Clutches, Brakes and FlywheelsIntroductionClutches, brakes and flywheels are rotating-machine components that control the transmission, absorption and storage of mechanical energy. A clutch connects two rotating members so that torque can be transmitted; a brake absorbs mechanical energy and reduces or stops motion; and a flywheel stores kinetic energy and releases it when the energy supplied by a prime mover is temporarily less than the energy demanded by the machine. These components occur in automobiles, machine tools, presses, turbines, compressors, conveyors, hoists and industrial automation systems. Their design requires simultaneous consideration of torque, power, friction, heat generation, strength, wear, vibration, controllability and safety. Learning ObjectivesAfter studying this chapter, the student should be able to: - Explain the operating principles and applications of clutches, brakes and flywheels.
- Derive and apply torque-capacity equations for friction clutches and brakes.
- Distinguish uniform-pressure and uniform-wear assumptions.
- Estimate heat generation and braking energy.
- Determine flywheel energy requirements and speed fluctuation.
- Select suitable materials, configurations and safety factors.
- Evaluate practical design limitations, maintenance requirements and modern developments.
1. Mechanical Energy, Torque and PowerFor a rotating shaft, torque and power are related by P = Tω = (2πNT)/60 where P is power in watts, T is torque in newton-metres, ω is angular speed in radians per second and N is rotational speed in revolutions per minute. The equation assumes steady rotational motion and consistent SI units. The work done by a torque through an angular displacement θ is W = ∫ T dθ For constant torque, W = Tθ. A clutch transmits this work from one shaft to another. A brake converts it mainly into heat. A flywheel temporarily stores it as rotational kinetic energy: E = 1/2 Iω² where E is kinetic energy in joules and I is mass moment of inertia in kg·m². This expression assumes a rigid body rotating about a fixed axis. 2. Clutches2.1 Function and operating principleA clutch provides a detachable connection between a driving shaft and a driven shaft. When engaged, friction or positive mechanical engagement causes both members to approach the same angular speed. During engagement, some relative motion is permitted so that torque rises progressively rather than instantaneously. A clutch must: - transmit the required torque without excessive slip;
- engage smoothly;
- dissipate engagement heat;
- occupy limited space and mass;
- release completely when required; and
- maintain acceptable life under wear and contamination.
Clutches are used for starting vehicles, changing gears, connecting machine tools and isolating driven equipment during overload or maintenance. 2.2 Classification| Type | Principle | Typical application |
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| Plate or disc clutch | Friction between annular surfaces | Automobiles, machinery | | Cone clutch | Friction between conical surfaces | Low-speed drives, older vehicles | | Centrifugal clutch | Engagement by centrifugal force | Small engines | | Electromagnetic clutch | Magnetic force clamps friction surfaces | Automation, printing machinery | | Hydraulic clutch | Fluid pressure operates friction members | Vehicle transmissions | | Dog clutch | Positive tooth engagement | Gearboxes where engagement occurs near equal speed | | Centrifugal-locking clutch | Speed-dependent mechanical engagement | Special drives |
Friction clutches permit engagement at different speeds, whereas dog clutches should generally be engaged only after speeds are synchronized. 2.3 Torque capacity of a plate clutchConsider an annular friction element with inner radius r_i, outer radius r_o, coefficient of friction μ and axial clamping force W. For an elemental ring of radius r and width dr, dA = 2πr dr If the normal pressure is p, the elemental friction force is μp dA and the elemental torque is dT = μp(2πr dr)r = 2πμp r²dr The total torque is obtained by integrating from r_i to r_o. For a clutch with n friction surfaces, T = 2πμn ∫[ri to ro] p r² dr The number n is the number of active friction interfaces, not merely the number of plates. For example, a single-disc clutch with two working faces has n = 2. #### Uniform-pressure theory Uniform pressure assumes p is constant over the entire contact area. This is a reasonable approximation for a new clutch when the elastic and geometric conditions distribute pressure nearly evenly. Since W = πn p(ro² − ri²) integration gives T = (2/3) μW [(ro³ − ri³)/(ro² − ri²)] The equivalent mean radius is therefore Rp = (2/3)[(ro³ − ri³)/(ro² − ri²)] and T = μW R_p. #### Uniform-wear theory In service, wear is approximately proportional to the product of pressure and sliding distance. If sliding speed is proportional to radius and wear is uniform, pr = C where C is a constant. The pressure is consequently higher near the inner radius. Integration gives T = μW [(ro + ri)/2] Thus the mean radius under uniform wear is Rw = (ro + ri)/2 Uniform-wear theory is commonly preferred for a used clutch. Uniform-pressure theory generally predicts a slightly larger torque capacity for the same dimensions and clamping force. 2.4 Single- and multiple-plate clutchesFor a single-disc clutch with two friction faces, n = 2. A multiple-plate clutch increases torque capacity without proportionally increasing diameter. If there are m plates in total, the number of active interfaces is commonly m − 1 when the arrangement alternates between driving and driven plates. The torque equation remains T = n μ W R with R selected from the applicable pressure or wear model. The axial force may be produced by coil springs, diaphragm springs, hydraulic actuators, pneumatic systems or electromagnetic devices. Diaphragm springs are widely used because they provide compact packaging and relatively uniform clamping force over the operating range. 2.5 Cone clutchesA cone clutch uses conical friction surfaces. Let α be the semi-cone angle, measured between the cone surface and the axis. For an idealized cone clutch, the axial force W_a produces a normal force approximately equal to W_a/sin α. Therefore, the friction torque may be written as T = μ(Wa/sin α)R where R is an appropriate mean friction radius. This relation assumes uniform contact and neglects edge effects. A small cone angle increases mechanical advantage but may cause self-locking or difficult release. Cone clutches are therefore less common where rapid disengagement is required. 2.6 Engagement dynamics and heatDuring engagement, the relative angular speed between driving and driven members produces frictional work. If the effective rotating inertia referred to the clutch is I_e and the relative angular speed changes from ω_1 to ω_2, the ideal relative kinetic-energy loss is Q = 1/2 Ie(ω1² − ω2²) This energy becomes heat in the friction materials, subject to losses and external work. Repeated engagement may produce a greater thermal load than steady torque transmission. The designer must check both instantaneous temperature rise and average heat-rejection capability. If a clutch transmits torque T during slip at relative angular speed ω_rel, the instantaneous friction power is Pf = Tωrel The relation applies while slipping and should not be confused with transmitted shaft power after full engagement. 2.7 Clutch design considerationsImportant design checks include: - torque capacity with an appropriate service factor;
- allowable contact pressure;
- friction coefficient over temperature and speed ranges;
- lining wear and compatibility with oil or water;
- thermal capacity and cooling;
- torsional vibration and judder;
- spline, hub and shaft strength;
- spring fatigue;
- release force and pedal or actuator travel; and
- protection against accidental engagement.
The design torque is often selected as Td = Ks Tw where T_d is design torque, T_w is working torque and K_s is a service factor. The value of K_s depends on shock, load variation, starting frequency, duty cycle and consequences of failure; it must be chosen from appropriate standards or manufacturer data rather than assumed universally. 3. Brakes3.1 Function and classificationA brake produces a resisting torque to reduce or control speed. Unlike a clutch, which normally transmits power between rotating members, a brake transmits torque from a rotating member to a stationary frame or another controlled member. Common types include drum brakes, disc brakes, band brakes, block brakes, internal-expanding brakes, electromagnetic brakes, hydraulic brakes and regenerative brakes. Brakes operate in three distinct modes: - Service braking: repeated control of speed.
- Emergency braking: rapid stopping under abnormal conditions.
- Holding braking: maintaining a stationary position, often on an incline.
A brake must be designed for torque capacity and thermal performance. A brake with adequate static torque may still fail during repeated high-energy stops because of overheating or fade. 3.2 Braking torque and stopping energyFor a rotating system with equivalent inertia I_e, the energy that must be absorbed during a stop from angular speed ω_1 to ω_2 is Eb = 1/2 Ie(ω1² − ω2²) + Wext where W_ext represents energy added by external forces during braking. For example, gravity may add energy when a vehicle travels downhill. If external work is absent, W_ext = 0. For a vehicle of mass m moving at speed v, translational kinetic energy is Ek = 1/2 mv² Because energy varies with the square of speed, doubling speed quadruples the kinetic energy that must be absorbed. 3.3 Disc brakesFor a disc brake with n active friction surfaces, friction coefficient μ, normal clamping force W and mean radius R, T = n μ W R The value of R depends on the pressure distribution. Under uniform wear, R = (ro + ri)/2 Under uniform pressure, R = (2/3)[(ro³ − ri³)/(ro² − ri²)] The equations assume concentric, planar contact and neglect pad flexibility, thermal distortion and pressure nonuniformity. Disc brakes offer good cooling and consistent performance, but they require actuator force and careful control of pad wear. Ventilated discs increase heat-transfer area and promote airflow. 3.4 Drum and block brakesFor a simple block brake with normal force N at radius r and coefficient of friction μ, Tb = μNr This expression assumes a single friction contact at a constant effective radius. The actuator force depends on the geometry of the lever and on whether friction assists or opposes application. In a drum brake, shoes press against the inner surface of a rotating drum. A leading shoe may experience self-energizing action: friction increases the normal force in the direction of application. This reduces actuator effort but can create sensitivity, uneven braking and locking if poorly controlled. 3.5 Band brakesFor a flexible band wrapped around a drum, the capstan relation is Tt/Ts = e^(μθ) where T_t is the tight-side tension, T_s is the slack-side tension, μ is the coefficient of friction and θ is the wrap angle in radians. The braking torque is Tb = (Tt − Ts)r This relation assumes a flexible band, uniform friction coefficient and impending sliding. It is not valid without modification when the band is rigid, friction is dynamic rather than limiting, or the contact pressure is substantially nonuniform. 3.6 Thermal design of brakesThe average heat-generation rate over a braking interval is Pavg = Eb/tb where t_b is braking time. During repeated cycles, the average thermal load over a complete cycle is Pcycle = Eb f where f is the number of braking events per second. The brake must reject heat through conduction, convection and radiation. Brake fade may result from friction-coefficient reduction, lining decomposition, gas formation, thermal expansion or fluid vaporization. Thermal analysis should consider: - maximum single-stop energy;
- stop frequency and duty cycle;
- rotor or drum temperature limits;
- heat partition between contacting members;
- cooling-air velocity;
- thermal gradients and distortion; and
- allowable lining temperature.
3.7 Stopping time and distanceIf constant braking torque T_b acts on a rotating system of inertia I_e, angular deceleration is α = Tb/Ie and stopping time from angular speed ω_0 is ts = Ieω0/Tb The angular stopping displacement is θs = Ieω0²/(2Tb) These equations assume constant torque and no external driving torque. In real systems, torque may vary with speed, temperature, pressure and control strategy. 4. Flywheels4.1 Purpose and distinction from a governorA flywheel reduces cyclic speed fluctuation by storing and releasing energy. It does not maintain a constant mean speed under sustained changes in load. That is primarily the function of a governor or control system. Flywheels are particularly important in reciprocating engines, punching machines, shears, presses and piston compressors, where torque supplied and torque demanded fluctuate during each cycle. The instantaneous energy change in a flywheel is related to torque imbalance: dE = (Td − Tl)dθ where T_d is driving torque and T_l is resisting or load torque. Integrating over the relevant angular interval gives the maximum fluctuation of energy ΔE. 4.2 Coefficient of fluctuation of speedIf ω_max and ω_min are the maximum and minimum angular speeds during a cycle, the coefficient of fluctuation of speed is Cs = (ωmax − ωmin)/ωmean where ωmean = (ωmax + ωmin)/2 For small speed fluctuations, the energy fluctuation is approximately ΔE = Iωmean² Cs Therefore, the required mass moment of inertia is I = ΔE/(ωmean² Cs) The approximation assumes C_s is small. The exact expression follows from ΔE = 1/2 I(ωmax² − ωmin²) 4.3 Flywheel sizingFor a thin-rim flywheel, most mass is concentrated near radius R, so I ≈ mR² where m is rim mass. The required rim mass is consequently m ≈ ΔE/(R²ωmean² Cs) This approximation neglects hub and spoke inertia. For a solid disc of mass m and radius R rotating about its central axis, I = 1/2 mR² A rimmed flywheel is generally more mass-efficient when a large inertia is required because mass placed farther from the axis contributes more strongly to I. 4.4 Flywheel stresses and safetyA rotating rim experiences centrifugal tensile stress. For a thin rotating ring, a simplified estimate is σt ≈ ρv² where σ_t is circumferential stress, ρ is material density and v is rim speed. Since v = ωR, the stress rises with the square of speed. The simplified relation neglects spoke flexibility, stress concentration, residual stress, keyways, shrink fits and geometric discontinuities. Actual flywheel design requires recognized design standards, material certification, balancing and overspeed testing. A flywheel must also be checked for fatigue caused by cyclic speed variation, thermal effects, attachment stresses and fracture containment. Cast iron, forged steel and composite materials may be used, depending on speed, stored energy, cost and safety requirements. 5. Worked Engineering ExamplesExample 1: Torque capacity of a single-disc clutchA single-disc clutch has outer friction radius 0.125 m, inner radius 0.075 m, coefficient of friction 0.30 and axial force 4000 N. Determine torque capacity using uniform-wear theory. There are two active friction surfaces. For uniform wear, R = (ro + ri)/2 = (0.125 + 0.075)/2 = 0.100 m Hence T = nμWR = 2 × 0.30 × 4000 × 0.100 = 240 N·m The clutch capacity is 240 N·m, assuming clean, fully engaged surfaces and constant μ. A service factor would be applied when selecting the clutch for a machine. Example 2: Braking energy and average powerA rotating system has equivalent inertia 8 kg·m² and is stopped from 1200 rpm. The stopping time is 4 s. Determine the energy absorbed and average braking power. First convert speed: ω = 2πN/60 = 2π(1200)/60 = 125.66 rad/s Energy absorbed is Eb = 1/2 Iω² = 1/2 × 8 × (125.66)² = 63,165 J Average power is Pavg = Eb/t = 63,165/4 = 15,791 W ≈ 15.8 kW The brake must absorb approximately 63.2 kJ per stop and dissipate an average of 15.8 kW during this stopping interval. Peak power may be higher if braking torque is not constant. Example 3: Flywheel inertiaA machine requires a maximum energy fluctuation of 1200 J. Its mean speed is 300 rpm and the permitted coefficient of fluctuation of speed is 0.02. Determine the required moment of inertia using the small-fluctuation equation. ωmean = 2π(300)/60 = 31.416 rad/s Therefore, I = ΔE/(ωmean² Cs) = 1200/[(31.416)² × 0.02] = 60.79 kg·m² The required inertia is approximately 60.8 kg·m². If the effective rim radius is 0.75 m, the thin-rim estimate of mass is m = I/R² = 60.79/(0.75²) = 108.1 kg The actual flywheel mass will be higher if hub and spoke contributions, manufacturing tolerances and safety margins are included. 6. Integrated Applied Case: Clutch and Brake Selection for a Press DriveBackgroundThis illustrative case concerns a mechanical stamping press driven by an electric motor through a flywheel. The press requires intermittent engagement for each stroke and must stop safely during an emergency. ProblemThe existing friction clutch overheats during rapid cycling, while the emergency brake occasionally exhibits fade. The machine operator also reports a long stopping time at high production rates. AnalysisThe clutch experiences high engagement energy because the press mechanism is initially stationary while the flywheel rotates. Its average transmitted torque is not the only design criterion; the energy per engagement and engagement frequency determine thermal loading. The brake must absorb the flywheel’s rotational energy during an emergency stop. Since energy is proportional to ω², stopping at maximum speed is particularly demanding. The design team calculates: - equivalent rotating inertia referred to the clutch and brake;
- maximum and average engagement frequency;
- clutch slip time and energy per cycle;
- emergency stopping energy;
- allowable lining and rotor temperatures;
- required torque with a suitable service factor; and
- structural strength of hubs, keys, shafts and mounting bolts.
Approach and solutionA multi-plate, oil-cooled clutch is selected to increase torque capacity and remove engagement heat. The clutch control system is tuned to reduce unnecessary slip while retaining acceptable shock levels. A ventilated disc brake with temperature-rated friction material replaces the original brake. Temperature sensors and a cycle counter are added to provide condition monitoring. The flywheel is retained because it smooths the cyclic power demand, but its maximum speed is limited through the motor-control system. A guarded enclosure is provided around the flywheel and brake assembly, and the emergency circuit is designed to produce a fail-safe brake application. OutcomeThe revised system reduces clutch temperature, improves repeatability of engagement and decreases emergency stopping time without exceeding the brake’s thermal limit. Preventive maintenance intervals are based on lining thickness, measured temperature history and cycle count rather than calendar time alone. Learning points- Torque capacity alone does not guarantee satisfactory clutch performance.
- Brake energy depends strongly on speed and must be evaluated for the worst case.
- Flywheels reduce cyclic speed fluctuation but increase stored energy and emergency-braking demand.
- Mechanical design, controls, guarding and maintenance must be treated as one safety system.
7. Materials, Friction and ManufacturingFriction materials should provide stable coefficient of friction, adequate wear resistance, thermal stability and compatibility with the environment. Organic linings are common for moderate temperatures; sintered metallic materials tolerate higher temperature and pressure; ceramic or carbon-based materials may be selected for specialized high-energy applications. Friction coefficients are not constants in practice. They vary with pressure, speed, temperature, surface condition, humidity, oil contamination and bedding history. Consequently, catalog values must be interpreted with the supplier’s test conditions and duty cycle. Rotors and drums require adequate hardness, thermal conductivity and resistance to distortion. Surface finish affects bedding and noise. Manufacturing tolerances, runout and parallelism are important because uneven contact can produce vibration, local overheating and accelerated wear. 8. Vibration, Noise and ControlClutch judder may result from torsional resonance, uneven friction, warped plates, contamination or nonlinear engagement. Brake squeal is associated with friction-induced vibration and structural modes. Remedies may include damping elements, improved stiffness, modified pad geometry, controlled surface finish and changes to the actuation profile. Modern electronically controlled systems regulate clutch pressure and brake force using speed, torque, temperature and position measurements. Antilock braking systems prevent wheel lock by modulating hydraulic pressure. Regenerative braking recovers part of the vehicle’s kinetic energy as electrical energy, reducing friction-brake heat, although friction brakes remain necessary for low-speed stopping, holding and emergency operation. 9. Safety, Standards and MaintenanceA stored-energy device can fail catastrophically if it fragments at speed. Flywheels require guarding, overspeed protection, balancing, inspection and appropriate containment. Brakes must be fail-safe where loss of braking could cause unacceptable risk. Clutches should prevent unintended engagement and should be capable of dissipating foreseeable overload energy. Designers should consult applicable machinery-safety, automotive, pressure, electrical and occupational-safety standards. The exact standard depends on the equipment, jurisdiction and application. Risk assessment should address normal operation, foreseeable misuse, maintenance, control-system failure and residual energy. Good maintenance practice includes: - measuring lining or pad thickness;
- checking cracks, scoring, runout and thermal damage;
- inspecting springs, splines, keys and fasteners;
- checking hydraulic or pneumatic leakage;
- monitoring temperature and engagement cycles;
- maintaining correct adjustment and alignment; and
- replacing friction components in matched sets when required.
10. Limitations and Design JudgmentThe equations in this chapter are engineering models, not complete descriptions of real components. They commonly assume rigid bodies, ideal contact, constant friction coefficient, uniform pressure or uniform wear, negligible deformation and known heat transfer. Real design requires experimental data, manufacturer curves, finite-element analysis where appropriate, dynamic simulation and prototype testing. A technically correct equation can still produce an unsafe design if the loading spectrum, environmental conditions or failure consequences are ignored. Designers should use conservative assumptions for uncertain inputs, document the basis for service factors and verify performance under worst credible conditions. Chapter SummaryClutches connect and disconnect rotating members, brakes absorb mechanical energy and control motion, and flywheels store and release rotational energy to reduce cyclic speed variation. Friction-clutch and brake capacity is commonly represented by T = nμWR, with the mean radius determined using either uniform-pressure or uniform-wear theory. Engagement and braking generate heat that must be assessed from energy and duty cycle, not torque alone. Flywheel sizing is based on the fluctuation of energy and allowable speed variation, with I = ΔE/(ω_mean²C_s) for small speed fluctuations. Practical design also requires analysis of strength, fatigue, thermal behavior, friction-material performance, vibration, guarding, control systems and maintenance. Safe design integrates analytical models with standards, testing and risk assessment. Key TermsClutch: A device that connects or disconnects rotating members. Brake: A device that produces resisting torque to reduce or stop motion. Flywheel: A rotating energy-storage device that limits cyclic speed fluctuation. Friction clutch: A clutch transmitting torque through frictional contact. Plate clutch: A clutch using flat annular friction surfaces. Cone clutch: A clutch using conical friction surfaces. Multiple-plate clutch: A clutch with several friction interfaces. Uniform pressure: An idealization in which contact pressure is constant over the friction area. Uniform wear: An idealization in which pressure multiplied by sliding radius is constant. Mean friction radius: The effective radius at which the total friction torque acts. Service factor: A multiplier accounting for shock, overload and operating severity. Brake fade: Reduction of braking effectiveness, commonly caused by overheating. Self-energizing brake: A brake in which friction assists the application force. Capstan equation: The relation between tight- and slack-side tensions in a wrapped flexible member. Equivalent inertia: A combined inertia referred to a selected rotating shaft. Coefficient of fluctuation of speed: Speed range divided by mean speed. Energy fluctuation: The maximum difference between supplied and absorbed energy during a cycle. Rim speed: The tangential speed at the outer radius of a rotating component. Friction material: A material designed to provide controlled friction, wear and thermal performance. Regenerative braking: Braking that converts part of mechanical energy into recoverable electrical energy. Review QuestionsShort Answer Questions- What is the principal functional difference between a clutch and a brake?
- Why is uniform-wear theory often preferred for a used friction clutch?
- State the expression for the kinetic energy of a rotating body.
- What is meant by brake fade?
- Why does a flywheel not replace a governor?
Long Answer Questions- Derive the torque-capacity equation for an annular friction clutch under uniform pressure.
- Explain the design and operating principles of single-plate and multiple-plate clutches.
- Discuss the thermal design requirements of a friction brake.
- Derive the flywheel inertia equation based on allowable coefficient of fluctuation of speed.
- Compare disc, drum, band and electromagnetic brakes with respect to operation and application.
Analytical / Application Questions- A two-face clutch has inner and outer radii of 60 mm and 100 mm, coefficient of friction 0.28 and axial force 3000 N. Determine its torque capacity using uniform-wear theory.
- A brake absorbs 45 kJ in 3 s during a stop. Determine the average heat-generation rate and discuss two design measures needed for repeated stops.
- A flywheel-driven press experiences a large increase in emergency stopping energy after its operating speed is raised by 20%. Explain quantitatively why the increase is greater than 20% and identify the design checks required.
Multiple Choice Questions with AnswersQuestion 1Which component primarily stores rotational kinetic energy to reduce cyclic speed fluctuation? A. Flywheel B. Brake C. Clutch D. Shaft coupling Correct Answer: A. Flywheel Question 2Under uniform-wear theory for an annular friction surface, which relation is used? A. p = constant B. pr = constant C. p/r = constant D. p + r = constant Correct Answer: B. pr = constant Question 3A single-disc clutch has two active friction surfaces. In the torque equation T = nμWR, what is n? A. 1 B. 2 C. The outer radius D. The axial force Correct Answer: B. 2 Question 4If a rotating system is stopped from angular speed ω to zero and external work is absent, the energy absorbed by the brake is A. Iω B. Iω² C. 1/2 Iω² D. 2Iω² Correct Answer: C. 1/2 Iω² Question 5If vehicle speed is doubled, its translational kinetic energy, with mass unchanged, becomes A. Half as large B. Twice as large C. Three times as large D. Four times as large Correct Answer: D. Four times as large Question 6For a flexible band brake at impending sliding, the capstan relation is A. T_t + T_s = e^(μθ) B. T_t/T_s = e^(μθ) C. T_t − T_s = e^(μθ) D. T_tT_s = μθ Correct Answer: B. T_t/T_s = e^(μθ) Question 7A flywheel primarily reduces A. Cyclic speed fluctuation B. Static shaft strength C. Bearing friction to zero D. The coefficient of friction Correct Answer: A. Cyclic speed fluctuation Question 8For a thin-rim flywheel, the approximate mass moment of inertia is A. I = m/R² B. I = 1/2 mR² C. I = mR² D. I = 2mR² Correct Answer: C. I = mR² Question 9Brake fade is most directly associated with A. Excessive cooling B. Loss of braking effectiveness due to thermal effects C. Increased flywheel inertia D. Perfectly constant friction Correct Answer: B. Loss of braking effectiveness due to thermal effects Question 10A clutch transmits 240 N·m at 1200 rpm. Its ideal transmitted power is approximately A. 3.0 kW B. 15.1 kW C. 30.2 kW D. 60.3 kW Correct Answer: C. 30.2 kW
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