Thermodynamics 1

BMM 2513 THERMODYNAMICS 1

Prof Dr Hj Shahrani Bin Hj AnuarDr. Ftwi Yohaness HagosDr. Thamir Khalil Ibrahim

1

Chapter 4. First law in closed system

This course focuses on the application of thermodynamics fundamentals in various engineering system including properties

of pure substance, perpetual motion machine, first law, second law and

entropy.

COURSE SYNOPSIS

BMM2513 THERMODYNAMICS 1

2

Chapter-4Topic 7. First law in closed system


1. Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors.

2. Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems.

3. Develop the general energy balance applied to closed systems.4. Define the specific heat at constant volume and the specific heat at

constant pressure.5. Relate the specific heats to the calculation of the changes in internal

energy and enthalpy of ideal gases.6. Describe incompressible substances and determine the changes in

their internal energy and enthalpy.7. Solve energy balance problems for closed (fixed mass) systems that

involve heat and work interactions for general pure substances, ideal gases, and incompressible substances.

OBJECTIVES

Topic 7. First law in closed systemChapter-4

LECTURE SCHEDULEWEEK TOPIC

1 Introduction and Basic Concept

2 System, boundary, work, heat

3 Energy & general energy analysis

4 Energy and environment impact

5 Pure substances, gas equations

6 Property table and chart

7 First law in closed system

WEEK TOPIC

8 First law steady flow system

9 Second law, heat engine, PMM

10 Second law, heat pump

11 Second law, Carnot cycle principles

12 Entropy, entropy relationships

13 Entropy, Isentropic efficiency

14 Presentation


4

Chapter-4Topic 7. First law in closed system

i. Definitions of Thermodynamicsii. Basic Applications of Thermodynamicsiii. System, Boundary and Surroundingiv. Closed & Open Systemv. Properties, Intensive and Extensive Propertiesvi. Equilibrium And Quasi-static Equilibriumvii. State, Path, Process and Cycleviii. Simple Compressible Substanceix. Pressure, atmospheric, gauge, absolutex. Temperature and Zeroth Law Of Thermodynamics


Thermodynamics concept

RecaptulateChapter-2: Energy, energy transfer & general energy analysis

v. Properties, Intensive and Extensive Properties Properties are any measurable characteristics of a system. eg.

Pressure p, temperature T, volume V, mass m and density ρ. Extensive properties are the mass-dependent properties of a

system. i.e. the properties that will vary proportionally with mass of the system. e.g. volume V

Intensive properties are the properties that are not dependent on mass. e.g. temperature T, density ρ.

NB: If any Extensive Property is divided by the mass we would also obtain an intensive property.


Thermodynamics concept

Chapter-2: Energy, energy transfer & general energy analysisRecaptulate

FIRST LAW OF THERMODYNAMICSEnergy can be neither created nor destroyed during a process but can change forms• Principle of energy conservation• Sound basis for studying the relationships among the various

forms of energy and energy interactions• All adiabatic processes between two specified states of a closed

system, the net work done is the same regardless of the nature of the closed system and the details of the process.


Essentially

W

Q

ΔUSingle ProcessQ – W = ΔU

W1

W2

W3

Q1Q2

Cyclic ProcessΣQ – ΣW = 0

ΣQ = ΣWΔU = 0



4.1 Moving Boundary WorkMoving boundary work (P dV work)

The expansion and compression work in a piston-cylinder device.

Work associated with a moving

boundary is called boundary work.

Quasi-equilibrium process A process during which the system

remains nearly in equilibrium at all times.

Wb is positive for expansionWb is negative for compression

Gas does a differential amount of work Wb as it forces the piston to move by a differential

amount ds.



4.1 Moving Boundary Work

The boundary work done during a process depends

on the path followed as well as the end states.

The area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system.

Cyclic Process



4.1 Moving Boundary WorkPolytropic, Isothermal, and Isobaric processes

where : C = constantsn = polytropic exponent)

Polytropic process

Polytropic process schematic and P-V diagram

QuestionWhat is the boundary work for

constant-volume process?

n = ∞

n = 0

n > 1

n = 1

P

V


Polytropic Process According Relationship PV n = C

¿𝑃2𝑉 2

𝑛𝑉 21−𝑛−𝑃1𝑉 1

𝑛𝑉 11−𝑛

1−𝑛

𝑊 12=𝑃2𝑉 2−𝑃1𝑉 1

1−𝑛 =𝑚𝑅 (𝑇 ¿¿2−𝑇 1)

1−𝑛 ;𝑛≠1¿

𝑊 12=∫1

2

𝑃𝑑𝑉=¿𝐶∫1

2 𝑑𝑉𝑉 𝑛 =𝑃1𝑉 1

𝑛∫1

2 𝑑𝑉𝑉𝑛 =

𝑃1𝑉 1𝑛(𝑉 2

1−𝑛−𝑉 11−𝑛)

1−𝑛 ¿

Boundary Displacement Work from state 1 to state 2 is given,




( 𝑃𝑉𝑇 )1=( 𝑃𝑉𝑇 )

2

𝑃1

𝑃2=𝑉 2

𝑉 1

𝑇 2=𝑇 1

isothermal

𝑊 12=𝑷𝟏𝑽𝟏 𝒍𝒏𝑽 𝟐

𝑽 𝟏=𝒎𝑹𝑻 𝟏 𝒍𝒏

𝑽 𝟐

𝑽 𝟏KJ 𝑊 12=𝑷𝟏𝑽𝟏 𝒍𝒏

𝑷𝟏

𝑷𝟐=𝒎𝑹𝑻 𝟏 𝒍𝒏

𝑷𝟏

𝑷𝟐K J

𝑊 12=∫1

2

𝑃𝑑𝑉=¿∫1

2 𝐶𝑉 𝑑𝑉=𝐶 𝑙𝑛 𝑉 2

𝑉 1=𝑷𝟏𝑽 𝟏 𝒍𝒏

𝑽𝟐

𝑽𝟏KJ ¿

Isothermal Process According Relationship PV = C


For ideal gas




Isobaric Process According Relationship P = C (n = 0)

𝑾 𝟏𝟐=∫1

2

𝑃𝑑𝑉=¿𝑃∫1

2

𝑑𝑉 ¿





𝑾 𝟏𝟐=𝑷 (𝑽𝟐−𝑽 𝟏 )KJ

Isochoric Process According Relationship V = C (n = ∞)


𝑾 𝟏𝟐=∫1

2

𝑃𝑑𝑉=¿𝟎KJ ¿

Constant volume or isochoric





4.2 Energy Balance For Closed Systems

Energy balance for any system undergoing any process

Energy balance in the rate form

The total quantities are related to the quantities per unit time is

Energy balance per unit mass basis

Energy balance in differential form

Energy balance for a cycle




Net Heat Input

Net Work Input

First-law relation for closed systems when sign convention is used

The first law cannot be proven mathematically, but no process in nature is known to have violated the first law, and this should be taken as sufficient proof.

Energy balance sign conventionHeat Input & Work Output +veHeat Output & Work Input -ve




NB: The first law cannot be proven mathematically, but no process in nature is known to have violated the first law, and this should be taken as sufficient proof.


Net Work and Net Heat in Cyclic ProcessesWhen any closed system is taken through a cycle, the net work

delivered to the surroundings is equal to the net heat taken from the surroundings .i.e

Process1−2 :𝑄12−𝑊 12=𝑈 2−𝑈 1

Process2−3 :𝑄23−𝑊 23=𝑈 3−𝑈 2

Process3−1 :𝑄31−𝑊 31=𝑈 1−𝑈 3

Adding for net value

Consider an arbitrary cycle an shown

Σ𝑄=Σ𝑊 ΣΔ𝑈=0

4.2 Energy Balance For Closed SystemsConstant-pressure Expansion/Compression Process

General analysis for a closed system undergoing a quasi-equilibrium constant-pressure process. Q is to the system and W is from the system.

H2 H1

where

𝑄−𝑊 h𝑜𝑡 𝑒𝑟=𝐻 2−𝐻 1 𝑸=𝑯𝟐−𝑯𝟏NB: Microscopic

Wother ≈ 0

BMM2513 THERMODYNAMICS 1Topic 7. First law in closed system

Chapter-4


4.2 Energy Balance For Closed SystemsConstant-pressure Expansion/Compression Process

HWU b

For a constant-pressure expansion or compression process

An example of constant-pressure process






4.3 Specific HeatsSpecific heat at constant volume, cv

The energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant

Specific heat at constant pressure, cp

The energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant

Constant-volume and constant-pressure specific heats cv and cp (helium gas).



4.3 Specific Heats

The equations in the figure are valid for any substance undergoing any process.

cv and cp are properties. cv is related to the changes in internal

energy and cp to the changes in enthalpy. A common unit for specific heats is kJ/kg·°C

or kJ/kg·K. Are these units identical?

True or False? cp is always greater than cv

Formal definitions of cv and cp.



4.4 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

23

Joule showed using this experimental apparatus that u=u(T)

For ideal gases, u, h, cv, and cp vary with temperature only.

Internal energy and

enthalpy change of an

ideal gas




Ideal-gas constant-pressure specific heats for some gases (Table A–2c for cp equations).

In ideal-gas tables, 0 K is chosen as the reference temperature.

At low pressures, all real gases approach ideal-gas behavior, and therefore their specific heats depend on temperature only.

The specific heats of real gases at low pressures are called ideal-gas specific heats, or zero-pressure specific heats, and are often denoted cp0 and cv0.

u and h data for a number of gases have been tabulated. These tables are obtained by choosing an arbitrary

reference point and performing the integrations by treating state 1 as the reference state.




Internal energy and enthalpy change when specific heat is taken constant at an average value

(kJ/kg)

(kJ/kg)




1. Easiest and accurate way is by using tabulated data of u and h if available

2. From cv or cp relations (Table A-2c) as a function of temperature and the integrate. It is inconvenient for hand calculations but quite desirable for computerized calculations. The results obtained are very accurate.

3. Using average specific heats is very simple and certainly very convenient when property tables are not available. Reasonably accurate if the temperature interval is not very large.

Three ways of calculating u and h

Three ways to calculate u


Relationship between Specific Heats and Gas Constant (Cp Cv R) From definition of enthalpy and substitute gas equation,

Its derivative with respective to temperature T becomes

Substitute definition for specific heats, becomes

Molar basis

Specific heat ratio

Specific ratio varies slightly with temperature 1.667 for monatomic gases He, Ar 1.4 for many diatomic gases e.g. air

at room temperature

cp of ideal gas determined from cv , R.





4.5 Internal Energy, Enthalpy and Specific Heats of Solids/Liquids

Substance whose specific volume (or density) is constantExamples: solids and liquids

Incompressible Substance



4.5 Internal Energy, Enthalpy and Specific Heats of Solids/LiquidsIncompressible Substance

Internal Energy Changes



4.5 Internal Energy, Enthalpy and Specific Heats of Solids/LiquidsIncompressible Substance

Enthalpy Changes

Enthalpy of compressed liquid

Definition

Derivative

Solid vΔP ≈ 0; Tcuh avg

LiquidsConstant-pressure process in heaters (ΔP = 0)

Constant-Temperature in pumps (ΔT = 0)

Tcuh avg

Pvh

More accurate than

Enthalpy Change


1. Moving boundary worki. Wb for an isothermal processii. Wb for a constant-pressure processiii. Wb for a polytropic process

2. Energy balance for closed systemsi. Energy balance for a constant-pressure expansion or

compression process3. Specific heats

i. Constant-pressure specific heat, cpii. Constant-volume specific heat, cv

4. Internal energy, enthalpy, and specific heats of ideal gasesi. Specific heat relations of ideal gases

5. Internal energy, enthalpy, and specific heats of incompressible substances (solids and liquids)


SUMMARY



Example 1: Gas in piston-cylinder device Calculate the work done when a gas expands from 350 kPa and 0.03 m3 to a final volume of 0.2 m3 according to the relationship PV1.5 = C.



Example 2: Gas in piston-cylinder device Calculate the change in specific volume of air when compress isothermally from 150 kPa and 20oC to 1100 kPa.



Example 3: Isothermal expansion of steam in cylinder-piston assemblyA mass of 1 kg of saturated dry steam is expanded isothermally in a cylinder-piston assembly from 40 bar (4 MPa) to 2 bar (0.2 MPa). Sketch the process in a clearly labeled P-V diagram and determine:-a) change in enthalpy in kJ,b) change in internal energy in kJ, andc) work done in kJ if the heat supplied to the steam is 857 kJ.



Example 3: Isothermal expansion of steam in cylinder-piston assembly (cont’d)



Example 4: Isothermal expansion of steam in cylinder-piston assemblyA piston–cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. Determine the final temperature of the steam.


We




P2V2-P1V1

Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy of air per unit mass, using (a) data from the air Table A–17, (b) the functional form of the specific heat (Table A–2c), and (c) the average specific heat value Table A–2b.




(a) Internal energy of air at T1 and T2 from Table A–17

Air cp(T) in Table A–2c expressed in third-degree polynomial

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Thermodynamics 1

Engineering