Question

Question 4 [25 marks] An electric motor drives a coaxial rotor through a single-plate clutch which has two pairs of driving surfaces each 300 mm external and 175 mm internal diameter. The variation of normal pressure with radius is given by a + b/r Pa and the maximum pressure is 150 kPa. The spring force on the system is 500 N. The mass of the motor armature and shaft is 750 kg and its radius of gyration is 240 mm; the rotor has a mass of 1500 kg and its radius of gyration is 200 mm. The motor is brought up to a speed of 600 rpm; the current is then switched off and the clutch suddenly engaged with the stationary rotor. If the coefficient of friction in the system is 0.35: (a) Determine the final speed of the motor and rotor. (b) Find the time taken to reach that speed. (c) How much energy is lost during slipping? The engineer at the plant wants to use this clutch to transmit power to a system that has a load torque of 150 Nm. Describe what would happen if this system was put into service (time of slip, final angular velocity). Would you recommend this clutch for use? Justify your response. [17+8 marks]

Answer 1

scluten. (1) Wz, Iz. الن HIH rotor Il mated (6) ut Will, angular speed (Initials and mement og Inertia y armature. W2, 12- angular speed and mof of potaz. -) Given) Da 3oomm, d=175mm Pa = 1 150kPa p=atblr. f = 500N (spring force) Iq = migh = 7500 (0.24)² = 1 Borgm? sq = mak2 = 15000X (0:2) = 60kgm² u=0.35 1 when unten is engaged W, = LAMI 20X100 62.83 sadls. Go wyco Now, Ild, = -mt { Taking awesign because it is all accelerating the I222- mit X mt Ti moter when engaged] Mim2 = Thoment due to engagement ALS -Me IL i-matta dee, di t=0, degraus at too با هماهر mel Itt der Csscanned with Carpetanner mat a ( 1 Iz Gawi C2=WL

(x) (i) wa deer der C Gawa Cz zwr at FICT)t th] - [(*)etwe W = I (We we) – m.( Iit 1 it It First calculating time to reach speed time relative speed will be zero. So, at that (1) 2-wal (1152) (I2 +11) m =t bhn the Mt = dau/ponar. - (i) "an Dh Pa 2a Sen prar (ii) atby P . a 150x103 125 200-3 a + bx103 X2 2150X10 1757 11.426 150x103 co C Scanned with CamScanner

Para b = 46.3X203 Dh . 15 P= load Fs spring force. youd LookSp (949) 0.0678 0115 S 225 (artb) ar 008755 26 C9 (0.15 0.067517) = 25 + 6co (0.15-0.0875) 2 5002 0.0466 at 0.3 92b az 378.748103 ba m= 22X0.358 s( at borar 0.15 2 21199 S 2 (artb.8) dr stfood CScanned with CamScanner

R*199* [a (ouse-0.0845* ]]+6 (03520* LY 2:199[-376.7480 & CreoxGo) -0.08782 de porno +46-347038 0.1$20.08752) = 21199 -347150+ 343.63) = 4.69 Nm. using eg. (1) t= w1-01 ( l, Iv) (1+1) Mi ta (6253-0) X (160X60) (180 t60) X 4.69 602.845) ang cin Energy cost during slipping = 1 TR I wil-we) (11+12) 2 x 60 X180 X (62-632-0) (240) & G 82 kJ (Titlu)m 18-849 (160+60) X 150 - If it is used to transmit 150Frm. time of slip as derived earlies will be t we-wp it (1 112) 62.83-0 t. dith Camscanner

- Yes I recommend this luteh since it transters more terque and takes less time to reach the velouty. Also, Is wit Iz ur = fit12) w. 180x61.83 to I (160+60) w wa 47112 rades) Ang CScanned with CamScanner