Thermodynamics

Question 1

SI-Units
Temperature 
Mass 
Length
Time
Velocity
Acceleration (ускорение)
Force
Pressure
Energy
Power

Answer 1

Temperature - Kelvin
Mass - kg
Length - m
Time - s
Velocity - m/s
Acceleration (ускорение) - m/s(2)
Force - kg*m/s(2) - N - Newton
Pressure - kg/(m*s(2)) - Pa - Pascal
Energy - kg*m(2)/s(2) - J - Joule
Power - kg*m(2)/s(3) - W - Watt

Question 2

Definitions:
system
surroundings
boundary

Answer 2

System: whatever we want to study
Surroundings: whatever is around the system but not part of it
Boundary: the interface between the system & the surroundings

Question 3

Laws of Thermodynamics: 1, 2

Answer 3

1. the overall energy of any closed system is
   conserved. No energy can be lost. This is equivalent to the energy balance.
2. The second law of thermodynamics introduces irreversibilities. Examples of such
   irreversibilities are:
   • friction (resulting from flow of matter), e.g.: pipes, turbines, compressors
   • heat transfer with finite temperature difference in boilers and heat exchangers as result of:
   convection, radiation
   • spontaneous chemical reactions
   • chemical reaction (combustion), e.g.: combustion chamber of a boiler, combustion chamber
   of gas turbines
   • mixing: quantities of the same substances at different temperatures different substances at
   the same temperature
   • Electric current flow through a resistance

Question 4

Energy Analysis of Thermodynamic Cycles

Answer 4

Energy balance of a cycle

ΔEcycle = ΔQcycle − Wcycle
Since we return to the orginial state there is
no net change in the cycle energy and thus

Wcycle = ΔQcycle
This applies to every thermodynamic cycle
independent of the processes

Question 5

Energy Analysis of Thermodynamic Cycles 2

Answer 5

we define the thermal efficiency

η = Wcycle/Qin

and if we fill in the above

η = Qin − Qout/Qin = 1 − Qout/Qin

From this we see that always η < 1

Question 6

Clausius Inequality

Answer 6

From Clausius and Kelvin-Planck we learn that we need a hot and a cold reservoir for any
thermodynamic cycles.

Clausius inequality:

ф(δQ/T)b ≤ 0

with δQ heat transfer and T absolute temperature at a given part of the system boundary b

Introducing an arbitary variable (произвольная переменная) S we can reformulate:

ф(δQ/T)b = −S

with:
S=0: no irreversibilites (необратимый процесс процесс, который нельзя провести в противоположном направлении через все те же самые промежуточные состояния. Все реальные процессы необратимы. Переход кинетической энергии макроскопического движения через трение в теплоту, то есть во внутреннюю энергию системы, является необратимым процессом. Законы необратимых процессов могут быть обоснованы с помощью методов электрокинетической теории тепла)
S>0: irreversibilites present
S<0: impossible

Let us call this variable „entropy“. (физическая величина, используемая для описания термодинамической системы, одна из основных термодинамических величин. Энтропия является функцией состояния и широко используется в термодинамике, в том числе технической (анализ работы тепловых машин и холодильных установок) и химической (расчёт равновесий химических реакций).