thermodynamics and engines

Question 1

first law of thermodynamics

Answer 1

describes how energy is conserved in a system through heating, cooling and doing work

Question 2

equation for first law of thermodynamics

Answer 2

Q = △U + W
U = internal energy

Question 3

positive and negative Q

Answer 3

+Q = energy transferred to system
-Q = energy transferred away from system

Question 4

non-flow process

Answer 4

a process which occurs in a closed system
no gas flows in or out the system

Question 5

condition to apply first law of thermodynamics to a closed system

Answer 5

gas has to be an ideal gas
means energy is only dependent on the temperature
need to assume that if any work is done there is a change in volume

Question 6

ideal gas equation

Answer 6

pV = nRT

p1 V1 / T1 = P2 V2 / T2

Question 7

positive and negative meaning for work done

Answer 7

+W = work done by the system
-W = work done on the gas

Question 8

isothermal

Answer 8

a process in which the temperature of the system remains constant
as internal energy is dependent on only temperature, △U = 0
so Q = W

Question 9

what does it mean when Q = W

Answer 9

supplying heat energy to the system will result in an equivalent amount of work being done by the gas (volume increases)
if work is done on system then an equivalent amount of energy is lost

Question 10

isothermal equations

Answer 10

pV = constant
p1 V1 = p2 V2

Question 11

adiabatic process

Answer 11

a process in which no heat is gain or lost in a system
Q = 0
so △U = -W

Question 12

what does △U = -W mean

Answer 12

any change in the internal energy is caused by work done on/by system

if work is done by system then internal energy will decrease by the equivalent amount vice versa

Question 13

adiabatic process equations

Answer 13

pV^γ = constant

p1 V1^γ = p2 V2^γ

Question 14

changes at a constant pressure equation

Answer 14

W = p△V
V1 / T1 = V2 / T2

Question 15

Changes at a constant volume

Answer 15

W = 0 so Q = △U
means all of the heat transferred to/out the system goes directly to increasing/decreasing internal energy
p / T = constant

p1 / T1 = p2 / T2

Question 16

p/V graph

Answer 16

arrow drawn to indicate in which direction the change is happening
area under = magnitude of work done

Question 17

isotherm

Answer 17

p ∝ 1/V
isothermal expansion = right
isothermal compression = left
the higher the temperature the further the curve is from the origin

Question 18

p-V diagrams for adiabatic processes

Answer 18

have a steeper gradient than a isotherm
area is larger than isotherm curve so more work is done to compress a gas adiabatically than isothermally
the gas does less work if it expand adiabatically instead on isothermally

Question 19

cyclic process

Answer 19

a system can go through different processes to form a loop

Question 20

net work done of cyclic curve

Answer 20

find the difference between work done by the system and work done to the system
work done per cycle = area of loop

Question 21

four stroke engines

Answer 21

engines which burn fuel once every four strokes of a piston

Question 22

internal combustion engines

Answer 22

contains cylinders filled with air
air is mixed with fuel, which is then burned, releasing a large amount of energy
the air fuel mixture is trapped by two tight fitting pistons which move up and down

Question 23

induction

Answer 23

piston starts at top of cylinder and moves down, increasing the volume of the gas above it
this sucks in a mixture of air and fuel in the open inlet valve
pressure remains constant, just below atmospheric pressure
indictaor diagram is straight horizontal line

Question 24

compression

Answer 24

inlet valve is closed and the piston moves back up
work is done on the gas increasing the pressure
just before the piston is at the end of the stroke, the spark plug creates a spark which ignites the air-fuel mixture
temperature and pressure increases at an almost constant volume

Question 25

expansion

Answer 25

hot air-fuel mixture expands and does work on the piston pushing it downwards work done by gas to expand is more than work done on gas to compress as it has a higher temperature just before the piston is at the end of the stroke the exhaust valve opens reducing the pressure

Question 26

exhaust

Answer 26

piston moves up cylinder and the burnt gas leaves through the exhaust valve pressure remains almost constant, just above atmospheric pressure

Question 27

four stroke diesel engines

Answer 27

same four strokes but have differences -in the induction stroke only air is sucked in, not a air-fuel mixture -dont have spark plug so in compression stroke, air is just compressed until it reaches a temperature high enough to ignite diesel fuel just before the stroke ends, diesel is sprayed into the cylinder through a fuel injector and ignites expansion and exhaust stroke is the same

Question 28

how the four stroke diesel engine induction diagram differs

Answer 28

flatter top showing the point at which diesel fuel is injected and heats up to combustion temperature

Question 29

theoretical indicator diagrams

Answer 29

petrol engine = otto cycle diesel engine = diesel cycle both models make the following assumptions- -same gas is continuously taken around the cycle. gas is pure air with a adiabatic constant of 1.4 -pressure and temperature chnages can be instantaneous -heat source is external -engine is frictionless

Question 30

theoretical cycle for the four stroke petrol engine

Answer 30

A) gas is compressed adiabatically B)heat is supplied while volume is kept constant C)gas is allowed to cool adiabatically D)system is cooled at a constant volume

Question 31

theoretical cycle for a four stroke diesel engine

Answer 31

A)gas is adiabatically compressed B)heat is supplied but pressure is kept constant C)gas is allowed to cool adiabatically D) system is cooled at a constant volume

Question 32

comparing theoretical and real life diagrams

Answer 32

engineers use this to show how well the engines are performing -corners of theoretical diagram are not rounded as its assumed same air is used continuously irl corners are rounded as inlet and exhaust valves let in new air and burnt gas out -irl heating doesn't take place at a constant volume as an increase in pressure and temperature wouldn't be instantaneous -theoretical model doesn't include the small amount of negative work caused by the loop between induction and exhaust lines as it assumes the same air circulates -engines have an internal heat source, not an external one, means temperature rise is not as large as the theoretical one as fuel used to to heat the gas is not completely burned in the cylinder -energy is needed to overcome friction caused by moving parts irl so net work done will always be less than theoretical model

Question 33

indicated power

Answer 33

net work done by cylinder in one second can think of indicated power of as maximum theoretical power generated by the gases in the engine cylinders

Question 34

indicated power equation

Answer 34

indicated power of engine = area of p-V loop x no. cycles per second x no. cylinders

Question 35

crankshaft

Answer 35

a piston moving up and down in an engine is connected to a crankshaft by a rigid rod this converts the up and down motion of the piston into a rotational motion the greater the pressure on the piston (hence the larger the force generating the up and down motion), the greater the rotational motion (torque) generated

Question 36

output (brake) power

Answer 36

the useful power output at the crankshaft P = Tw

Question 37

friction power

Answer 37

friction occurs between many moving parts of an engine, for example between the piston and the cylinder, at the bearings and when the valves are opened and closed work needs to be done to overcome friction in the engine the power needed to do this is called friction power. This means that the brake power of the engine is less than the indicated power friction power = indicated power - brake power

Question 38

input power

Answer 38

input power is the amount of heat energy per unit time it could potentially gain from burning fuel input power = calorific value x fuel flow rate

Question 39

the calorific value

Answer 39

the calorific value of fuel tells you how much energy the fuel has stored in it per unit volume

Question 40

mechanical efficiency

Answer 40

affected by the amount of energy lost through moving parts (e.g. through friction) its the ratio of the power generated by cylinders to the power output at the crankshaft mechanical efficiency = brake power / indicated power

Question 41

thermal efficiency

Answer 41

describes how well heat energy is transferred into work thermal efficiency = indicated power / input power

Question 42

overall efficiency

Answer 42

brake power / input power

Question 43

second law of thermodynamics

Answer 43

heat engines must operate between a heat source and a heat sink

Question 44

heat engines

Answer 44

convert heat energy into work no engine can transfer all heat energy its supplied into useful work though some heat will always end up increasing the temperature of the engine if the engine reaches the temperature of the heat source then no work is done

Question 45

heat sink

Answer 45

a region which absorbs heat from the engine, if not then the engine will reach the temperature of the heat source and no work will be done

Question 46

heat engines explained

Answer 46

heat energy transferred to engine from heat source = Qh some of this energy is transferred into useful work but some enrgy must be transferred into the heat sink, Qc which has a lower temperature means heat engines cant be 100% effecient

Question 47

heat engine efficiency

Answer 47

W / Qh = Qh - Qc / Qh

Question 48

maximum theoretical efficiency

Answer 48

by assuming the perfect conditions max theoretical efficiency = TH -TC / TH in practise real heat engines efficiency is lower than its theoretical maximum as energy is lost due to friction and the energy needed to move them also fuel doesnt burn entirely so the value of QH is lower in practice

Question 49

maximising heat engines efficiency

Answer 49

to maximise efficiency as much of the input heat must be transferred usefully however heat engines are very inefficient as there is a lot f waste heat which is transferred to surroundings and lost

Question 50

combined heat and power plants

Answer 50

CHPs try to limit energy waste by using waste heat for other uses like heating nearby houses and buildings

Question 51

reversed heat engines

Answer 51

operate between hot and cold reserviours like other engines heat is taken from the cold space and put in the hot space heat natuurally flows from hot to cold so work has to be done to do this

Question 52

input/output of work for heat engines

Answer 52

heat engine = output of work reverse heat engine = input of work

Question 53

refridgerators

Answer 53

aims to extract as much heat as possible from cold sace and exert into the suroundings for each joule of work done work is done to transfer heat away from cold space via pipes on the back of the appliance

Question 54

heat pumps

Answer 54

cold space is outside hot space inside aims to pump as much heat as possible to hot space per joule of work done used to heat rooms and water inside homes

Question 55

coefficient of performance

Answer 55

reverse heat engines are judged on how well they can transfer heat based on the amount of work done on them a measure of how well this work is converted into heat transfer

Question 56

COP example

Answer 56

COP of 4 transfers 4J of heat energy for every 1J of work done a COP greater than 1 means the amount of energy transferred is more than the work done needed to transfer it the higher the COP the lower the running costs

Question 57

COP for refridgerators

Answer 57

as its heat is removed from cold space important COP = QC / W = QC / QH - QC maximum theoretical effieciency COP COP = TC / TH - TC

Question 58

COP for heat pumps

Answer 58

as heat is transferred to hot space thats important COP = QH / W = QH / QH - QC max theoretical effeciency COP COP = TH / TH - TC