• Author / Uploaded
• Irvin Biala

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DC MOTORS 𝑉𝑠ℎ

525. A simplex lap wound armature has 560 conductors and carries a current of 125 Amperes per armature current path. If the flux per pole is 20mWb. Calculate the electromagnetic torque developed by the armature? 𝑃𝑍𝑂𝑁𝐼𝑎 60𝑎 𝑍𝑂𝐼𝑎 580∗20∗10−3∗125 = 2𝜋𝑚 2𝜋∗1

Pd = EbIa = T=

= 230.77 𝑁𝑚

526. A 6 pole lap wound shunt motor takes 300 A when the speed is 500 rpm. The flux per pole is 0.06 Wb and the armature has 870 turns. Neglecting the shunt field current, calculate the brake horsepower of the motor. Assume a constant loss of 4%. 𝑃𝑍𝑂𝑁 𝑍500∗870∗0.06 = = 435 𝑉 60𝑎 60∗1 𝐸𝑏𝐼𝑎 435∗300 Pd = 746 = 746 = 175ℎ𝑝

Eb =

Pbrake = 0.96*175 = 168 hp 527. A 220 V shunt motor driving its normal load draws an armature current of 50 A from a 220 V DC voltage. The armature resistance of this motor including brushes is 0.25 ohm. How much armature current will this motor draw from a 220 V DC source when driving the same load with the field adjusted to maintain the same speed. Eb1Ia1 = Eb2Ia2 (Vs - IaRa)Ia1 = (Vs - IaRa)Ia2 (220 – 50*0.25)50 = (200 - Ia*0.25)Ia2 Ia2 = 55.76 A 528. A DC shunt motor has a full load rating of 15 hp, 230 V, 57.1 A, 1400 rpm. The armature circuit resistance is 13 0.13 ohm and the field circuit resistance is 115 ohm. Neglecting the effect of armature reaction, determine the no load line current.

230

Ish = 𝑅𝑠ℎ = 115 = 2𝐴 Ia = 57.1 – 2 A = 55.1 A Pa = 55.12 * 0.13 = 394.68 W Psh = 230*2 = 460 W Pin = 230*57.1 = 13 133 W Pout = 15 * 746 = 11 190 W Pstray = 13 133- 11 190 – 394.68 -460 = 1 088.32 W 1088.32 = 230IaNL – IaNL2Ra IaNL = 4.745 ImNL = 4.745 + 2 = 6.745 A 529. At rated load, a shunt motor draws an armature current of 50A from a 230 V DC source mains while running at 1000 rpm. At no load the armature current drops to % A. If the effect of armature reaction has weaken the flux by 2 % from no load to full load. Determine the speed of the motor at no load. Assume the armature resistance to be 0.15 ohm. Eb1 = Vs – Ia1Ra1 = 230 – 50*0.15 = 222.5 V Eb2 = Vs – Ia2Ra2 = 230 – 5*0.15 = 229.25 V 𝐸

N = ks∅ N2 = N1

𝐸𝑏2∅1 = 𝐸𝑏1∅2

229.25∗0.98∅2 222.5∅2

10001

= 1010 rpm

530. The nameplate rating of a shunt motor is 150 hp, 600 V, 205 A, 1700 rpm. The resistance of the shunt field circuit is 240 ohms and the total armature circuit resistance is 0.15 ohm. Calculate the speed regulation of the motor. 𝑉𝑠ℎ

600

Ish = 𝑅𝑠ℎ = 240 = 2.5𝐴 Ia = 205 – 2.5 A = 202.5 A Pa = 202.52 * 0.15 = 6150.937 W Psh = 600*2.5 = 1500 W Pin = 205*600 = 123 000 W Pout = 150 * 746 = 111900 W Pstray = 123000 – 111900 – n6150.937 – 1500 = 3449 W 3449 = 600IaNL – IaNL2*0.15 IaNL = 5.757 %NR = 1788.07 – 1700 / 1700 = 5.18 %

531. A DC shunt motor runs at 600 rpm on a 240 V supply while drawing a line current of 30A. Its armature and field resistances are 0.5 ohm and 120 ohm respectively. (Viedor, 21018) resistance should be placed in series with the armature circuit in order to reduce the speed to 400 rpm. Assume no changes in the armature or field current. 𝑉𝑠ℎ

Ia = Im – Ish = 30 -2 = 28 A Eb1 = Vs – Ia1Ra1 = 240 – 28*0.5 = 226 V 𝑁2 𝑁1

400 600

= 226

532. A 120 V shunt motor has an armature equivalent of 0.5 ohm and a field resistance of 60 ohm. At full load, the motor takes 10A and the speed is 1000 rpm. At what speed must this motor maybe driven as a generator to deliver 10A to an external load at 120 V. 𝑉𝑠ℎ 𝑅𝑠ℎ

=

120 60

=2

Ia = 10 -2 = 8 A E = 120 – 8*0.5 = 116 V Ia = 10 + 2 = 12 A Eg = 120 + 12*0.5 = 126 V 𝑁𝑚 𝐸𝑏

Ng = Eg

126 116

= 1000

Im =

𝑃𝑚 𝑉𝑠

=

8500 = 240

35.416 A

𝐸𝑚

238.37

Nm = Ng 𝐸𝑔 = 450271.39 = 395.25 rpm

= 150.667 V

Eb2 = Vs – Ia2Ra2 150.667 = 240 – 28(0.5 + R) R = 2.69 ohms

Ish =

Ia = 625 + 2.823 = 627.823 A Eg = 240 + 627.823*0.05 = 271.39 V

Ia = 35.416 – 2.823 = 32.593 A Eb = 238.37 V

240

Ish = 𝑅𝑠ℎ = 120 = 2𝐴

Eb2 = Eb1

𝑃𝐿 150 000 = = 625 𝐴 𝑉𝐿 240 𝑉𝑙 240 Ish = 𝑅𝑠ℎ = 85 = 2.823 𝐴

IL =

= 1086.2 rpm

533. A belt driven 150 kW shunt wound DC generator is running at 450 rpm and is supplying full load to a bus at 240 V. At what speed will it run if the belt breaks and the machine continues to run taking 8.5 kW from the busbar? The armature and field resistances are 0.05 ohm and 85 ohms respectively. The brush contact drop is 1.5 V per brush. Neglect armature reaction.

534. A shunt motor draws a current of 40 A from 120 V source and runs at 1200 rpm at rated load. The armature and field circuit resistances are 0.1 ohms and 60 ohms respectively. Determine the speed of the motor at half load. Assume a brush drop of 2V at rated load and 1 V at half load. 𝑉𝑠

Ish = 𝑅𝑠ℎ =

120 60

=2𝐴

Ia = 40 -2 = 38 A Eb = 114.2 V Ia2 = 38/2 = 19 A Eb2 = 117.1 V 𝐸2 𝐸1

N2 = N1

117.1 114.2

= 1200

= 1230 rpm

535. A DC shunt motor has a full load rating of 15 hp, 230 V, 57.1 A, 1400 rpm. The armature circuit resistance is 0.13 ohm and the field circuit resistance is 115 ohms. Neglecting the effect of armature reaction, determine the no load speed. Ish =

𝑉𝑠 𝑅𝑠ℎ

=

230 115

=2𝐴

Iafl = 57.1 – 2 = 55.1 A Pafl = 55.12 * 0.13 = 394.68 W Psh = 230 * 2 = 460 W Pin = 230 * 57.1 = 13 133 W Pout = 15 *746 = 11 190 W Pstray = 13 133 – 11 190 – 394.68 – 460 = 1 088.32 W 1088.32 = 230 IaNL – IaNL2*0.13 IaNL = 4.745 A 𝐸𝑛𝑙

229.383

Nnl = Nfl𝐸𝑓𝑙 = 1400222.837 = 1441.126 rpm

536. A 220 V shunt motor delivers 40hp on full load at 950 rpm and has an efficiency of 88 %. The armature and field resistances are 0.3 ohm and 110 ohms, respectively. Determine the starting resistance such that the starting armature current does not exceed 1.36 times the full load current.

torque at 1200 rpm, what value of resistance must be added to the armature circuit? 𝑉𝑠

600

Ish = 𝑅𝑠ℎ = 240 = 2.5 𝐴 Ia1 = 205 – 2.5 = 202.5 A Eb1 = 600 – 202.5(0.15) Eb1 = 569.625 V 𝑁2

Pin = 40 * 746 / 0.88 = 33 909 kW Im = Pin / vs = 33 909.1/220 = 154.13 A Ish = 220/110 = 2 A Ia = 154.13 – 2 = 152.13 A 𝑣𝑠

R =0.704 ohm 537. A 120 V DC motor rated at 5hp has a full load efficiency of 86 %. The field and armature resistance are 60 ohms and 0.75 ohm, respectively. What starter resistance will be required to limit the armature current at starting 200 % of its rated armature current? 5∗746 = 0.86

4337.21 W

Im = Pin / Vs = 4337.21/ 120 = 36.14 A Ish = 120/60 = 2 A Ia = 36.14 – 2 = 34.14 A 𝑣𝑠 Ias = 𝑅+𝑅𝑎 R = 1 ohm 538. A 10 hp 230 V DC motor of 85 % full load efficiency is located 450 ft from the supply mains, If the motors starting current is 1.75 times the full load current, what is the smallest cross sectional area of copper wire required when the allowable voltage drop in the feeder at starting is limited to 24 V? P

R=

𝑉𝑠−𝐸𝑏2 𝐼𝑎2

Ra

R = 0.827 ohm

220

Ias = 𝑅+𝑅𝑎 = 1.6(152.13)

P=

1200

Eb2 = Eb1𝑁1 = 569.6251700 = 402.088

10∗746 = 0.85 =

8776.47W

Im = 8776.47/ 230 = 38.15 A I = 175Im = 1.75 * 38.15 = 66.76 A R = Vdrop/I = 24/66.76 = 0.359 ohm A = 10.37 * 2* 450 / 0.359 = 25 997.2 CM 539. The nameplate rating of a shunt motor is rated 150 hp, 600 V, 205 a, 1700 rpm. The resistance of the shunt field circuit is240 ohms and the total armature circuit resistance is 0.15 ohms. If the motor is to be deliver full load

540. A 120 V shunt motor draws a current of 77 A at rated load. The armature and shunt filed resistances are 0.2 ohms 60 ohms respectively. A multi tap starter rheostat is inserted in the armature circuit to limit the starting current. If the rheostat has a resistance of 1.732 ohm, determine the percentage tap used such that the starting current will be limited to 150 % of its full load value? Ish =

𝑉𝑠 𝑅𝑠ℎ

=

120 60

=2𝐴

Ia = 77 – 2 = 75 R=

𝑉𝑠−𝐸𝑏2 𝐼𝑎2

Ra

R = 0.866 ohm %Tap = R / Rrheostat = 0.866/1.732 = 50% 541. A 10 hp power 220 V DC shunt motor has an armature and field resistance of 0.25 ohm and 100 ohms respectively. The full load efficiency is 83 %. Determine the value of the starting current will not exceed 200 percent of full load value. Pin = Pout/ eff = 10*746/0.83 = 8987.952 W Im = Pin/Vs = 9897.952/220 = 40.854 A Ish = vs/Rsh = 220/100 = 2.2 A Ia = 40.854 – 2.2 = 38.654 A R=

𝑉𝑠−𝐸𝑏2 𝐼𝑎2

Ra

R = 2.59 ohm 542. A 50 hp, 550 V shunt motor draws a line current of 4.5 A at no load. The shunt field resistances is 275 ohms and the armature resistance exclusive of brushes, is 0.3 ohms, the brush drop at full load is 2 V. At full load the

motor draws a line current of 84 A. Calculate the efficiency at full load. Ish = Vsh/Rsh = 550 /275 = 2A Ianl = 4.5 – 2 = 2.5 Pstray = 550*2.5 – 2.52*0.3 = 1373 W Iafl = 84.2 – 2 = 82 A Pinfl = 550*84 = 46200 W Pafl = Iafl2*Ra = 822*0.3 = 2017.2 W Pshf; = Vsish = 550*2 = 1100 W Pb = 2*82 = 164 W Ploss = 4654.325 W Eff = 46200 – 4654.325 / 46200 = 89.92 % 543. A 100 V shunt motor is developing 6 hp while running at an overall efficiency of 86%. The armature and shunt field resistances are 0.06 ohm and 50 ohms respectively. Determine stray power losses. Pout = 6*746 = 4476 W Pin = 4476/0.86 = 5204.65 Im = 5204.65/100 = 52.04 A Ish = 100/50 = 2 A Ia = 52.04 – 2 = 50.04 A Pa = 50.042 * 0.06 = 150.24 W Psh = 100*2 = 200 W Pstray = 378.41 W

546. A certain shunt motor has an armature resistance of 0.05 ohm. It draws 50 A at a terminal voltage of 120 V. Assume other miscellaneous losses at at 1%. Determine the output horsepower of the motor. Pin = VsIm = 120*50 = 6 000 W Pa = Ia2*Ra = 502*0.05 = 125 W Pout = Pin – (Pa = Pmisc) = 6000 – (125 + 60) = 5815 W Pout = 5815 / 746 = 7.79 hp Pout = 7.8 hp 547. The armature and field resistances of a shunt machine are 0.2 ohm and 110 ohms respectively. While running as a generator, the generated emf is 250 V at 1200 rpm. If the machine is run as shunt motor, it takes 4 A at 220 V. At a certain load the motor takes 30 A at 220 v. However, on load, the armature reaction weakens the field by 2%. Find the motor speed at this load. 𝑉𝑠

Ia = 4 – 2 = 2 A Eb1 = 220 – 2*0.2 = 219.6 V Eb = kN 𝐸𝑏1 𝐸𝑔 219.6 1200 250 =

𝑁1 = 𝑁𝑔 𝑁1 =

544. The shaft power of a shunt motor is 7.8 hp. It draws 50 A from 120 V. The field winding draws 1.2 A. What is the efficiency of the motor?

220

Ish = 𝑅𝑠ℎ = 110 = 2 𝐴

1054.08 rpm

Ia2 = 30 – 2 = 28 A Eb2 = 220 – 28*0.2 = 214.4 V 𝐸

∅1

214.4∅1

𝑁2 = 𝑁1 𝐸𝑏2 ∅2= 104.08 219.6∗0.98∅1 𝑏1

Eff = 7.8*746 / 120*50 = 96.98 % 545. A 500 V shunt motor running at 720 rpm takes an armature current of 50 A. The equivalent armature resistance is 0.4 ohm. What resistance must be placed in series with the armature to reduce the speed to 540 rpm while maintaining the same torque? Eb1 = Vs – IaRa = 500 – 50*0.4 = 480 V Eb = kN Eb2 = Eb1*N2/N1 = 480 *540/720 = 360 V R = 500-360 / 50 - 0.4 = 2.4 ohms

𝑁2 = 1050 rpm 548. A shunt motor, which has a field resistances of 220 ohms and an armature resistance of 0.8 ohm takes 26 A from a 260 V supply when running at 500 rpm on full load. In order to control the speed of the motor a 1.2 ohm resistor is connected in series with the armature. Calculate the speed at which the motor will run when supplying full load torque.

Ish =

𝑉𝑠 𝑅𝑠ℎ

=

260 220

= 1.18 𝐴

Ia = 26 – 1.18 = 24.82 A Eb1 = 260 – 24.82*0.8 = 240.14 V Eb2 = 200 – 24.82(0.8 + 1.2) = 210.36 𝐸

𝑁2 = 𝑁1

𝐸𝑏2 𝐸𝑏1

= 1000

105 108

𝑁2 =884 rpm

210.36

𝑁2 = 𝑁1 𝐸𝑏2=500 240.14= 436 rpm 𝑏1

549. The input and output powers of a 200 V, shunt motor are 30 and 28 kW respectively. Determine the efficiency of the motor if the output power is reduced to 10kW. 𝑃𝑚 30000 = = 150 𝑉𝑠 200 𝑉𝑠 200 Ish = 𝑅𝑠ℎ = 40 = 5 𝐴

Im =

A

Ia = 150 – 5 = 145 A Pa = 1452 * 0.0125 = 262.6125 W Psh = 200 * 5 = 1000 W Pstray = 30 000 – 10 000 – 262.8125 – 1000 = 737.187 W 200 Ia = 10 000 + Ia2*0.0125 + 737.187 Ia = 53.87 A Eff = 10 000/ 200*58.87 = 84.936 % 550. A 220 V series motor takes 15A and runs at 700 rpm. Assume the armature resistance including the resistance of the series windings is 0.8 ohm. At what speed will it run if a 2 ohm resistance is connected in series with supply voltage? Assume the motor is taking the same current. Eb = 220 – 15*0.8 = 208 V Eb2 = 220 – 15*2.8 = 178 V 𝑁2 = 𝑁1

Eb1 = 110 – 20*0.1 = 108 V Eb2 = 110 – 50*0.1 = 105 V

𝐸𝑏2 𝐸𝑏1

= 700

178 208

𝑁2 =599 rpm 551. The equivalent armature resistance of a series motor is 0.1 ohm. When connected across a 110 V mains, the armature takes 20 A and its speed is 1000 rpm. Determine the speed of the motor when the armature takes 50 A from the same mains, with the field increased by 10 %.

552. A 400 V series motor has a field resistance of 0.2 ohm and an armature resistance of 0.1 ohm. The motor takes 30 A of current at 1000 rpm while developing full load torque. What is the motor speed when this motor develops 60% of full load torque. 𝑇2

I𝑎2 = Ia1√𝑇1 = 30√

0.6𝑇1 = 𝑇1

23.24 A

Eb1 = 400 -30*0.3 = 391 V Eb2 = 400 – 23.24 *0.3 = 393.028 V 𝐸

𝑁2 = 𝑁1 𝐸𝑏2 = 1000*393.028*30 / 391 * 23.24 = 𝑏1

1297.6 rpm 553. A 10 hp, 230 V, 1200 rpm series motor having rated load efficiency of 85.5 %. The armature resistance including brushes is 0.3 ohm. The field resistances is 0.25 ohm. Assuming the flux varies directly as the armature current, what value of resistance should be placed in series with this motor when starting, in order that the starting current maybe limited to a value that will exert a starting torque equal to 150 % of its rated torque? Pin = 10*746 / 0.855 = 8725.146 W Im = 8725.146 / 230 = 37.93 A T = kIa2 T1/T2 = Ia12/Ia22 Ia2 = 46.45 A R = 230/46.45 – 0.28 = 4.4 A 554. The field and armature resistances of a 220 V series motor are 0.2 ohm and 0.1 ohm respectively. The motor takes 150 A of current while running at 700 rpm. If the total iron and friction losses are 350 W, determine the motor efficiency.

Pin = 220 * 30 = 6 600 W Pa = 50.042 * 0.06 = 150.24 W Psh = 100*2 = 200 W Pstray = 620 W Eff = 6600 -620 / 6600 = 90.6 % 555. A 100 V series motor is used to drive a load through a pulley. This machine has an armature resistance of 0.2 ohm and a series field resistance of 0.25 ohm. When a torque of 25 Nm is applied to the pulley, the speed is 600 rpm. If stray power losses at this load is 300 W, calculate the armature current. 2𝜋𝑇𝑁 60

Pout =

=

2𝜋∗600∗25 = 60

1570.796 W

Pd = 1570.796 + 300 = 1870.796 W Pd = EbIa Pd = VsIa – Ia2(Ra + Rse) 1870.769 = 100Ia – Ia2(0.2 + 0.25) Ia = 20.62 A 556. A 400 V series motor, having an equivalent armature circuit resistance of 0.5 ohm, takes 44 A of current while running at 650 rpm. What is the motor speed for a line current of 36 A. Eb1 = 400 – 44*0.5 = 378 V Eb2 = 400 – 36*0.5 = 382 V 𝑁1𝐼𝑎1

N2 = 6𝑁2𝐼𝑎2 =

650∗382344 = 378∗36

802.85 rpm

557. On full load a 500 V series motor takes 100 A and runs at 820 rpm. The armature and series field resistances are 0.1 and 0.04 ohm respectively. What will be its speed when developing half full load toque and with a 0.08 ohm diverter connected across the series field winding. Assume flux is proportional to the field current. Eb1 = 500 – 100*0.5 = 486 V 𝑇1 100∗100 = 0.5𝑇1 𝐼𝑎2( 𝐼𝑎2∗0.08 ) 0.08+0.04

𝐼𝑎2 = 88.6 A 𝑁2 =

𝑁1𝐸2𝐼1 820∗489∗100 = 486∗57.733 = 𝐸1𝐼2

1429 rpm

558. A 400 V series motor working with unsaturated field is taking 60 A and running at 840 rpm. The total resistance of the motor is 0.1 ohm. At what speed will the motor run when developing half full load torque. Eb1 = 400 – 60*0.01 = 394 V T = kIa2 𝑇2Ia1 = 𝑇1

I𝑎2 = √

0.5𝑇1∗602 𝑇1

√

= 42.43 A

Eb2 = 400 – 42.43*0.1 = 395.76 V 𝐸

395.76∗60

𝑁2 = 𝑁1 𝐸 𝑏2𝐼𝑎1 = 840 394∗42.43 = 1193 rpm 𝐼𝑎2 𝑏1

559. A long shunt compound motor takes a current of 42 A from a 230 V source. The generator parameters are: Ra = 0.1, Rse = 0.2 and Rsh = 50 ohms. If the friction and windage losses amount to 400 W, determine the overall efficiency of the machine. 𝑉𝑠

Ish = 𝑅𝑠ℎ =

230 50

= 4.6 𝐴

Ia = 42 – 4.6 = 37.4 A Pa = 37.42 * 0.1 = 139.876 W Pse = 37.4 2 *0.2 = 279.752 W Psh = 4.6 *230 = 1058 W Plosses = 1139.876 + 279.752 + 1058 + 400 = 1877.628 W Pin = 230*42 = 9660 W Eff = 9660 – 1877.628 / 9660 = 80.56 % 560. A 230 V, long shunt machine has the following parameters; Ra = 2.0 ohm, Rsh = 460 ohm and Rse = 0.25 ohm. When the machine is run at no load at its normal speed and rated voltage, the armature draws 0.6 A. Determine the armature current drawn if the machine delivers an output of 5 bhp. Pstray = 20*0.6 – 0.62*(2 + 0.25) = 137.2 W At loaded Condition Pout = 5*746 = 3730 W Ish =

𝑉𝑠 𝑅𝑠ℎ

=

230 460

= 0.5 𝐴

Im = Ia + 0.5 Pa = Ia2Ra = Ia2*2 Psh = 230*0.5 = 115 W Pin = 230* (Ia + 0.5) = 230 Ia + 115 230 Ia + 115 = 3730 + Ia2 + 0.25 Ia + 115 +137.2 Ia = 21.22 A

561. A 25 hp, 250 V long shunt compound motor takes 5 A when running light and runs at 1150 rpm. Armature resistances, 0.15 ohm, shunt field resistances, 200 ohm and series field resistances, 0.05 ohm. At rated load the motor takes 86.5 A and runs at 720 rpm. Determine the ration of torque developed from no load to full load.

armature resistance is 0.10 ohm and the series field resistance is 0.08 ohm. What is the stray power loss? Vsh = Vs – Im*Rse = 220 – 6*0.08 = 219.52 V Ish = 219.52 / 200 = 1.1 A Ia = Im – Ish = 6 – 1.0976 = 4.9 A Pa = Ia2*Ra = 4.9 2*0.10 = 2.4 W Pse = Im2*Rse = 62*0.08 = 2.88 W

562. A 200 V shunt motor has armature and shunt field resistances of 0.15 ohm and 125 ohm respectively. When running light, it takes 6 A and runs at 1200 rpm. As eries winding of 0.05 ohm is added to make it long shunt cumulatively compounded. This winding increases the flux per pole by 25% when the motor is taking its full load current of 100 A. Neglecting armature reaction, solve for the operating speed as a compound motor. 𝑉𝑠

200

Ish = 𝑅𝑠ℎ = 125 = 1.6 𝐴 Ia = 6 – 1.6 = 4.4 A Eb1 = 200 – 4.4*0.15 = 199.34 V Ia2 = 100 – 1.6 = 98.4 A Eb2 = 200 – 98.4*(0.15 + 0.05) = 180.32 V N2 =N1

𝐸𝑏2 𝑂1 𝐸𝑏1𝑂2

= 1200

180.32 𝑂1 199.34∗1.25𝑂1

N2 = 868.4 rpm

Psh = 219.52*1.1 = 241.47 W Pin = 220*6 = 1320 W Pstray = 1320 – 2.4 – 2.88 -21.47 = 1073.25 W

565. The pony break test of an elevator door drive shunt motor, the ammeter and voltmeter measuring the input read 34 A and 220 V. The speed of the motor is found to be 910 rpm and the balance on a 2ft brake arm read 27.2 pounds. The tare weight of the arm is found to be positive 2.3 pounds. Determine the efficiency of the motor at this load. T = (scale reading – dead weight) * length of arm T = (27.2 – 2.3) * 2 = 49.8 lbft 2𝜋𝑇𝑁

563. A 220 V DC motor has an armature resistance of 0.4 ohm, a shunt field resistance of 175 ohms and a series field resistance of 4 ohms. This machine is connected as a short shunt compound motor to 220 V mains. Calculate the power developed by the armature current is 30A. Neglect the brush drop. Vsh = Eb + IaRa Ish = 0.005714Eb + 0.06857 Im = Ia + Ish = 30 + 0.005714 Eb + 0.06857 Eb = Vs – ImRse – IaRa Eb = 220 – (0.005714Eb + 30.06857)*4 – 30*0.4 Eb = 85.76 V Pd = EbIa = 85.766 * 30 = 2572.8 W 564. The input current to a 220 V, short circuit compound motor at no load is 6A. The shunt field circuit resistance is 220 ohms, the

2𝜋∗910∗49.8𝑁 = 8.628 33 000 220∗34 = 10.026 hp 746

Hpout = 33 000= Hpin =

𝑉𝑠𝐼𝑚 == 746

hp

Eff = 8.628/ 10.026 = 86.056 %

566. The pony break test of an elevator door drive shunt motor, the ammeter and voltmeter measuring the input read 34 A and 220 V. The speed of the motor is found to be 910 rpm and the balance on a 0.61 m brake arm read 11.9 kg. The dead weight of the arm is found to be 1.09 kg. What is the output hp of the door motor. T = (scale reading – dead weight) * length of arm T = (11.9 – 1.09)(9.81) * 0.61 = 64.688 Nm Hp = 2πTN/44760 = 2π*910*64.688/44760 = 8.26 hp

567. A shunt motor with an armature and field resistance of 0.055 and 32 ohms, respectively, is to be tested for its mechanical efficiency by means of a rope brake. When the motor is running at 1400 rpm, the longitudinal pull on the 6 inch diameter pulley is 57 pounds. The readings of the line ammeter and voltmeter are 35 and 105 respectively. Calculate the efficiency. T = force on the rope * radius of the pulley T = 57 * 3/12 = 14.25 lbft Hp out= 2πTN/33 000 Hp out= 2π*1400*14.25/33 000 = 3.798 hp Hp = 105*35 / 746 = 4.926 hp Eff = 3.798/4.926 = 77.1 %

568. A shunt motor was tested by means of a pony brake having a length arm of 3.5 feet and a tare weight of 5.7 pounds. The current drawn by the machine from a 240 volt line was 50.9 A when the scale reading was 24 lbs and the speed of the motor was 1215 rpm. Calculate the rotational losses of the motor. The armature and shunt field resistances of the machine are 0.25 ohm and 120 ohm, respectively. T = (scale reading – dead weight) * length of arm T = (24 – 5.7) * 3.5 = 64.05 lbft Hp out= 2πTN/33 000 Hp out= 2π*1215*64.05/33 000 = 14.817 hp 𝑉𝑠

240

Ish = 𝑅𝑠ℎ = 240 = 1 𝐴 Ia = 50.9 – 1 = 49.9 A Pa =Ia2*Ra = 49.9 2 * 0.1 = 249 W Psh = 240 * 1 = 240 W Pin = 240*50.09 = 12216 W Pstray = 12216 – 14.817*746 – 249 – 240 = 673.5 W

569. Calculate the force that will be exerted on the scale in a pony brake test when a 20 hp, 1400 rpm motor is operating at full load. The length of the brake arm is 3ft. and the tare weight of the brake is 3.75 lb.

T = (scale reading – dead weight) * length of arm T = (F – 3.75)*3 Hp out= 2πTN/333 000 F = 28.76 lbs