thermo

isobaric work

-p(v2-v1)

“assuming no stagnation pressure losses”

means can use bernoulli in pipe even though no streamline

combustion chemical reaction

CH₄ + 2O₂ → CO₂ + 2H₂O

mass flow rate (thermo)

m = (work output rate=power output) / (Wnet)

c_v equation

c_v = c_p - R

adiabatic equation

P₁V₁^gamma = constant

OTTO cycle PV diagram

otto cycle cycles

1-2: adiabatic compression (isentropic)

2-3: heat added at constant volume

3-4: adiabatic expansion (isentropic)

4-1: heat rejected at constant volume

pros of hydrocarbon fuels

pro: high energy density

con: pollution

isochoric work (constant volume)

0

adiabatic work

first one in formula book

isothermal work

second in formula book

isothermal change in internal energy

U = 0 because no change in temp

enthalpy change in ideal gas

He - Hi = Cp (Te - Ti)

work in turbine

W < 0 because work developed from gas or liquid passing though blades

Pump/Compressor Work

W > 0
compressor - work done on a gas passing them through to raise pressure

pump - work input give energy to liquid passing through

gamma equaiton

gamma = Cp / Cv

first law

Qh - Qc = W

thermal efficiency

output/input in terms of Qh, Qc and W

corollary 1

impossible to make a system that can transfer hear from a cooler to hotter body without work

• fridges and hear pumps

corollary 2 and 3

• reversible cycles

• reversible cycles/engines (engines) are more always efficient than a heat engine

• all reversible engines have the same efficiency

corollary 4

its impossible to reduce temp of a system to zero kelvin in finite number of operations\

• entropy, temp relation dS = dQrev/T - at T=0 S=0 = perfect order, very difficult to reach

• Efficiency rev = 1 - Tc / Th (carnot efficiency)

corollary 5

• carnot efficiency eq. = max efficiency achieved when heat transferred only between Th, Tc

HENCE

• A multi-reservoir engine exchanges heat at intermediate temperatures
  → reduces effective temperature difference

• Smaller temperature differences → less work extracted per unit heat

THEREFORE

• multi reservoir engines still reversible but less efficient

corollary 6 (clausius inequality)

• heat transferred/temperature ≤ 0

  • = 0 → reversible cycle

  • < 0 → irreversible cycle

• measures how ideal a cycle is

• leading to entropy eq. in corollary 4

  • for irreversible ΔS > ΔQ/T (entropy of universe always increasing)

corollary 7

• FOR REVERSIBLE PROCESS between final states ΔS = ∫ΔQ/T

• heat is path dependent (how you get from state 1 to 2 matters)

• entropy is path independent only what state 1 and 2 is matters

isentropic (reversible adiabatic process)

PT^γ/γ-1 = constant

T-S diagram

• magnitude of net heat transfer for any reversible cycle = area enclosed by cycle

corollary 8 - entropy change in any process

• First term: entropy transfer with heat

• Sgen : entropy generated due to irreversibilities (friction etc.)

• Reversible: Sgen = 0

• Irreversible: Sgen > 0

  • system entropy stays same but change in entropy of surroundings changes

internal energy formula

ΔU = mCv(ΔT)

COP formula

B = Qh or Qc / W (depedns if fridge or heater)

thermal efficiency eq. and terms

• n = Wnet/Qh = 1 - Qc/Qh

• Qh = Wnet + Qc

• Qh = heat input

• Qc = heat rejected (output)

reversible process

very slow so stays near equilibrium, no losses, gas pressure inside = pressure of surroundings

irreversible process

• S_gen = 0

• “free expansion” means irreversible and that P_ext = 0

• W = -P_ext dV

carnot cycle efficiency

n = W_net / Q_in = T_H - T_C / T_H = 1 - Q_out/Q_in

carnot cycle (all reversible)

1-2: isothermal expansion (T_H)

2-3: Adiabatic expansion

3-4: Isothermal compression (T_C)

4-1: Adiabatic compression

joule brayton cycle (all reversible)

1-2: adiabatic compression

2-3: heat addition at constant pressure

3-4: adiabatic expansion

4-1: heat rejection at constant pressure

efficiency for joule brayton

n = -W_net / Q₂₃

W_net for joule brayton

W₃₄ + W₁₂ = cp(T₃ - T₄) + cp (T₂ - T₁)

r_p

pressure ratio = p₂ / p₁ = p₃ / p₄

bar to pascal

1 Pa = bar x10⁵

turbine and compressor

modelled as joule brayton cycle

• left (1-2) is compressor (isentropic compression)

• right (3-4) is turbine (isentropic expansion)

power to drive part

(mass flow rate)(specific to part work)

• work for compressor = cp[T₂-T₁]

• work for turbine = cp[T₃-T₄]

• work for whole cycle W_net = W₁₂ + W₃₄

heat input between turbine and compressor (Ideal)

Q = mass flow rate * cp[T₃ - T₂]

isentropic joule cycle efficiency compressor/whole cycle

compressor: T_2s - T₁ / T₂ -T₁

whole cycle efficiency: W_net / Q₂₃

isentropic joule cycle efficiency turbine

turbine: T₃ - T₄ / T₃ - T_4s

• s bit on bottom

• smaller - bigger

the actual temperature at exit from compressor

T₂ not T_2s

What part drives the other in joule cycle

turbine produces shaft work that drives compressor (T-S diagram just about thermodynamic states)

JOULE CYCLE WITH NOZZLE

• W₁₂ = W₃₄

• for thrust of nozzle use SFEE to find V₅

  • then F = (mass flow rate)(V₅)

• ideal nozzle T_5s/T₄ = T₅ / T₄

JOULE CYCLE WITH POWER TURBINE

JOULE CYCLE WITH REHEAT

• HP turbine drives compressor

• W_c = - W_HP

• - W_LP = W_NET = cp(T₅ - T₆)

JOULE CYCLE WITH AFTERBURNING

bar to pascal

1 bar = 10⁵ Pa

turbine efficiecny

actual/ideal (s)

compressor efficiecny

ideal (s) / actual

for temperature pressure ratios equaiton

always use s version of top temperature eg. T_2s not T₂

Joule cycle horizontal bits

constant pressure

JOULE CYCLE WITH HEAT EXHANGER

ε = actual temp rise / max temp rise = T₃ - T₂ / T₅ - T₂

• max pos value ε = 1 so T₃ = T₅

JOULE CYCLE WITH INTERCOOLING

• ΔT = T₄ - T₁

• trying to minimise ΔT

• dont sum pressure, multiply

• (R_{p (machine)})^_1/n n = number of stages

reciprocating internal combustion engines

piston moves up and down with each motion being called a stroke. connecting rod and cranshaft are required to convert to rotational motion

• not SFEE problem

“suck” = induction stroke piston down

“squeeze” = compression stroke, valves close piston travels up cylinder

“bang”= expansion stroke combustion happens forces piston down

“blow” = exhaust stroke, valve open piston travel up, gases expelled

otto cycle efficiency

n = Wnet/ Q₂₃ (Q_in)