chapter 3: fundamentals of mechanics and...

1/11/00 Electromechanical Dynamics 1

Chapter 3: Fundamentals of Mechanics and Heat


Force

• Linear acceleration of an object is proportional to the applied force:where– F = force acting on an object [N]

– m = mass of the object [kg]

– a = acceleration of the object [m/s2]

amF =

mF

x(t)


Torque

• Torque is produced when a force exerts a twisting action on an object, tending to make it rotate

• Torque is the product of the force and the perpendicular distance to the axis of rotation:where– T = torque [Nm]

– F = applied force [N]

– r = radius [m]

– φ = angle of applied force

• Example– calculate the braking force needed for a motor with a 1 m

diameter braking drum that develops a 150 Nm starting torque

φsin⋅⋅= rFT

r

axis of rotation

FT

φ


Work

• Work is done whenever a force F moves an object a distance d in the direction of the force:where– W = work [J]

– F = force [N]

– d = distance [m]

• Example– calculate the work done

on a mass of 50 kg thatis lifted to a height of 10 m

dFW =


Work

• Work is performed on a rotating object by a torque when there is an angular rotation:where– T = torque [N m]

– δ = angular displacement [radians]

• Example– calculate the work performed by an electric motor that

develops a 100 Nm torque at 1750 rpm on a pulley that lifts a mass in 25 s

δTW =


Power

• Power is the rate of which work is performedwhere– P = power [W]

– W = work [J]

– t = time to do the work [s]

• Common units are kW and hp– 1 hp = 746 W = 0.746 kW

• Example– calculate the power developed

by an electric motor that lifts a mass of 500 kg to a height of 30 m in 12 s

t

WP =


Power

• The mechanical power output of a motor depends on the torque and rotational speed:where– P = mechanical power [W]

– T = torque [N m]

– n = speed [rpm]

• In more general terms: – ω = speed [radian/s]

• Example– Calculate the power output on a

motor rotating at 1700 rpm during a prony brake test where the two spring scales indicate 25 N and 5 N, respectively

55.960

2 TnTnP ≈= π

TP ω=


Transformation of Energy

• Forms of energy include:– mechanical energy (potential and kinetic)

– thermal, chemical, and atomic energy

– electrical energy (electric and magnetic)

• Energy can be transformed from one form to another– the term “machine” is the generic term for those devices that

convert power from one form of energy into another

– conservation of energy: can not be created or destroyed

– conservation of power: power in plus stored released energy equals power out plus energy stored and power losses


Machine Efficiency

• Whenever energy is transformed, the output is always less than the input because all machines have losses:where– Po = output power

– Pi = input power

– Ploss = power losses

• The efficiency of a machine is defined as:where– η = percent efficiency

• Alternate forms of the definition:

%100×=i

o

P

Pη

%100

%100

×−=

×+

=

i

lossi

losso

o

P

PP

PP

P

η

η

lossoi PPP +=

PLoss

MachinePi Po


Kinetic Energy

• Kinetic energy is stored in moving objects– energy must be added to an object to make it move

• For objects with linear motion:– Ek = kinetic energy [J]

– m = object’s mass [kg]

– v = object’s velocity [m/s]

• For objects with rotational motion:– J = moment of inertia [kg m2]

– ω = angular velocity [rad/s]

221 vmEk =

221 ωJEk =

m

v

J

ω


Inertia, Torque, and Speed

• To change the speed of a rotating object, a torque must be applied for a period of time

• The rate of change of the speed (angular acceleration) depends upon the inertia as well as the torque:where– ∆ω = change in angular velocity

– ∆t = time interval of applied torque

– T = torque

– J = moment of inertia

• Example– a flywheel with an 10.6 kg m2 inertia turns at 60 rpm. How long

must a 20 Nm torque be applied to increase the speed to 600 rpm?

J

Ttn

J

Tt

∆=∆

∆=∆

55.9

ω


Speed of a Motor / Load System

• An electric motor applies a torque on the shaft

• A load applies a counter-torque on the shaft

• The net torque will accelerates or decelerate the shaft:

Motor Load

Tm Tldω

ldmnet TTT −=


Speed of a Motor / Load System

• Torque-speed characteristics of an electric induction motor and a fan load

20kN m

10

0

Tor

que

0 900 1800 rpmSpeed

Max torque

operating pointzero net torqueconstant speed

motor

fan


Directional Flow of Power

• Power supplied to the mechanical system– applied torque is in the same direction as rotation

• Power absorbed from the mechanical system– applied torque is in the opposite direction of rotation

Motor Load

Tm TldωMotor Load

Tm Tldω

Power

Power


Heat

• Heat is a form of energy and the SI unit is the joule– energy of vibrating atoms/molecules

– thermal potential is expressed as a temperature

• Thermal energy systems are analogy to DC circuits– heat [J]→ electrical charge

– temperature [K,°C] → voltage

– heat flow [W] → current

– thermal mass [J/ °C] → capacitance

– thermal conductivity [W/(m °C)] → conductance

– thermal insulation→ resistance


Temperature

• The temperature depends upon the received heat, mass, and material characteristics:where– Q = change in the quantity of heat [J]

– m = mass of object [kg]

– c = specific heat [J/(kg °C)]

– ∆t = change in temperature [K, °C]

• Example– for a water heater, calculate the heat required to raise the

temperature of 200 L of water from 10°C to 70°C assuming no losses (cH2O = 4180 J/kg °C; 1 LH2O = 1 kg)

tcmQ ∆=

mQ

∆t


Temperature

• Kelvin temperature scale is a measure of the absolute value

• Thermal mass is the mass of object times specific heat


Heat Transfer by Conduction

• By heating one end of a metal bar– its temperature rises due to increase atomic vibrations

– the vibrations are transmitted down the bar

– the temperature at the other end of the bar rises

• Thermal conduction is similar to the flow of electric current:where– P = heat transmitted [W]

– (t1 - t2) = temperature difference across object [°C]

– d = thickness of object [m]

– A = cross sectional area [m2]

– λ = thermal conductivity [W/(m °C)]

d

AttP

λ)( 21−=

d

λ

A

t1 t2

PP


Thermal Convection

• A continual current of fluids that provide cooling is called natural convection– Fluids, like air, oil, and water, in contact with hot surfaces

warm up and become lighter

– lighter fluids rise

– cooler fluids replace the rising fluids

– the warm fluids cool and sink

• The convection process can be accelerated by employing a fan or pump to create a rapid circulation– called force convection


Heat Loss by Convection

• The heat lost by natural air convection is:where– P = heat loss by convection [W]

– A = surface area of the object [m2]

– t1 = surface temperature of the object [°C]

– t2 = ambient temperature of the surrounding air [°C]

• Example– a totally enclosed motor has an

external surface area of 1.2 m2

– when operating at full-load, the surface temperature rises to 60°Cin an ambient of 20°C

– calculate the heat loss by natural convection

( ) 25.1

213 ttAP −=


Heat Loss by Convection

• The heat loss by forced air convection is:where– P = heat loss by convection [W]

– Va = volume of cooling air [m3/s]

– t1 = temperature of the incoming (cool) air [°C]

– t2 = temperature of the outgoing (warm) air [°C]

• Example– a fan rated at 3.75 kW blows 240 m3/min of air through a 750

kW moter to carry away the heat

– if the inlet temperature is 22°C and the outlet temperature is 31°C, estimate the losses in the motor

( )211280 ttVP a −=


Radiant Heat

• Radiant heat energy (electromagnetic waves-infrared spectrum) can pass through empty space or vacuum

• All objects radiate heat energy as a function of temperature

• All objects absorb radiant energy from other surrounding objects

• An object reaches a temperature equilibrium point when– it is the same temperature as that of its surroundings

– it radiates as much energy as it receivesand the net radiation is zero


Radiant Heat Loss

• The heat that an object looses by radiation:where– P = heat radiated [W]

– A = surface area of object [m2]

– T1 = object’s temperature [K]

– T2 = temperature of surrounding objects [K]

– k = constant that depends on the nature of the object’s surface

• Example– calculate the heat loss by radiation for the motor from the

natural convection example, which has an enamel surface

( )42

41 TTAkP −=

Type of surface k [W/(m2 K4)polished silver 0.2 × 10-8

bright copper 1.0 × 10-8

oxidized copper 2.0 × 10-8

aluminum paint 2.0 × 10-8

oxidized iron 4.0 × 10-8

insulation 5.0 × 10-8

enamel paint 5.0 × 10-8


Homework

• Problems:3-5, 3-9, 3-12, 3-17, and 3-20

chapter 3: fundamentals of mechanics and...

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