ME 20000THERMODYNAMICS 1 SPRING 2020 SECTIONS 1.1–3.15

Riley BartaPerry Teaching Fellow

1

DO YOU KNOW YOUR UNITS?

2

VITAL TO OBTAINING ALL POINTS, CAN HELP CHECK EQUATION USE

Pressure

• 1 bar = 0.1 MPa = 100 kPa = 105 Pa

Temperature:

• T(K) = T(℃) + 273.15

• ΔT K = ΔT(℃)

Force:

• 1 N = 1 kg × 1 m/s2

Energy:

• 1 J = 1 N × m = 1 Pa × 1 m3

Power:

• 1 W = 1 J/s, 1 kW = 103 W

BASIC CONCEPTS

3

What are the types of systems?

What is a property?

What are the types of properties?

What is state?

What is state principle?

ENERGY INTERACTIONS

4

ENERGY INTERACTIONS

(CONT’D)Closed system energy balance (i.e. for piston-cylinder, rigid tanks, etc.)

𝑊 includes boundary work plus any other work interactions (electrical

work, shaft work, spring work, etc.)

Δ𝐸 = Δ𝑈 + Δ𝐾𝐸 + Δ𝑃𝐸 = 𝑄 −𝑊

Sign Conventions:

W > 0 (positive) work done by the system (work output)

W < 0 (negative) work done on the system (work input)

Q > 0 (positive) heat transfer into system (heat input)

Q < 0 (negative) heat transfer out of system (heat output)

PROPERTY DETERMINATION

5

ENERGY INTERACTIONS

(CONT’D)

Look for two independent intensive properties to fix the state of the pure

substance

1.) Start in either saturated temperature or saturated pressure table to

determine the phase

2.) Determine the phase as saturated liquid (SL), saturated liquid-vapor

mixture (SLVM), saturated vapor (SV), compressed liquid (CL), or

superheated vapor (SHV)

3.) Based on Step 2, use appropriate set of tables to fix the state of the

pure substance

Which previously independent, intensive properties become dependent

on/in the dome?

PROPERTY DETERMINATION

6

ENERGY INTERACTIONS

(CONT’D)

Known properties: 𝑇, 𝑣 , or 𝑇, 𝑢 , or 𝑇, ℎ OR 𝑃, 𝑣 , or 𝑃, 𝑢 , or 𝑃, ℎ

IF: 𝑣 > 𝑣𝑔, or 𝑢 > 𝑢𝑔, or ℎ > ℎ𝑔 at the given 𝑇 or 𝑃, then the state is SHV.

We must use an appropriate superheated vapor table.

IF: 𝑣𝑓 < 𝑣 < 𝑣𝑔, or 𝑢𝑓 < 𝑢 < 𝑢𝑔, or ℎ𝑓 < ℎ < ℎ𝑔 at the given 𝑇 or 𝑃, then the

state is SLVM. We must compute quality 𝑥 , and use quality to compute

other properties.

IF: 𝑣 < 𝑣𝑓, or 𝑢 < 𝑢𝑓, or ℎ < ℎ𝑓 at the given 𝑇 or 𝑃, then the state is CL. We

must use an appropriate compressed liquid table, if we can find one.

CL approximation:

𝑥 =𝑣 − 𝑣𝑓

𝑣𝑔 − 𝑣𝑓=

𝑢 − 𝑢𝑓

𝑢𝑔 − 𝑢𝑓=

ℎ − ℎ𝑓

ℎ𝑔 − ℎ𝑓≡

𝑚𝑔

𝑚𝑓 +𝑚𝑔

PHASE CHANGE – T-𝒗 DIAGRAM

7

ENERGY INTERACTIONS

(CONT’D)

Critical

point

𝑣

𝑇 𝑃1

𝑃2 > 𝑃1

𝑇𝑐

𝑃𝑐Sat. Vapor line

Sat. Liquid line

CL

states

SHV

statesSLVM

states

𝑃0 < 𝑃1

Sat. vapor at 𝑃0

Sat. liquid at 𝑃0

Line of constant

pressure

PHASE CHANGE – P-𝒗 DIAGRAM

8

ENERGY INTERACTIONS

(CONT’D)

𝑣

𝑃𝑐𝑇𝑐

Sat. Vapor line

Sat. Liquid line

CL

state

s

SHV

statesSLVM

states𝑃

Critical

point

𝑇1𝑇2 > 𝑇1

𝑇0 < 𝑇1

Line of constant

temp

SPECIFIC HEATS

9

Heat (energy) required to rise the temperature of a unit mass of a

substance by one degree.

Two types of specific heats:

• Specific heat at constant volume 𝑐𝑣

• Specific heat at constant pressure 𝑐𝑝

Specific heat at constant volume (𝒄𝒗)

Specific heat at constant pressure (𝒄𝒑)

𝑐𝑣 =𝜕𝑢

𝜕𝑇𝑣

𝑐𝑝 =𝜕ℎ

𝜕𝑇𝑝

INCOMPRESSIBLE SUBSTANCES

10

Substances where volume changes associated with pressure changes

are negligible.

Can be a valid assumption for liquids and solids, but not gases

𝑐𝑝 = 𝑐𝑣 = 𝑐

If we assume constant specific heats:

𝑢2 − 𝑢1 = 𝑐 𝑇2 − 𝑇1

ℎ2 − ℎ1 = 𝑐 𝑇2 − 𝑇1 + 𝑣 𝑃2 − 𝑃1

IDEAL GAS ASSUMPTION

11

When can we model a substance as an ideal gas?

What is the difference between ideal gas and superheated vapor phase?

What is the ideal gas equation of state?

What single, intensive property can we use to determine other properties

with an ideal gas?

IDEAL GAS ASSUMPTION

12

𝑅: specific gas constant 𝑅 =ത𝑅

𝑀, 𝑀 = molecular weight (kg/mol)

𝑃𝑣 = 𝑅𝑇

USE ABOLUSTE PRESSURE

AND TEMPERATURE – Pascals

(Pa) and Kelvin (K)

• ത𝑅 is the universal gas contant, ത𝑅 = 8.314kJ

kmol−K

𝑃𝑉 = 𝑚𝑅𝑇

𝑐𝑝 = 𝑐𝑣 + 𝑅 𝑐𝑣 =𝑅

𝑘 − 1 𝑐𝑝 =𝑘𝑅

𝑘 − 1

• Specific heat ratio, 𝑘 = 𝑐𝑝/𝑐𝑣 Always greater than 1!

• 𝑐𝑝 and 𝑐𝑣 can be expressed in terms of 𝑘 and 𝑅 for ideal

gases

IDEAL GAS ASSUMPTION

13

IDEAL GAS MODEL (CONT’D)• Use tabulated data: most accurate, ideal gas tables

• Use constant specific heats: note that 𝑐𝑣 ≠ 𝑐𝑝 for ideal gases

Δ𝑢 = 𝑢2(𝑇2) − 𝑢1(𝑇1)

Δℎ = ℎ2(𝑇2) − ℎ1(𝑇1)

values at 𝑻𝟏 and 𝑻𝟐 from ideal

gas tables

Δ𝑢 ≈ 𝑐𝑣(𝑇2 − 𝑇1)

Δℎ ≈ 𝑐𝑝(𝑇2 − 𝑇1)

Only when 𝒄𝒗 and 𝒄𝒑 are

constant or almost constant

for the temperature range

Computing changes in 𝑢 and ℎ for ideal gases

COMMON ELEMENTS

14

FROM ALL SPECIAL PROBLEMS AND SOLVED EXAMPLES THUS FAR

Assumptions:

1. Closed system

2. Quasi-equilibrium process

3. Frictionless piston

4. ∆KE ≈ ∆PE ≈ 0

5. …

Basic equations:

1.

2. → special case: cycles

3.

4.

5. …

W pdV E KE PE U Q W

,V m U mu

,f g f f g fx u u x u u

BOUNDARY WORK

15

MATCH CORRECT TERMS

Process Path Work equation

Isobaric

Isochoric

Product of pressure and

volume remains constant

Polytropic

W p V

0W

2 2 1 1

1

p V p VW

n

21 1

1

lnV

W p VV

𝑃 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑃𝑉𝑛 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑃𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

WORK

16

• Remember to include all kinds of work done in the energy balance:

•

• Shaft does work on system, system does boundary work

100 kPaatmp

boundary shaftW W W

2

1

V

boundary

V

W pdV

2

1

shaftW d

EXAMPLE 1

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DRAW P-V DIAGRAM 100 kPaatmp 100 kPaatmp

1 120 kPap 2 ?p

2

4000 N

0.2 m

weightpiston

A

p

V

atmp

1 2p p

mg

A

1V2V

1 2p p

Q

EXAMPLE 2

18

DRAW P-V DIAGRAM 100 kPaatmp

3 200 kPap

2

4000 N

0.2 m

weightpiston

A

Which additional

assumption is

needed here?

Hint: related to

gas mass

atmp

mg

A3p

R

p

V

1 ?p

2 3V V1V

1 2p p

3p

Q

EXAMPLE 3

19

DRAW P-V DIAGRAM 100 kPaatmp

2 160 kPap 2

4000 N

0.2 m

weightpiston

A

1 80 kPap

Which additional

assumption is

needed here?

Hint: related to

gas mass

2 0k x x

A

mg

A2p

atmp

p

V1V 2V

1p

2p

Q

EXAMPLE 4

20

DRAW P-V DIAGRAM 100 kPaatmp

1 100 kPap 2 ?p

rigid tank

p

V

1p

1 2V V

2p

Would pressure

measured at the

top face and

bottom face of

the tank be the

same?

Q

PROBLEM 1A

21

• 100 kg of water at 10 bar and 179.9 ℃ fills up a rigid tank. What is

the volume of the tank?

a) 0.1127 m3

b) 194.4 m3

c) Insufficient information

PROBLEM 1B

22

• For a simple, compressible substance, pressure and temperature are

known. How many additional properties are required to fix the state

of the substance?

a) 0

b) 1

c) 2

d) Insufficient information

PROBLEM 1C

23

• Which of the following most closely approaches the ideal gas

condition for a substance?

a) P >>> PCritical ; T >>> Tcritical

b) P >>> PCritical ; T <<< Tcritical

c) P <<< PCritical ; T >>> Tcritical

d) P <<< PCritical ; T >>> TCritical

PROBLEM 1D

24

• Which assumptions are required to apply W = PΔV? For a process?

Circle all that apply.

a) Ideal Gas

b) Incompressible

c) Constant Pressure

d) Quasi-Equilibrium

e) Neglect Changes in PE and KE

PROBLEM 1E

25

• Enthalpy of water at a high pressure and treated as a superheated

vapor depends only on temperature: (T/F)

• Enthalpy of water vapor at low pressure and treated as and ideal gas

depends only on temperature: (T/F)

• For a given temperature, enthalpy of a compressed liquid treated as

an incompressible is higher than enthalpy of the saturated liquid:

(T/F)

• For an incompressible substance, cp is always higher than cv: (T/F)

• For any substance and phase, cp – cv = R: (T/F)

PROBLEM 1F

26

• For any general substance undergoing a constant temperature

process in a constant mass system, is PV = Constant always true?

Justify using property diagram and/or equation.

PROBLEM 2

27

• Water with an absolute pressure of 10 bar and a quality of 0.25

(State 1) is expanded in a closed piston-cylinder device along a path

𝑃𝑣 = Constant until the absolute pressure drops to 1 bar (State 2).

– Find the quality at State 2 [%]

– Calculate the specific work during the process [kJ/kg]

– Determine the heat transfer per unit mass of water during the

process [kJ/kg]

– Show the process on a P-v diagram relative to the vapor dome

and isotherms for the two states. Label states, axes, and direction

of the process

PROBLEM 2

28

• Given

• Find

• Assumptions

• Basic Equations

PROBLEM 2

29

• Solution:

PROBLEM 2

30

PROBLEM 2

31

PROBLEM 3

32

• Two rigid tanks are connected by a pipe and a valve. The left tank with a volume of 0.5 m3 initially contains air at an absolute pressure of 2 bar and an absolute temperature of 400 K. The right tank has an initial volume of 0.25 m3 and initially contains air at an absolute pressure of 3 bar and an absolute temperature of 350 K. The valve in the pipe connecting the two tanks is opened and the air is allowed to mix until the final temperature and pressure are the same in each tank. Heat transfer of 90 kJ occurs from the air to the surroundings during the process. Mair = 28.97 kg/kmol, don’t interpolate

• Given:

• Find:

– T2 [K]

– P2 [bar]

PROBLEM 3

33

• Assumptions:

• Basic Equations:

• Solution:

PROBLEM 3

34

PROBLEM 3

35

PROBLEM 3

36

PROBLEM 3

37

38

39

40

VL = 0.5 m3

PL = 2 bar

TL = 400 K

VR = 0.25 m3

PR = 3 bar

TR = 350 K