Thermodynamics

Question 1

Closed system

Answer 1

• Non flow process
• sum of heat-sum of work=0
• change in heat-change in work=change in internal energy
• no mass transfer across boundary

Question 2

Reversible process

Answer 2

When fluid changes it state it undergoes a process (no change in entropy)

If fluid is in thermodynamic equilibrium during a process it is reversible process.

Question 3

Non reversible

Answer 3

If fluid is not in thermal equilibrium during a process then defined as a non reversible process

Question 4

Thermodynamic equilibrium

Answer 4

Need to be in thermal, mechanical and chemical equilibrium.

When no further changes can take place in system if it were isolated from its surroundings

Question 5

Thermodynamics

Answer 5

• study of heat and work
• state of system determined by properties at that time
• work (movement of a force) and heat (temperature difference) are not properties and are dependent on the states.

Question 6

1st law of thermodynamics

Answer 6

•Total amount of energy stays the same (conservation of energy)

Engineering POV-net work can never be more than heat supplied

Question 7

2nd law

Answer 7

Heat cannot flow from cold to hot only vice versa

Engineering POV-net work must always be less than heat supplied

Heat cannot be completely converted into work (as entropy) lost to surroundings

(Special case it will only flow from cold to hot if you put work into the system like an air conditioner)

Question 8

3rd law

Answer 8

There is an absolute minimum temp absolute 0

Question 9

Zeroth law

Answer 9

Thermal equilibrium

If two systems and in equilibrium with a third system then the two then those two in thermal equilibrium with each other

Question 10

Enthalpy

Answer 10

Sum of the internal energy and the product of volume and pressure in a system. (h=u+pv)

Heat content of a system at constant pressure.

Heat content in the system

Even if something is adiabatic that means no transfer of heat.
DQ-DW=DU (DQ=0)
However that does not mean h=0 as there is still heat in the system

Question 11

entropy

Answer 11

Measure of the amount of energy unavailable to do work. Measure of uncertainty/disorder

If you increase temp you increase entropy as more energy put into the system excited the molecules and the amount of random activity. As a gas expands in a system entropy increases.

In irreversible system entropy always goes up

Question 12

Constant pressure process

Answer 12

•change in volume results to work being done by the system (W=pdV)
(Q-pdV=dU)
(dh=du+pdv)

•Q=h2-h1 (only in reversible process)

Question 13

Constant volume process

Answer 13

•work=0 as no change in volume (dQ=du)

h=u+pv

Question 14

Adiabatic process (no heat exchange)

Answer 14

•Q(heat)=0 hence -W=u2-u1

Look at isentropic section for additional equations

Question 15

Polytropic process (changing p and v)

Answer 15

•changing pressure and volume

P1(V1)^n=P2(V2)^n
P1V1/T1=P2V2/T2
W=(P1V1-P2V2)/(n-1)
W=(mR(T2-T1))/1-n   (R=287)
Q=(u2-u1)+(P1V1-P2V2)/(n-1)

dh=cp dt
du=cv dt

du=cdT like Q=mcT but unit in j/kg

PV=mRT

Question 16

Isentropic process

Answer 16

When a process is reversible and adiabatic.

P1(V1)^γ=P2(V2)^γ
P1V1/T1=P2V2/T2
W=(P1V1-P2V2)/(γ-1)
Q=(u2-u1)+(P1V1-P2V2)/(γ-1)
dh=cp dt 
du=cv dt
du=cdT like Q=mcT but unit in j/kg
PV=mRT (don’t get R mistaken with R0)
γ=Cp/Cv      R=Cp-Cv        Cv=R/(1-γ)
W=mR(T1-T2)/(γ-1)

Question 17

Isothermal process (constant T)

Answer 17

•constant temperature du=0

ds=dq/T unit j/k (entropy) reversible process

Tds=du+p dv

dh=du+ d(pv)

Tds=dh- v dp

Question 18

Steady flow rate

Answer 18

Q-W=mass flow rate(specific enthalphy+velocity of flow+elevation)

Question 19

Nozzle and diffuser

Answer 19

Velocity changes due to change of area

Question 20

Throttle

Answer 20

Change in pressure at certain peaks

Question 21

Gibbs-dalton law

Answer 21

•the pressure of a mixture of gases is equal to the sums of the pressures of the individual constituents when each occupies a volume equal to that of the mixture at the temperature of the mixture.

Sum of the individual pressure when the constituent occupies the container on its own

Pt=P1+P2+P3+..

Question 22

Mole fraction

Answer 22

Pi=Xi * P total

Pi is the partial pressure
Xi is the mole fraction

Xi=(na/n total)

Pi=na(RT/V total)

Constant temp and volume