11: Engineering Physics Flashcards
First Law of Thermodynamics (4)
- Q = ΔU + W
- Q is energy transferred to the system by heating
- ΔU is the increase in internal energy
- W is work done by the system
Ideal Gas Equation
p V = n R T
Isothermal Changes (4)
- Where the temperature of the system is constant
- The gas’s absolute temperature determines its internal energy
- So ΔU = 0, meaning Q = W
- Hence, supplying heat energy to the system results in an equivalent amount of work being done by the gas so its volume increasing
Isothermal Change Equation
p V = constant
Adiabatic Changes (3)
- Where no heat is transferred in or out of the system
- So Q = 0 meaning ΔU = -W
- Hence, any change in the internal energy of the system is caused by work done by/on the system
Adiabatic Change Equation
p V^γ = constant
Constant Pressure Change Equation
V / T = constant
Constant Volume Changes (3)
- If volume is constant, no work is done by/on the system
- So W = 0 meaning Q = ΔU
- Hence, all heat energy transferred to the system goes into increasing its internal energy
Constant Volume Change Equation
p / T = constant
p-V Diagrams for Isothermal Changes (3)
- p-V diagram for a compression, where work is done on the system:
Textbook p539 Figure 2 - p-V diagram for an expansion, where work is done by the system:
Textbook p539 Figure 2 - The higher the temperature of the process, the further from the origin the p-V diagram is
p-V Diagrams for Adiabatic Changes (3)
- The p-V curves for adiabatic processes have a steeper gradient than isothermal processes
- More work is done to compress gas adiabatically than isothermally:
Textbook p540 Figure 4 - The gas does less work if it expands adiabatically instead of isothermally
Textbook p541 Figure 5
p-V Diagram for Constant Volume Changes (3)
- p-V diagrams for changes with constant volume are straight vertical lines
- No work is done as volume doesn’t change and there is no area under the line
- If a system is heated at constant volume, its pressure will increase:
Textbook p541 Figure 6
p-V Diagram for Constant Pressure Changes (3)
- p-V diagrams for constant pressure changes are straight horizontal lines
- The work done is the area of the rectangle under the graph – W = p ΔV
- Textbook p541 Figure 7
Work Done =
Area below the graph
Work Done per Cycle =
Area of loop
Four-Stroke Petrol Engine Cycle (4)
- Induction
- Compression
- Expansion
- Exhaust
Induction (3)
- The piston moves down, increasing the volume of the gas (air-fuel mix) as the inlet valve is open
- The pressure of the gas remains constant just below atmospheric pressure
- Indication diagram: Textbook p544 Figure 1
Compression (5)
- The inlet valve is closed and the piston moves up the cylinder
- This does work on the gas, increasing the pressure
- Just before the piston is at the end of this stroke, the spark plug creates a spark, igniting the gas
- The temperature and pressure increase at almost constant volume
- Indication diagram: Textbook p544 Figure 3
Expansion (4)
- The hot gas expands and does work on the piston, pushing it down
- The work done by the gas is greater than the work done to compress it as it is now at a higher temperature
- Just before the piston is at the end of this stroke, the exhaust valve opens, reducing the pressure
- Indication diagram: Textbook p545 Figure 4
Exhaust (3)
- The piston moves up the cylinder and the burnt gas leaves through the exhaust valve
- The pressure remains almost constant
- Indication diagram: Textbook p545 Figure 5
Four-Stroke Diesel Engine Cycle (5)
- In the induction stroke, only air is pulled into the cylinder
- Diesel engines don’t have a spark plug, so in the compression stroke, the air is compressed until its temperature is high enough to ignite the fuel
- Just before the end of the stroke, diesel is sprayed into the cylinder through a fuel injector and ignites
- The expansion and exhaust strokes are the same as a petrol engine
- The indicator diagram has a flatter peak at the start of the expansion stroke, showing the point where fuel is injected and heats up to combustion temperature:
Textbook p545 Figure 7
Assumption of Theoretical Cycles (5)
- The same gas is taken continuously around the cycle
- The gas is pure air with an adiabatic constant γ = 1.4
- Pressure and temperature changes can be instantaneous
- The heat source is external
- The engine is frictionless
Petrol Engine Cycle Theoretical Diagram (5)
- A: The gas is compressed adiabatically
- B: Heat is supplied at constant volume
- C: The gas cools adiabatically
- D: The system cools at constant volume
- Textbook p546 Figure 8
Diesel Engine Cycle Theoretical Diagram (5)
- A: The gas is compressed adiabatically
- B: Heat is supplied at constant pressure
- C: The gas cools adiabatically
- D: The system cools at constant volume
- Textbook p546 Figure 10
Comparison of Theoretical and Real Diagrams (7)
- Curved corners: because valves take finite time to open and close
- No constant volume process: because piston would have to stop
- Compression and expansion not adiabatic curves: because energy is lost by heat transfer
- The cycle is open because engine needs to
draw in air and expel exhaust - Heating not at constant pressure: because fuel injection and combustion cannot be exactly controlled
- Area of diagram is less because energy is lost by heat
transfer and incomplete combustion - Pressure not as high because incomplete combustion
Indicated Power =
(Area of p–V loop) x (no. of cycles per second) x (no. of cylinders)
Output or Brake Power Equation
P = T ω
Friction Power =
Indicated power - brake power
Input Power =
Calorific value x fuel flow rate
Engine Efficiency (5)
- Mechanical efficiency is affected by the amount of energy lost through moving parts
- Mechanical efficiency = brake power / indicated power
- Thermal efficiency describes how well heat energy is transferred into work
- Thermal efficiency = indicated power / input power
- Overall efficiency = brake power / input power
Impossibility of Engine Working only by First Law (4)
- No engine can transfer all the heat energy it is supplied into useful work
- Some heat ends up increasing the temperature of the engine
- If the engine temperature reaches that of the heat source, then no heat flows and no work is done
- So, engines cannot work by only the first law as Q and W would both be 0
Second Law of Thermodynamics
The need for a heat engine to operate between a source and a sink
Diagram of Heat Engine Operating between Heat Source & Heat Sink
Textbook p552 Figure 1
Heat Engine Efficiency =
W / Q_H = (Q_H - Q_C) / Q_H
Heat Engine Maximum Theoretical Efficiency =
(T_H - T_C) / T_H
Reason for Lower Efficiencies of Practical Engines
There is usually a lot of waste heat, which is transferred to the surround area and lost
Combined Heat & Power Schemes
They use waste heat to heat houses and supply heat to businesses nearby as well as generate electricity to use and supply to the national grid
Refrigerators (4)
- The cold space is inside the refrigerator and the hot space is the refrigerator’s surroundings
- A refrigerator extracts heat energy from the cold space
- Work is done to transfer heat energy via pipes on the back of the appliance
- Refrigerators keep enclosed spaces cool so they can keep perishable food fresh
Heat Pumps (3)
- The cold space is outdoors and the hot space is inside a house
- A heat pump pumps heat into the hot space
- They are used to heat rooms and water in homes
Reversed Heat Engine Diagram
Textbook p555 Figure 1
Coefficient of Performance
The amount of heat energy transferred per unit of work done
Refrigerator Coefficient of Performance (3)
- It’s the heat removed from the cold space, that’s important for a refrigerator
- COP_ref = Q_C / W = Q_C / (Q_H - Q_C)
- Maximum theoretical COP_ref = T_C / (T_H - T_C)
Heat Pump Coefficient of Performance (3)
- It’s the heat transferred to the hot space, that’s important for a heat pump
- COP_hp = Q_H / W = Q_H / (Q_H - Q_C)
- Maximum theoretical COP_hp = T_H / (T_H - T_C)
Moment of Inertia
A measure of how much an object resists a change to its rotational speed
Moment of Inertia for a Point Mass
I = m r²
Moment of Inertia for an Extended Object
I = Σ m r²
Factors Affecting Moment of Inertia (2)
- The magnitude of the object’s mass
- The distribution of the object’s mass about its centre of rotation
Angular Kinetic Energy Equation
Eₖ = 1/2 I ω²
Factors Affecting Energy Storage Capacity of a Flywheel (3)
- Increase its mass to increase its moment of inertia and so kinetic energy stored
- Increase its angular speed to increase its kinetic energy stored
- Use a wheel with spokes or a heavier rim so more mass is concentrated further from the axis of rotation, increasing its moment of inertia and kinetic energy stored
Flywheel
A heavy wheel, which has a high moment of inertia in order to resist changes to its rotational motion
Uses of Flywheels in Machines (3)
- Flywheels are charged as they are spun, turning input torque into rotational kinetic energy. If it keeps spinning at the same rate, it stores the energy for later use
- Just enough power is continuously input to overcome frictional torque, keeping the flywheel fully charged
- When extra energy is needed, the flywheel decelerates, transferring some of its kinetic energy to another part of the machine
Uses of Flywheels for Smoothing Torque & Speed (4)
- Flywheels are used in machines to smooth engine and load (torque due to resistance forces) torque
- In systems where the power supplied can vary, flywheels keep the angular velocity of rotating components constant. It uses each spurt of power to charge and delivers the energy smoothly
- In systems where the force to exert can vary, flywheels are used. If the load torque is too high, the flywheel decelerates, releasing extra energy
- If the engine torque is greater than the load torque, the flywheel charges by accelerating to store the spare energy
Angular Displacement
The angle through which a point has been rotated
Angular Velocity
The angle a point rotates through per unit time
Angular Speed
The magnitude of angular velocity
Angular Acceleration
The rate of change of angular velocity
Angular Velocity Equation
ω = Δθ/Δt
Angular Acceleration Equation
α = Δω/Δt
Torque Equations
- T = F r
- T = I α
Angular Momentum Equation
angular momentum = I ω
Conservation of Angular Momentum
Assuming no external torques act, the total angular momentum of a system remains constant
Angular Impulse =
Change in angular momentum
Angular Impulse Equation
T Δt = Δ(I ω)
Rotational Dynamics Work Done Equation
W = T θ
Rotational Dynamics Power Equation
P = T ω
