Practice Questions -3 Solutions

pdf

School

University of West Alabama *

*We aren’t endorsed by this school

Course

162

Subject

Mechanical Engineering

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by CountTigerMaster889

Deakin University CRICOS Provider Code: 00113B n k k E A A   '     2 60  n  T EI P A   A A T R I E V   BASIC FORMULAS F F I K  

Deakin University CRICOS Provider Code: 00113B The armature of a brushed motor has a resistance of 0.25  and when it is connected to a 125V dc bus, it draws an armature current of 60A. Calculate the back EMF generated in the armature conductors of the motor. E=V – Ia Ra =125-(60 x 0.25) = 110 V

Deakin University CRICOS Provider Code: 00113B A dc shunt motor having an armature resistance of 0.25  and a brush contact voltage drop (BD) of 3V receives an applied voltage across its armature terminals of 120V. Calculate the armature current when: I. The speed produces a counter EMF of 110V at a given load. II. The speed drops (due to application of additional load) and the counter EMF is 105V. I. Ia =V – (E + BD) / Ra =120-(110+3)/0.25 = 28A II. Ia =V – (E + BD) / Ra =120-(105+3)/0.25 = 48A

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Deakin University CRICOS Provider Code: 00113B A 220V DC shunt motor draws 10A at 1800rpm. The armature resistance is 0.2  and field winding resistance is 440  . ( a )What is the torque? A V R V I F T F 5 . 0 440 220     A A A I I I F L A 5 . 9 5 . 0 10         V A V R I V E A A T 218 2 . 0 5 . 9 220          kW A V EI P A 07 . 2 5 . 9 218      s rad n / 188 60 1800 2 60 2       m N s rad kW P T     0 . 11 / 188 07 . 2  F A L I I I   A A A T R I E V  

Deakin University CRICOS Provider Code: 00113B A 220V DC shunt motor draws 10A at 1800rpm. The armature resistance is 0.2  and field winding resistance is 440  . (b) What will be the speed and line current at a torque of 20 N-m (if field current is constant)? 16 . 1 / 188 218     s rad V E K K E A A    A m N K T I I K T A A A A 3 . 17 16 . 1 20        A A A I I I F A L 8 . 17 5 . 0 3 . 17         V V R I V E A A T 217 2 . 0 3 . 17 220       rpm x n s rad V K E A 3 10 79 . 1 2 60 / 187 16 . 1 217          F A L I I I   A A A T R I E V  

Deakin University CRICOS Provider Code: 00113B A 10hp, 115V Dc series motor takes 40A at its full load speed of 1800rpm. What is the torque at 30A?   s rad n / 188 60 1800 2 60 2         kW hp W hp P 46 . 7 746 10           m N s rad kW P T T P      6 . 39 / 188 46 . 7   A F F A A A I I K K I K T    A F I I  2 A F A I K K T    025 . 0 40 6 . 39 2 2     A m N I T K K A F A    m N A I K K T new A F A new     2 . 22 30 025 . 0 2 2

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Deakin University CRICOS Provider Code: 00113B A 440-V, shunt motor has armature resistance of 0.8 Ω and field resistance of 200 Ω. Determine the back e.m.f. when giving an output of 7.46 kW at 85 percent efficiency. Motor input power = 7.46 × 103 /0.85 W Motor input current = 7460/0.85 × 440 = 19.95 A ; If = 440/200 = 2.2 A Ia = 19.95 − 2.2 = 17.75 A ; Now, Eb = V − Ia. ra Eb = 440 − (17.75 × 0.8) = 425.8 V

Deakin University CRICOS Provider Code: 00113B A 25-kW, 250-V, d.c. shunt machine has armature and field resistances of 0.06 Ω and 100 Ω respectively. Determine the total armature power developed when as a motor taking 25 kW input. Input current I = 25,000/250 = 100 A ; If = 250/100 = 2.5 A ; Ia = Il – If = 97.5 A Eb = 250 − (97.5 × 0.06) = 250 − 5.85 = 244.15 V Power developed in armature = Eb Ia = 244.15 × 97.5/1000 = 23.8 kW

Deakin University CRICOS Provider Code: 00113B A 220-V d.c. shunt motor runs at 500 r.p.m. when the armature current is 50 A. Calculate the speed if the torque is doubled. Given that Ra = 0.2 Ω Ta ΦIa . Since Φ is constant, Ta I a Ta1 I a1 and Ta2 I a2 Ta2 /Ta1 = I a2 /I a1 2 = I a2 /50 or I a2 = 100 A Now, N2 /N1 = Eb2 /Eb1 − since Φ remains constant. Eb1 = 220 − (50 × 0.2) = 210 V Eb2 = 220 − (100 × 0.2) = 200 V N2 /500 = 200/210 N2 = 476 r.p.m.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Deakin University CRICOS Provider Code: 00113B The armature winding of a 200-V, 4-pole, series motor is lap-connected. There are 280 slots and each slot has 4 conductors. The current is 45 A and the flux per pole is 18 mWb. The field resistance is 0.3 Ω; the armature resistance 0.5 Ω and the iron and friction losses total 800 W. The pulley diameter is 0.41 m. Find the pull in newton at the rim of the pulley

Deakin University CRICOS Provider Code: 00113B The armature circuit resistance of a 18.65 kW 250-V series motor is 0.1 Ω, the brush voltage drop is 3V, and the series field resistance is 0.05. When the motor takes 80 A, speed is 600 r.p.m. Calculate the speed when the current is 100 A