Y2: Option C - Engineering Physics Flashcards

Question

What is shown by a concave 'Angular velocity-time'

Answer 1

Decreasing α

Answer 2

A pair of forces that cause no resultant linear motion, but which cause an object to turn

Answer 3

The turning effect of a force (or couple) on an object

Answer 4

T = Fr F: force r: perpendicular distance to the axis of rotation ∴ T = Fr cosθ

Answer 5

T = Iα T = Fr, and F=ma ∴ T = mar a = αr ∴ T = α(mr^2) I = mr^2 ∴ T = Iα

Answer 6

W = Tθ For a linear system, W = Fs For a rotation, s = θr ∴ W = Frθ T = Fr ∴ W = Tθ

Answer 7

P = Tω Power = rate of change of energy (Work done) ∴ P = ΔW/Δt W = Tθ ∴ P = Δ(Tθ)/Δt ω = Δθ/Δt ∴ P = Tω

Answer 8

The opposing torque experienced by a rotating system due to the friction of it's components T(net) = T(applied) - T(friction)

Answer 9

A heavy wheel with a high moment of inertia, to resist changes in rotational motion

Answer 10

Flywheels are charged as they as spun, with the input torque stored as rotational kinetic energy. (just enough power is constantly supplied to keep the flywheel fully changed until the energy is required)

Answer 11

- Mass: (Ek ∝ I), and (I ∝ m) ∴ Ek ∝ m - Mass distribution: (Ek ∝ I), and (I ∝ r^2) ∴ Ek ∝ r^2 - Angular speed: Ek ∝ ω^2

Answer 12

- May be impractical eg. large, heavy wheel may not fit into the machine - May damage the machine eg. Increasing ω will increase the centrifugal force, potentially damaging the mechanism

Answer 13

- Usually carbon fibre (although they have a lower mass, can reach much higher speeds without causing damage) - Friction is reduced (eg. lubrication, levitating wheel with superconductors, work in a vacuum, etc)

Answer 14

- Smoothing torque: If input power varies, flywheels uses spurts of energy to change up, and supply constant torque to the system as they decelerate - Smoothing angular velocity: A charged flywheel will maintain the angular velocity of rotating components when the power supply stops - Supplying additional energy: If the system is required to exert varying forces, a constant power input can charge a flywheel which will decelerate to release spurts of energy

Answer 15

- Potter's wheel - Regenerative braking - Power grids - Wind turbines - Riveting machine

Answer 16

Machines designed to store as much energy as possible (as much Ek as the ω, r and m of the wheel allows)

Answer 17

- They are very efficient - They last a long time without degrading - Recharge time is short - React and discharge quickly - Environmentally friendly

Answer 18

- Much larger and heavier than other storage methods - Pose a safety risk as the wheel could break apart at high speeds (protective casing increases weight) - Energy lost through friction - Can oppose direction changes in moving objects

Answer 19

The product of the moment of inertia and angular velocity of a rotating object ∴ Angular momentum = Iω (Angular version of mv)

Answer 20

When no external force is applied (torque, friction, etc), the total angular momentum of a system remains constant ∴ I1ω1 = I2ω2

Answer 21

The change in angular momentum = Δ(Iω)

Answer 22

TΔt = Δ(Iω) T = Iα α = = Δω/Δt ∴ T = Δ(Iω)/Δt (⇒ ∴ T = rate of change of angular momentum) ∴ TΔt = Δ(Iω)

Answer 23

The movement of the centre of mass of an object

Answer 24

The rotation of an object about it's centre of mass

Answer 25

A combination of rotational and translational motion

Answer 26

Ep = Ek(trans) + Ek(rot) ∴ Ep = 1/2(mV^2) + 1/2(Iω^2)

Answer 27

V = √((2Ep)/(m+(I/r^2))) Ep = Ek(trans) + Ek(rot) ∴ Ep = 1/2(mV^2) + 1/2(Iω^2) ω = V/r ∴ Ep = 1/2(mV^2) + 1/2(I(V/r)^2) ∴ V = √((2Ep)/(m+(I/r^2)))

Answer 28

A volume of space filled with a gas

Answer 29

- Open system: Gas can flow in/out - Closed system: Gas can't enter or escape

Answer 30

Heat supplied to a system either increases the internal energy of the system or enables it to do work ∴ Q = ΔU + W Q: Energy transferred to the system by heating ΔU: Change in internal energy W: The work done by the system

Answer 31

- If energy is transferred to the system (system is heated), Q is POSITIVE - If energy is transferred away from the system (system is cooled), Q is NEGATIVE

Answer 32

- If the internal energy of the system increases, ΔU is POSITIVE - If the internal energy of the system decreases, ΔU is NEGATIVE

Answer 33

Changes to a gas in a closed system, as air can't flow in or out

Answer 34

(p1V1)/(T1) = (p2V2)/(T2) For an ideal gas, pV = nRT R is constant (Molar gas constant = 8.31 JK^-1mol^-1) n is fixed as change occurs in closed system ∴ pV/T = Constant ∴ p1V1/T1 = p2V2/T2

Answer 35

A change that occurs at a constant temperature (Assumed for slow changes)

Answer 36

Q = W For an ideal gas, U ∝ T T = constant ∴ ΔU = 0 ∴ Q = 0 + W ∴ Q = W If energy is applied to a system, it will result in an equivalent amount of work done (and vice versa)

Answer 37

p1V1 = p2V2 (Boyle's Law)

Answer 38

A change that occurs with no heat transfer in our out of the system (Usually assumed for Fast changes, as there isn't enough time for heat to be transferred)

Answer 39

ΔU = -W No heat is transferred ∴ Q = 0 ∴ 0 = ΔU + W ∴ ΔU = -W Any work done will result in an opposite change to the internal energy (and vice versa)

Answer 40

p1(V1)^𝛾 = p2(V2)^𝛾 𝛾: adiabatic constant

Answer 41

W = pΔV Work done = Force x distance ⇒ W=FΔx Pressure = Force / Area ⇒ F=pA ∴ W = pAΔx AΔx = ΔV ∴ W = pΔV

Answer 42

V1/T1 = V2/T2 (Charles' Law)

Answer 43

Q = ΔU If volume is constant, no work is done (W=0) ∴ Q = ΔU + 0 ∴ Q = ΔU Any heat transferred to or from the system directly changes the internal energy of the gas

Answer 44

P1/T1 = P2/T2 (Pressure Law)

Answer 45

- Graph showing the relationship between pressure and volume during the change - Arrow can be drawn on the curve to indicate the direction of the change

Answer 46

Isotherm = Curve on pV diagram for isothermal change - Shows p ∝ 1/V, as PV=constant - T is constant, but curve approaches closer to origin if T is lower (as work done is less)

Answer 47

The curve for an adiabatic change has a steeper gradient than for an isothermal change

Answer 48

- More work is required for adiabatic compression than for isothermal compression - More work is required for isothermal compression than for adiabatic compression (larger area under pV curve = More work done)

Answer 49

Isothermal: pV = constant ∴ let pV = k ∴ p = kV^-1 ∴ dp/dV = -kV^-2 k = pV ∴ dp/dV = -(pV)V^-2 ∴ dp/dV = -p/V Adiabatic pV^𝛾 = constant ∴ let pV^𝛾 = k ∴ p = kV^-𝛾 ∴ dp/dV = -𝛾kV^-𝛾-1 k = pV^𝛾 ∴ dp/dV = -𝛾(pV^𝛾)V^-(𝛾+1) ∴ dp/dV = -𝛾p/V Therefore, the gradient of an adiabatic curve is steeper than an isothermal curve, by a factor of 𝛾

Answer 50

Vertical line ∴ W = 0 (no area under graph)

Answer 51

Horizontal line (Work done as volume changes)

Answer 52

If a system undergoes different processes one after another, and returns to the original temperature and pressure before begining the processes again. This is shown by a closed loop on a pV graph

Answer 53

The difference between the work done by the system and the work done on the system ∴ Shown by the area inside the loop on a pV graph

Answer 54

An internal combustion engine where fuel is burned once every 4 strokes (1 stroke = up or down)

Answer 55

1) Induction - Piston starts at the top and moved down, drawing in an air-fuel mix through the inlet valve - Constant pressure (just below atmospheric) 2) Compression - Inlet value closes and piston moves up, compressing the air (Increased p, decreased V) - Just before the end of the stroke, the spark plug ignites the fuel causing a sudden temp and pressure increase, at almost constant volume 3) Expansion - The hot gas expands, and does work on the piston pushing it down (decreased p, increased V) - Work done by gas to expand is greater than work done on gas to compress it, as temp is higher (∴ Net output work drives engine) - Just before the end of the stroke, the exhaust valve opens and the pressure decreases 4) Exhaust - The piston moves up and the burned gas leaves through the exhaust valve - Constant pressure (just above atmospheric)

Answer 56

Undergo the same 4-stroke cycle with several slight differences for diesel: - Only air drawn in during induction - No spark plug: Air compressed until temp high enough for ignition, then diesel sprayed in by fuel injectors - No sharp peak before expansion, as injected fuel heats up, increasing the volume at an almost constant pressure, before expansion occurs

Answer 57

A pV graph showing the changes experienced by the system during a 4-stroke engine cycle

Answer 58

pV graph for the cycle of a 4-stroke engine, if it were to occur under perfect conditions

Answer 59

The theoretical cycle for a 4-stroke petrol engine

Answer 60

The theoretical cycle for a 4-stroke diesel engine

Answer 61

- The same gas is taken continuously around the cycle - Pressure and temperature changes are instantaneous - Heat source is external - Engine is frictionless

Answer 62

1) Gas is compressed adiabatically 2) Heat is supplied whilst volume is constant (increase p) 3) Gas is allowed to cool adiabatically 4) system is cooled at constant volume

Answer 63

1) Gas is compressed adiabatically 2) Heat is supplied whilst pressure is constant (increase V) 3) Gas is allowed to cool adiabatically 4) system is cooled at constant volume

Answer 64

- Corners of theoretical indicator diagram are not rounded, as it is assumed that the same air is cycled continuously - In real engine, volume is not constant during heating, as this could only occur is p and T changes were instantaneous - Theoretical loop doesn't include negative work between exhaust and induction, as it assumes same air is cycled again - Real temperature rise is less than theoretical, as engines have an internal heat source, and the fuel is not completely burned (∴

Answer 65

The net work done by the engine cylinders in 1 second (∴ = Max theoretical power generated by gases in engine cylinder)

Answer 66

Indicated power = (Area of loop)x(No. of cycles per second)x(No. of cylinders)

Answer 67

The amount of heat energy per unit time that could potentially be released from burning fuel

Answer 68

Input power = Calorific value x Fuel flow rate (⇒ Calorific value = How much energy the fuel has stored per unit volume)

Answer 69

The useful power output at the crankshaft

Answer 70

P = Tω (rotational power)

Answer 71

The power needed to overcome the friction of an engine

Answer 72

Friction power = Indicated power - Output power

Answer 73

Ratio of the energy available from the fuel, to the energy generated by the engine (Affected by how well energy is transferred into work)

Answer 74

Thermal efficiency = Indicated power / Input power

Answer 75

Ratio of the power generated by the cylinders to the power output at the crankshaft (Affected by the amount of energy lost due to friction between moving parts)

Answer 76

Mechanical efficiency = Output power / Indicated power

Answer 77

Ratio of the energy available from the fuel, to the power output at the crankshaft (Accounts for both mechanical and thermal efficiency)

Answer 78

Overall efficiency = Output power / Input power (∴ = Mechanical efficiency x Thermal efficiency)

Answer 79

Engines that convert heat energy into work

Answer 80

It is impossible for heat from a high temperature source to produce an equal amount of work ∴ All heat engines must must operate between a heat source and a heat sink, as some heat must increase the temp of the engine and then be disspursed (otherwise engine temp would reach source temp and work would stop) ∴ Q(H) = W + Q(C) Q(H): Heat transfer from the heat source W: Work done by the engine Q(C): Heat loss to surroundings (transfer to heat sink)

Answer 81

The ratio of the heat transferred from the source, to the useful output work (Affected by the difference between the source and sink temperature)

Answer 82

Efficiency = Q(H)-Q(C) / Q(H) Efficiency = W/Q(H) W = Q(H)-Q(C) ∴ Efficiency = Q(H)-Q(C) / Q(H)

Answer 83

T(H)-T(C) / T(H) Ratio of Q(H):Q(C) = Ratio of T(H):T(C) Efficiency = Q(H)-Q(C) / Q(H) ∴ Max theoretical Efficiency = T(H)-T(C) / T(H) as T = max energy transfer

Answer 84

- Energy lost to overcome friction - Fuel doesn't burn entirely etc.

Answer 85

To maximise the efficiency of a heat engine, as much input energy as possible must be transferred usefully (∴ Q(C) low as possible) However, some heat is always lost to the surroundings (Q(C) must be >0), so to avoid wasting energy, some 'combined heat and power (CHP)' plants use released heat for other purposes, such as heating local houses, generating electricity, etc.

Answer 86

An engine that transfers heat energy from a cold space to a hot space - Heat naturally flows from hot to cold, ∴ work is required to transfer - There will always be some energy in the cold space to be transferred, as the temp as above absolute 0

Answer 87

According to the second law of thermodynamics (Q(H) = W + Q(C)), there must always be a transfer of energy between a hot and cold space. If 100% of the energy from the hot space was converted into work (or vice versa for reverse engines), the engine temp would eventually be the same as source temp, so no work would be done T(C) > 0 Max theoretical Efficiency = T(H)-T(C) / T(H) ∴ Efficiency < 1

Answer 88

A refrigerant moves through coils, travelling inside and outside of the fridge: - The liquid refrigerant outside the fridge is decompressed, decreasing the pressure and therefore decreasing it's temperature as it flows into the fridge. - It then begins to absorb energy from it's surroundings as it travels through the evaporator coils inside the fridge, decreasing the temperature of the surroundings as the refrigerant heats up to become a gas. - As it leaves the fridge, the gaseous refrigerant is compressed, increasing the pressure, and therefore the temperature. - It then begins to travel through the condenser coils outside the fridge, where it releases it's energy to the surroundings, cooling as it returns to a liquid - The cycle then repeats as energy is absorbed from in the fridge and released to the surroundings, due to the work done during the compression and decompression stages

Answer 89

Work similar to refrigerators, as the refrigerant is sequentially compressed and decompressed to transfer energy from the cold space (outside the house), to the hot space (inside the house) However, for a heat pump: - Heat given out by compressed refrigerant usually heats a water tank - A fan will usually amplify the absorption of heat in the cold space

Answer 90

A measure of how well work is transferred into heat energy by a reverse heat engine: - Ratio of Q:W - ∴ = 1/efficiency, so shows how well work causes energy transferer, (whereas shows how well energy transfer causes work, in regular heat engines) - Can be greater than 1

Answer 91

COP(ref) = Q(C) / (Q(H) - Q(C)) For a refrigerator, the heat removed from the cold space determines how well it works ∴ COP(ref) = Q(C)/W W = Q(H) - Q(C) ∴ COP(ref) = Q(C) / (Q(H) - Q(C))

Answer 92

At max theoretical efficiency: COP(ref) = T(C) / (T(H) - T(C))

Answer 93

COP(hp) = Q(H) / (Q(H) - Q(C)) For a heat pump, the heat transferred to the hot space determines how well it works ∴ COP(hp) = Q(H)/W W = Q(H) - Q(C) ∴ COP(hp) = Q(H) / (Q(H) - Q(C))

Answer 94

At max theoretical efficiency: COP(hp) = T(C) / (T(H) - T(C))