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friction clutches - Arten Freios e Embreagens Industriais

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Table of ContentsHIGH INERTIA START OR STOP:A clutch or brake used to startor stop heavy, rotating drums,rolls or flywheels in a specificlength of time.Units selected should havesufficient heat sink capacity toabsorb the energy produced ineach start or stop.◗ “AIR CHAMP”PRODUCT CONSIDERATIONS:◗ Starts:“Air Champ” ®◗ SELECTION CONSIDERATIONS:Rotational inertiaTorqueHeat sink capacityPeak input rate capacity**The peak input rate capacity may be the limitingfactor in high inertia starts or stops even thoughthe Heat Sink Capacity is sufficient. The peak inputrate capacity is the rate at which a clutch or brakeabsorbs heat at the <strong>friction</strong> interface. The generationof heat results from slip at the interfaces duringan acceleration period. This heat generally willnot reach or exceed the Peak Input Capacity unlessthe acceleration time exceeds clutch or brake transienttime (time to reach the air pressure setting).Problems only occur when the start-up time isgreater than the response time of the clutch orbrake. The peak input generation rate during such astart is evaluated from an estimate of the speed differencebetween the facing and the <strong>friction</strong> disc atthe the end of the transient period and the torquevalue expected at the air pressure setting.The safe Peak Input Rate Capacity of a clutch orbrake with cast iron plates and organic <strong>friction</strong> liningsis approximately .9 horsepower per squareinch of interface area.◗ EXAMPLESelect a clutch to bring a roll with an inertiaload (WK 2 ) of 1800 lb. ft 2 from stop to 490RPM within 2 seconds. The selected clutchmust have a Heat Sink Capacity which exceedsthe energy in ft. lbs. produced at each start.FUNCTION EXAMPLES◗ PEAK INPUT RATE CAPACITY:Calculate the speed change (∆N 1 ) during thetransient period. (Assume 50% torque and atransient time of .1 second.)T (t)∆N 1 = = 13500 (.1) =19 RPM.039 (WK2) .039 (1800)The speed difference (N d ) between the facingand <strong>friction</strong> disc at the end of the transientperiod is the difference between full speed(RPM) and the speed change (∆N1).N d = RPM - N 1 = 490 - 19 = 471 RPMThe peak thermal input in HP is calculatedfrom the formula:HP = N d (T) = 471 (27000) = 202 HP63000 63000Where: T = Torque at the set air pressureThe effective interface area of a DFE 1650-S is227 square inches. (Data from the catalog tables.)The DFE 1650-S peak input capacity is:227 x .9 = 204 HPThe DFE 1650-S has sufficient capacity toabsorb the heat generated during the accelerationperiod. If the peak thermal input hadexceeded the peak input capacity, the starttime (t) would have to increase.D02147473D01147472◗ Stops:DPC9T–15TDFE, QFE1150–2500CALIPER BRAKESBD, VC500, SPC◗ TORQUE:T = .039 (WK2 ) RPM =t.039 (1800) 490) = 17,199 in. lbs.2Where; T = Torque in inch poundsWK 2 = Inertia load in lb ft 2RPM = Speed at the clutcht = Time in secondsWith a service factor of 1.5 the requiredtorque is:T = 1.5 x 17,199 = 25,799 in. lbs.◗ ENERGY PER CYCLE:Increasing the response time, by using a controlvalve with a small orifice, or adding an aircavity between the control valve and the clutch,increases the transient time. This reduces thethermal peaks that create damaging thermalgradients within the <strong>friction</strong> plates.◗ CONTROL CIRCUIT:A 3-way valve control circuit is used to affecta relatively slow start.3–WAY CONTROL–N.C.–ENGAGES CLUTCHOR BRAKE WHEN ACTUATOR SWITCH ISCLOSED.D03147474Ec = .00017 (WK 2 )(∆RPM) 2 =.00017 (1800) (490) 2 = 73,470 ft. lbs.REGULATORPROD. NO. 9400013-WAY CONTROL–N.C.PROD. NO. 949002D02147473DFE, QFE1150–2500Where: E c = Energy absorbed eachstart in ft. lbs.WK 2 = Inertia load in lb. ft 2RPM = Speed at the clutch◗ UNIT SELECTED:A DFE 1650-S, Product Number 964003.Torque capacity = 27,000 in. lbs. at 60 PSI.Heat Sink capacity = 870,000 ft. lbs.FILTERPROD. NO. 939101LUBRICATORPROD. NO. 939201383

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