chapter 3: fundamentals of mechanics and...
TRANSCRIPT
1/11/00 Electromechanical Dynamics 1
Chapter 3: Fundamentals of Mechanics and Heat
1/11/00 Electromechanical Dynamics 2
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)
1/11/00 Electromechanical Dynamics 3
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
1/11/00 Electromechanical Dynamics 4
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 =
1/11/00 Electromechanical Dynamics 5
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 =
1/11/00 Electromechanical Dynamics 6
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 =
1/11/00 Electromechanical Dynamics 7
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=
1/11/00 Electromechanical Dynamics 8
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
1/11/00 Electromechanical Dynamics 9
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
1/11/00 Electromechanical Dynamics 10
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
1/11/00 Electromechanical Dynamics 11
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
1/11/00 Electromechanical Dynamics 12
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 −=
1/11/00 Electromechanical Dynamics 13
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
1/11/00 Electromechanical Dynamics 14
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
1/11/00 Electromechanical Dynamics 15
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
1/11/00 Electromechanical Dynamics 16
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
1/11/00 Electromechanical Dynamics 17
Temperature
• Kelvin temperature scale is a measure of the absolute value
• Thermal mass is the mass of object times specific heat
1/11/00 Electromechanical Dynamics 18
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
1/11/00 Electromechanical Dynamics 19
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
1/11/00 Electromechanical Dynamics 20
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 −=
1/11/00 Electromechanical Dynamics 21
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 −=
1/11/00 Electromechanical Dynamics 22
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
1/11/00 Electromechanical Dynamics 23
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
1/11/00 Electromechanical Dynamics 24
Homework
• Problems:3-5, 3-9, 3-12, 3-17, and 3-20