 24; // in Rev/sec +V = 240; // in Volts +x= 2; + +//calculation: +//generated e.m.f., E, at initial load, is given by +E1 = V - Ia1*(Ra + Rf) +//When the current is increased to 30 A, the generated e.m.f. is given by: +E2 = V - Ia2*(Ra + Rf) +//E proportional to (Phi*n) +//E1/E2 = Phi1*n1/Phi2*n2 +//Phi2 = x*Phi1 +n2 = E2*n1/(E1) + +printf("\n\n Result \n\n") +printf("\n (a)generated e.m.f., E is %.1f V ",E1) printf("\n (b)speed of motor, n2, %.2f A ",n2)
\ No newline at end of file diff --git a/608/CH21/EX21.28/21_28.sce b/608/CH21/EX21.28/21_28.sce index d0bef77dc..08fab7bec 100755 --- a/608/CH21/EX21.28/21_28.sce +++ b/608/CH21/EX21.28/21_28.sce @@ -1,27 +1,27 @@ -//Problem 21.28: A 500 V shunt motor runs at its normal speed of 10 rev/s when the armature current is 120 A. The armature resistance is 0.2 ohm. (a) Determine the speed when the current is 60 A and a resistance of 0.5 ohm is connected in series with the armature, the shunt field remaining constant. (b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value by increasin resistance in the field circuit.
-
-//initializing the variables:
-Ia1 = 120; // in A
-Ia2 = 60; // in A
-Ra = 0.2; // in ohm
-n1 = 10; // in rev/sec
-R = 0.5; // in ohm
-x = 0.8;
-V = 500; // in Volts
-
-//calculation:
-//back e.m.f. at Ia1
-E1 = V - Ia1*Ra
-//at Ia2
-E2 = V - Ia2*(Ra + R)
-//E1/E2 = Phi1*n1/Phi2*n2
-Phi2 = Phi1
-n2 = Phi1*n1*E2/(Phi2*E1)
-//Back e.m.f. when Ia2
-E3 = V - Ia2*Ra
-Phi3 = x*Phi1
-n3 = Phi1*n1*E3/(Phi3*E1)
-
-printf("\n\n Result \n\n")
-printf("\n (a)speed n2 is %.2f rev/sec", n2)
+//Problem 21.28: A 500 V shunt motor runs at its normal speed of 10 rev/s when the armature current is 120 A. The armature resistance is 0.2 ohm. (a) Determine the speed when the current is 60 A and a resistance of 0.5 ohm is connected in series with the armature, the shunt field remaining constant. (b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value by increasin resistance in the field circuit. + +//initializing the variables: +Ia1 = 120; // in A +Ia2 = 60; // in A +Ra = 0.2; // in ohm +n1 = 10; // in rev/sec +R = 0.5; // in ohm +x = 0.8; +V = 500; // in Volts + +//calculation: +//back e.m.f. at Ia1 +E1 = V - Ia1*Ra +//at Ia2 +E2 = V - Ia2*(Ra + R) +//E1/E2 = Phi1*n1/Phi2*n2 +//Phi2 = Phi1 +n2 = n1*E2/(E1) +//Back e.m.f. when Ia2 +E3 = V - Ia2*Ra +//Phi3 = x*Phi1 +n3 = n1*E3/(0.8*E1) + +printf("\n\n Result \n\n") +printf("\n (a)speed n2 is %.2f rev/sec", n2) printf("\n (b)speed n3 is %.2f rev/sec", n3)
\ No newline at end of file |