MSE Energy

isolated system, no exchange of mass or energy so delta S >= 0

so no matter what we do in our universe S >= 0, it can not diminish; just stay the same for reversible processes

so even if I do a cycle, the entropy of the system is equal at beginning and end, but the entropy of the environment has increased (e.g. work or heat exchange)

no heat exchange => integral becomes 0, so change in entropy is only due to increased entropy (irreversibility) of transformation

S gen > 0: irreversible, real transformation

S gen = 0; reversible ideal transformation

S gen < 0: impossible transformation => check this for perpetuum mobiles :)

which one creates less entropy? => the lower the better because less irreversibility

energy is constant but entropy will go to S_max

so more energy will be transformed into entropy and the universe will die in a "warm death", e.g. everything has the same temperature, no gradients to push exchange anymore

(T2 / T1) = (p2/p1)^(k-1)/k) = (v1/v2)^(k-1)

valid for ideal gas, isentropic process, constant specific heat

heat engine converts thermal energy into work while operating in a cycle

work you get out will always be lower than the heat you put in

1 high temperature reservoire (boiler) gives heat

2 turbine gets work out

3 condenser: energy sink/ low temperature reservoir, gets heat out

4 pump: invests work

it is impossible for a device that operates in a cycle to work with only one high temperature reservoir and produce a net amount of work

=> you always need a system that recieves discharded heat (e.g. you need to cool down the expanded gas)

work out / energy input (because we do not care about the heating of the environment)

nu = 1 - Q_out/Q_in    = 1 - Q_cold/Q_hot = 1 - Q_condensor/Q_boiler

we cannot have 100% efficiency becase we cannot work in a thermodynamic cycle with QC = 0

most efficient thermodynamic cycle, physical limit

you can compare e.g. a nuclear power plant to the carnot cycle and compare efficiency

the carnot thermal efficiency depends only on temperature of the two reservoirs. the bigger the T-difference, the better the efficiency

nu = 1 - T_C / T_H  ; also called Carnot factor theta

1: impossible to construct an engine with two T reservoirs that is miore efficient than a reversible engine operating at the same temperatures  nu real < nu reversible

2: all engines that do the Carnot cycle between two given temperatures have the same efficiency nu_reversible

additional comment: no heat engine with a cycle can convert all of its heat input completely into work

refridgerator

extract heat from a low temperature reservoir and discharge it in high temperature reservoir, cool down the low temperature reservoir

has a compressor instead of a turbine, required input is work in compressor, desired output is Q_C

same as refridgerator (with a compressor), just different target

transfers heat from low T reservoir to high T reservoir. Target: heat up high T reservoir

invest W in compressor, get -Q_H to heat up resrevoir

Q_C is "free" from the environment

efficiency would be bigger than 1 because W_net_in < Q_H

measure for thermal efficiency

refridgerators: COP_R = QC/W = QC/(QH-QC)

for reversible refridgerators: COP_rev = 1 / (TH/TC - 1)

heat pump: COP_HP = COP_R + 1

mechanical work and heat are equivalent from a countable point of view

heat transfer transformation into work can happen only making use of two heat sources

you cannot have a cycle who only cause heat transfer from cool body to hot body

unless you supply external work (no refridgerator or heat pump without external work input)

reversible: theroy, ideal, no generated entropy; irreversible: real process, entropy generated

work producing (e.g. thermal engine) at equilibrium, reversible, produces max amount of work possible

work consuming (refridgerator/ heat pump) at equilibrium, reversible, requires minimal amount of work possible

1: conservation/ transformation of energy => quantity of energy

2: direction and limited extend of heat to work transformations, "quality" of energy and process direction

continuous

ambient air goes in; axial compressor with rotating blades; constant mass flow, distance between blades gets smaller, so gas = more dense, compressed air = more enthalpy to harvest; continuous mass flow = flow only in one direction

compressed gas in combustion chamber

drives a turbine

exhaust burned gas

so extraction of enthalpy from fuel and transforms it into mechanical energy, no torque, just hot gas expanding and seed up by a nozzle, can also just be thrust instead of big turbine (e.g. aeronautical)

ambient air in: T low, p low, h low

compression: T medium, p high, h medium

combustion: T high, p high, h high (because increased T)

expansion (turbine or just thrust) : T medium, p low, h low (because we exploit enthalpy)

tubine and compressor are connected by a shaft, so turbine loses some of the generated energy to drive the compressor

replace combustion chamber and atmosphere by two heat exchangers in order to make a closed cycle and just assume that combustion heat is a heat source from outside

no heat losses, no friction or pressure losses, all processes are reversible

operating fluid is just air (as perfect gas)

1 2: compress air, isentropic (reversible, adiabatic)

2 3: heat exchanger, takes up heat, at constant pressure

3 4: turbine is driven by expansion, isentropic,

4 1: heat exchanger, release heat into environment, at constant pressure (usually atmospheric)

equal to work done by machine

ideal, adiabatic, reversible, isentropic

BWE = Wc /Wt = compressor vs turbine, how much the compressor takes out compared to toal work, can be 40% to 50% because gas compression is a lot of work

between both heat exchangers p2/p1 = p3/p4 = PR

nu thermal = 1- (1/PR)^(k-1)/k, with k = cp/cv

velocity at output is higher than at input, thrust; the more mass the higher the thrust

part of the energy of exhaust gas is transferred to air before it enters the combustion chamber, by a regenerator, so less heat is needed in combustion chamber/ heat exchanger

T4 (output turbine) must be bigger than T2 (output compressor)

nu_regenerator = (Tx - T2)/(T4-T2)

practical values at around 60% to 80%

staging compressors (compressor, intercooler, compressor)

because if we increase pressure and T in compressor, v increases too. if we take out excess heat, it becomes more efficient

we have to heat up the gas afterwards, but the compressor needs less energy from the turbine, so more is available

the higher the max T of the cycle, the better the thermal efficiency possible (Carnot)

stage the turbines (combustion, turbine, combustion to reheat, turbine)

but now there is more heat needed but you take out more work

compression and expansion (Turbine) are not isentropical anymore.

in compression, the T becomes slightly higher than with ideal; entropy increases slightly (/)

in expansion, T is slightly higher at the end too, entropy increases too (\)

so the real compressor heats up more, that would be an advantage.

losses (dissipation = entropy generation in compressor and turbine)

pressure loss in the combustor (dissipation)

turbine exhaust pressure higher than atmospheric

heat losses

turbine: irreversible process, generate entropy, so then pressure > atmospheric, this is necessary so that air actually goes out of the turbine + often you have components afterwards that have additional pressure losses (e.g. silencer, filter)

input

some losses due to unburned fuel and thermal losses in combustion chamber

major losses due to conversion of thermal energy into mechanical energy (remember thermal efficiency in Carnot)

some mechanical losses in compressors and turbines and auxiliary apparatuses

scope is rising the pressure, everything else is a loss

(velocity decreases, pressure rises, temperature increases)