Processes of ideal gas, Constant Volume Process, Constant Pressure Process, etc. PDF

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Thermodynamics lesson 5
Processes of ideal gasses. In this lesson different processes of ideal gasses are discussed at different thermodynamic conditions....

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Course name: THERMODYNAMICS 1 Course Code: ME 313

Module: Processes of Ideal Gases

Prepared by: Engr. Marco A. Sollano

INTRODUCTION Study of the processes of ideal gas are significant in defining the behavior of a gas undergoing a change in a state whether it is nonflow for a closed system and or steady flow for an open system. Although there is no perfect gas that conforms perfectly to the ideal gas equations, we use the equations because it does a good job in describing the relationship between pressure, volume, amount, and temperature of a gas. In this module, different processes will be discussed enough to define thermodynamic behavior of common gasses at different conditions. There are five processes involved in any ideal gas system undergoing a change in state, namely: 1. Isometric or Isochoric Process (V = c) 2. Isobaric or Isopiestic Process (P = c) 3. Isothermal Process (T = c) 4. Isentropic or Adiabatic Process (S = c) 5. Polytropic Process (PVn = c)

OBJECTIVES: At the end of the module the students will be able to: 1. Identify the different processes of ideal gases. 2. Apply the principles involving the different processes of ideal gas to any thermodynamic system undergoing a change of state. 3. Interpret the different principles and equations governing any process of ideals gases. 4. Solve problems to any process of ideal gas.

SUBJECT MATTER

1. CONSTANT VOLUME PROCESS (V = c) An Isometric or Isochoric process a reversible constant volume process. A constant volume process may be reversible or irreversible. In this process, the working substance is contained in a rigid vessel. The curve for an isometric process is called isomer or isochore.

Constant Volume Heating or Cooling. The constant volume heating process is represented on p-v and T-s diagram as shown in Fig. 1 (a), (b) respectively.

(a.)

(b.) Figure 1: Isometric Process

A. Relation between p and T. 𝑻𝟐 𝒑𝟐 = 𝒑𝟏 𝑻𝟏 B. Nonflow work. 𝟐

𝑾𝒏 = ∫ 𝐩𝐝𝐕 = 𝟎 𝟏

C. Change in internal energy. 𝜟𝑼 = 𝒎𝒄𝐯 (𝑻𝟐 − 𝑻𝟏 )

D. The heat transferred.

𝑸 = 𝒎𝒄𝐯 (𝑻𝟐 − 𝑻𝟏 )

E. The change of enthalpy. 𝜟𝑯 = 𝒎𝒄𝐩(𝑻𝟐 − 𝑻𝟏 ) F. The change of entropy.

𝑻𝟐 𝑻𝟏 G. Reversible steady flow constant volume 𝜟𝑺 = 𝒎𝒄𝐯 𝒍𝒏

𝑸 = 𝜟𝑼 + 𝜟𝑲 + 𝜟𝑾𝒇 + 𝑾𝒔 + 𝜟𝑷

𝑾𝒔 = −(𝜟𝑾𝒇 + 𝜟𝑲 + 𝜟𝑷)

𝑾𝒔 = −𝜟𝑾𝒇 = 𝑽(𝒑𝟏 − 𝒑𝟐)

(𝜟𝑷 = 𝟎, 𝜟𝑲 = 𝟎)

𝟐

− ∫ 𝑽𝒅𝑷 = 𝑾𝒔 + 𝜟𝑲 𝟏

−𝑽(𝒑𝟐 − 𝒑𝟏 ) = 𝑾𝒔 + 𝜟𝑲

𝑽(𝒑𝟏 − 𝒑𝟐) = 𝑾𝒔 + 𝜟𝑲

𝑽(𝒑𝟏 − 𝒑𝟐 ) = 𝑾𝒔 H. Irreversible nonflow constant volume process.

(𝜟𝑲 = 𝟎)

𝑸 = 𝜟𝑼 + 𝑾𝒏 For reversible nonflow, WNF = 0 For irreversible nonflow, WNF is not equal 0 WNF = nonflow work WSF = steady flow work

EXAMPLE 1. Ten cu. ft. of air at 300 psia and 400oF is cooled to 140 oF at a constant volume. What are (a) the final pressure, (b) the work done, (c) the change of the internal energy, (d) the heat transferred, (e) the change of enthalpy, and (f) the change of entropy?

Given;

V = 10ft3 P1 = 300 psi absolute T1 = 400 + 460 = 860oR T2 = 140 + 460 = 600oR

Solution

a.) Finding for the final pressure 𝑇2

𝑝2 = 𝑝 1 𝑇1 𝑝2 =

𝑝1 𝑇2 (300)(600) = 𝟐𝟎𝟗 𝒑𝒔𝒊𝒂 = 860 𝑇1

b.) Work WNF =0 c.) Solving for the change in internal energy 𝜟𝑼 = 𝒎𝒄𝐯 (𝑻𝟐 − 𝑻𝟏 )

; cv of air = 0.1714 Btu/lbm-oR

Solving first for the mass of air, 𝑚=

𝑝1 𝑉1 𝑅𝑇1

=

(300)(144)(10) (860)(53.34)

= 𝟗. 𝟒𝟏𝟕𝒍𝒃𝒎

; R of air = 53.34 ft-lb/lbm-oR

Then for the internal energy, 𝛥𝑈 = 𝑚𝑐v(𝑇2 − 𝑇1) = (9.417) (0.1714) (600-860) = -420 Btu d.) Solving for the heat transferred 𝑄 = 𝑚𝑐v (𝑇2 − 𝑇1)

; Since Q = 𝛥𝑈 for constant volume process

𝑸 = 𝜟𝑼 = −𝟒𝟐𝟎 𝑩𝒕𝒖 e.) Solving for the change of enthalpy 𝛥𝐻 = 𝑚𝑐p (𝑇2 − 𝑇1 ) = (9.417) (0.24) (600-860) = -588 Btu

; cp of air = 0.24 Btu/lbm-oR

f.) Solving for the change of entropy 𝛥𝑆 = 𝑚𝑐v 𝑙𝑛

𝑇2 𝑇1

= (9.417) (0.1714) 𝑙𝑛

600

860

= -0.581 Btu/oR

Note: For cooling process the value of heat is always negative.

2. CONSTANT PRESSURE PROCESS (P = c) An isobaric or isopiestic process is an internally reversible process of a substance during which the pressure remains constant. In this process, the boundary of the system is inflexible as in a V = c process. The curve for an isobaric process is called isobar.

(a)

(b) Figure 2: Isobaric process

a. Relation between V and T. 𝑽𝟐 𝑻𝟐 = 𝑽𝟏 𝑻𝟏 b. Nonflow work. 𝟐

𝑾𝒏𝒇 = ∫ 𝐩𝐝𝐕 = 𝐩(𝑽𝟐 − 𝑽𝟏) 𝟏

c. Change in internal energy.

(c)

𝜟𝑼 = 𝒎𝒄𝐯 (𝑻𝟐 − 𝑻𝟏 )

d. The heat transferred.

𝑸𝑝 = 𝒎𝒄𝐩 (𝑻𝟐 − 𝑻𝟏 )

e. The change of enthalpy. 𝜟𝑯 = 𝒎𝒄𝐩(𝑻𝟐 − 𝑻𝟏 ) f. The change of entropy. 𝜟𝑺 = 𝒎𝒄𝐩 𝒍𝒏 g. Steady flow constant pressure

𝑻𝟐 𝑻𝟏

𝑸 = 𝜟𝑷 + 𝜟𝑲 + 𝜟𝑾𝒔𝒇 + 𝜟𝑯

𝑾𝒔𝒇 = −(𝜟𝑲 + 𝜟𝑷) 𝑾𝒔𝒇 = −𝜟𝑲

(𝜟𝑷 = 𝟎, ) 𝟐

− ∫ 𝑽𝒅𝑷 = 𝑾𝒔𝒇 + 𝜟𝑲 𝟏

𝟎 = 𝑾𝒔𝒇 + 𝜟𝑲

𝑾𝒔𝒇 = −𝜟𝑲

EXAMPLE 2. A perfect gas has a value of R = 319.2 J/kgm -oK and the specific heat ratio k = 1.26. If 120kJ are added to 2.27kgm of this gas at constant pressure when the initial temperature is 32.2 oC, find (a) T2, (b) change in enthalpy, (c) change in internal energy, (d) the work for a nonflow process. Given;

k = 1.26 m = 2.27kgm R = 319.2 J/kgm-K Qp = 120kJ T1 = 32.2 + 273 = 305.2K

Solution

a.) Solving for the final temperature 𝑘𝑅

𝑐𝑝 =

1.26(0.3192)

=

𝑘−1

1.26−1

= 𝟏. 𝟓𝟒𝟔𝟗

𝒌𝑱

𝒌𝒈𝒎−𝑲

Since the heat added are given, therefore Qp = mcp (T2 − T1 )

120 = 2.27 (1.5469) (T2 − T1 ) 𝐓𝟐 = 𝟑𝟑𝟗. 𝟒 𝑲

b.) Solving for the change in enthalpy Since for constant pressure process, 𝛥𝐻 = 𝑄𝑝 Therefore, 𝜟𝑯 = 𝒎𝒄𝐩(𝑻𝟐 − 𝑻𝟏) = 𝑸𝒑 = 𝟏𝟐𝟎 𝒌𝑱 c.) Solving for the change in internal energy

𝑘=

𝐶𝑝

𝐶𝑣

𝐶𝑝

Cv = 𝑘 =

1.5469 1.26

𝒌𝑱

= 𝟏. 𝟐𝟐𝟕𝟕 𝒌𝒈𝒎−𝑲

𝛥𝑈 = 𝑚𝑐v(𝑇2 − 𝑇1) = 2.27 ( 1.2277)( 339.4 − 305.2) = 𝟗𝟓. 𝟑 𝒌𝑱

d.) Solving for the work of the nonflow process 2

𝑾𝒏𝒇 = ∫ pdV = p(𝑉2 − 𝑉1) 1

= 𝑝(𝑉2 − 𝑉1 ) = (mRT2/p2 -mRT1/p1) = m R (T2 – T1) = 2.27 (0.3192) (339.4 – 305.2) = 24.78 kJ Note; R of air for Metric System = 0.3192 kJ/kg - K 3. ISOTHERMAL PROCESS or CONSTANT TEMPERATURE PROCESS ( T = C) An isothermal process is an internally reversible constant temperature process of a substance. The curve for an isothermal process is called isotherm. Isothermal processes can occur in any kind of system that has some means of regulating the temperature, including highly structured machines, and even living cells. Some parts of the cycles of some heat engines are carried out isothermally (for example, in the Carnot cycle). In the thermodynamic analysis of chemical reactions, it is usual to first analyze what happens under isothermal conditions and then consider the effect of temperature. Phase changes, such as melting or evaporation, are also isothermal processes when, as is usually the case, they occur at constant pressure. Isothermal processes are often used and a starting point in analyzing more complex, non-isothermal processes. Isothermal processes are of special interest for ideal gases. This is a consequence of Joule's second law which states that the internal energy of a fixed amount of an ideal gas depends only on its temperature.

Figure 3: Isothermal process

a. Relation between p and V and. P1 V1 = P2 V2

b. Nonflow work. 2

2CdV

Wnf = ∫1 pdV = ∫1

V

= C[ln(V2) -ln(V1)]

Wnf = C ln (V2/V1) = P1V1 ln (V2/V1)

c. Change in internal energy. 𝜟𝑼 = 𝟎 d. The heat transferred. 𝑸 = 𝜟𝑼 + 𝑾𝒏𝒇 = P1V1 ln (V2/V1) = mRT1 ln(P1/P2) e. The change of enthalpy. 𝜟𝑯 = 𝟎 f. The change of entropy.

𝐐 = 𝐦𝐑 𝐥𝐧(𝐏𝟏/𝐏𝟐) 𝐓 g. Steady flow isothermal 𝜟𝑺 =

𝑄 = 𝛥𝑃 + 𝛥𝐾 + 𝛥𝑊𝑠𝑓 + 𝛥𝐻

Wsf = Q − (ΔK + ΔP) 𝑾𝒔𝒇 = 𝑸

(𝜟𝑷 = 𝟎, 𝜟𝑲 = 𝟎) 𝟐

− ∫ 𝑽𝒅𝑷 = 𝑾𝒔𝒇 + 𝜟𝑲 𝟏

From pV = C, pdV + Vdp = 0, dp = -pdV/V Cross multiply then integrate both sides becomes, 2 pdV ) = ∫ pdV − ∫ VdP = − ∫ V (− V 1 1 1 2

2

P1V1 ln (V2/V1) = WSF + 𝛥𝐾 WNF = WSF (𝛥𝐾 = 0)

EXAMPLE 3. During an isothermal process at 88oF, the pressure on 8 lbm of air drops from 80 psia to 5 2 psig. For an internally reversible process, determine (a) the ∫1 pdV and the work of a nonflow 2

process, (b) the − ∫1 Vdp and the work of a steady flow process during which 𝛥𝐾 = 0, (c) Q, (d) ΔU and ΔH, and (e) 𝛥𝑆. Given; m = 8lbm P1 = 80 psia P2 = 5 + 14.7 = 19.7 psia T = 88 + 460 = 548 oR

Solution

2

a.) Since WNF = WSF = ∫1 pdV = P1 V1 ln (V2/V1) = mRT ln(P1/P2) = [(8)(53.34)(548)/778][ln(80/19.7)] = 421.2 Btu 𝟐

2

WNF = WSF = ∫𝟏 𝐩𝐝𝐕 = 421.2 Btu

b.) − ∫1 VdP = P1 V1 ln (V2/V1) = 421.2 Btu c.) Solving for Q Since, Q = Wnf Therefore,

Q = Wnf = 421.2 Btu

d.) 𝛥𝑈 = 0 , 𝛥𝐻 = 0 e.)

𝛥𝑆 =

Q

421.2 = 548 = 𝟎. 𝟕𝟔𝟖𝟔 𝐁𝐭𝐮/𝐑 T

4. ISENTROPIC PROCESS (S=C) An isentropic process is an idealized thermodynamic process that is both adiabatic and reversible. Adiabatic simply means no heat gain nor heat loss. It is a process in which the entropy of the fluid remains constant. This will be true if the process the system goes through is reversible and adiabatic. The work transfers of the system are frictionless, and there is no net transfer of heat or matter. Such an idealized process is useful in engineering as a model of and basis of comparison for real processes. This is idealized as reversible processes do not occur in reality; thinking of a process as both adiabatic and reversible would show that the initial and final entropies are the same, thus, the reason it is called isentropic (entropy does not change). These processes are named based on the effect they would have on the system (ex. isovolumetric: constant volume, isenthalpic: constant enthalpy). Even though in reality it is not necessarily possible to carry out an isentropic process, some may be approximated as such.

Figure 4: Graph of an isentropic process

a.) Relationship among p, V, and T. i.

Relation between p and V P1 V1k = P2 V2k = C

ii.

Relation between T and V From P1 V1k = P2 V2k = C and P1V1/T1 = P2V2/T2, we have T2/T1 = [V1/V2]k-1

iii.

Relation between T and p T2/T1 = [P2/P1](k-1)/k

b.) Nonflow work From pVk = C, p = CV-k 2

2

WNF = ∫1 pdV = ∫1 𝐶V-kdV = C ʃV-kdV Integrating and simplifying, 𝑷𝟐𝑽𝟐 − 𝑷𝟏𝑽𝟏 𝒎𝑹(𝑻𝟐 − 𝑻𝟏) WNF = = 𝟏−𝒌 𝟏−𝒌 c.) The change of internal energy ΔU = mcv (T2 – T1) d.) The heat transferred Q=0 e.) The change of enthalpy ΔH = mcp (T2 – T1) f.) The change of entropy ΔS = 0 g.) Steady flow isentropic Q = ΔP + ΔK + ΔH + Ws Ws = −ΔP − ΔK − ΔH Ws = −ΔH

ΔP = 0, ΔK = 0

2

− ∫ Vdp = 𝑊𝑠 + 𝛥𝐾 1

Let: 𝟏

𝑪 = 𝐩𝒌 𝐕,

𝟐

− ∫ 𝐕𝐝𝐩 = ∫ 𝑪𝒑𝟏/𝒌 𝒅𝒑

𝟏

𝑽 = 𝑪𝐩𝒌

𝟏

𝟐

− ∫ 𝐕𝐝𝐩 = 𝟏

𝟐 𝒌(𝒑𝟐𝑽𝟐 − 𝒑𝟏𝑽𝟏) = 𝒌 ∫ 𝒑𝒅𝑽 𝟏−𝒌 𝟏

Therefore, 𝟐

Wsf = k∫𝟏 𝒑𝒅𝑽 = kWnf EXAMPLE 4. An adiabatic expansion of air occurs through a nozzle from 828 kPa and 71oC to 138 kPa. The initial kinetic energy is negligible. For an isentropic expansion, compute the specific volume, temperature and speed at the exit section. Given: P1 = 828 kPa T1 = 71 + 273 = 344 K P2 = 138 kPa

Solution Solving for the exit temperature, T2/T1 = [P2/P1](k-1)/k T2 = T1 [P2/P1](k-1)/k = 344[138/828](1.4-1)/1.4 T2 = 206 K – 273 T2 = – 67oC Solving for the initial specific volume, V1 = RT1/P1 V1 = (0.28708)(344)/828 = 0.1193m3/kgm Solving for the specific volume at exit, V2 = V1[P1/P2]1/k V2 = (0.1193) [828/138]1/1.4 = 0.429m3/kgm Solving for the speed at the exit section, Δh = cp (T2 – T1) = 1.0062(206 – 344)

; cp of air = 1.0062 kJ/kg-K

Δh = -138.9 kJ/kgm

Q = ΔPE + ΔKE + Δh + WSF ΔKE = -Δh = 138,900 J/kgm ΔKE = KE2 – KE1 = v22/2k v22 = 2k(ΔKE) = 2 (1kgm-m/Ns2)( 138,900 N-m/kgm) = 277,800 m2/s2 v2 = 527.1 m/s

5. POLYTROPIC PROCESS (pV n = c) The term "polytropic" was originally coined to describe any reversible process on any open or closed system of gas or vapor which involves both heat and work transfer, such that a specified

combination of properties were maintained constant throughout the process. In such a process, the expression relating the properties of the system throughout the process is called the polytropic path. For an ideal gas, this polytropic path simplifies to PVn = C. Polytropic process is an internally reversible process during which PVn = C and P1V1n = P2V2n = PiVin where n is any constant. n is not equal to 1, 0 or k.

Figure 5. Graph of the polytropic process of an ideal gas

a.) Relationship among p, V, and T. i. Relation between p and V P1 V1n = P2 V2n = C ii. Relation between T and V From P1 V1n = P2 V2n = C and P1V1/T1 = P2V2/T2, we have T2/T1 = [V1/V2]n-1

iii. Relation between T and p T2/T1 = [P2/P1](n-1)/n

b.) Nonflow work From pVn = C, p = CV-n 2

2

WNF = ∫1 pdV = ∫1 𝐶V-ndV = C ʃV-ndV Integrating and simplifying, 𝑷𝟐𝑽𝟐 − 𝑷𝟏𝑽𝟏 𝒎𝑹(𝑻𝟐 − 𝑻𝟏) WNF = = 𝟏−𝒏 𝟏−𝒏 c.) The change of internal energy ΔU = mcv (T2 – T1) d.) The heat transferred Q = ΔU + Wnf 𝑚𝑅(𝑇2 −𝑇1 ) 1−𝑛

= mcv (T2 – T1) +

= m [(cv – ncv + R)/1-n] (T2 – T1) = m [(cp – ncv )/1-n] (T2 – T1) = mcv[ (k-n)/(1-n)] (T2 – T1)

; R = cp - cv ; k = cp/cv ; cp = k cv

Q = mcn(T2 – T1) Where: cn = cv[ (k-n)/(1-n)] = the polytropic specific heat e.) The change of enthalpy ΔH = mcp (T2 – T1)

f.) The change of entropy ΔS = Q/T = mcn ʃdT/T ΔS = mcn ln(T2/T1)

g.) Steady flow polytropic Q = ΔP + ΔK + ΔH + Ws Ws = Q − ΔP − ΔK − ΔH

ΔP = 0, ΔK = 0

Ws = Q − ΔH 2

− ∫ Vdp = 𝑊𝑠 + 𝛥𝐾 1

Let: 𝟏

𝑪 = 𝐩𝒏 𝐕,

𝟐

− ∫ 𝐕𝐝𝐩 = ∫ 𝑪𝒑𝟏/𝒏 𝒅𝒑

𝟏

𝑽 = 𝑪𝐩𝒏

𝟏

𝟐

− ∫ 𝐕𝐝𝐩 = 𝟏

𝟐 𝒏(𝒑𝟐𝑽𝟐 − 𝒑𝟏𝑽𝟏) = 𝒏 ∫ 𝒑𝒅𝑽 𝟏−𝒌 𝟏

Therefore, 𝟐

Wsf = n∫𝟏 𝒑𝒅𝑽 = nWnf EXAMPLE 5. Compress 4kg/s of CO2 gas polytropically (pV1.2 = C) from p1 = 103.4 kPa, T1 = 60oC to T2 = 227oC. Assuming ideal gas action, find p2, W, Cn, ΔS (a) as nonflow process, (b) as a steady flow process where ΔPE = 0, ΔKE = 0. Given:

p1 = 103.4 kPa T1 = 60 + 273 = 333K T2 = 227 + 273 = 500K 𝑚󰇗 = 4 kgm/s

Solution a) For nonflow process P2 = P1 [T2/T1]n/(n-1) = (103.4) [500/333]1.2/(1.2 – 1) P2 = 1184.9 kPa 𝑊󰇗 nf = 𝑚󰇗R (T2 – T1)/(1 – n) = (4)(0.18896)(500-333)/(1 – 1.2)

𝑾󰇗nf = - 631.13 kJ/s

Solving for the polytropic specific heat to get Q and ΔS cn = cv[ (k-n)/(1-n)] = 0.6561[(1.288-1.2)/(1 – 1.2)] cn = -0.2887 kJ/kg - K

; cv for CO2 = 0.6561 kJ/kg - K ; cp for CO2 = 0.8452 kJ/kg - K

𝑄󰇗 = mcn (T2 – T1) = 4 (-0.2887) (500 - 333) 𝑸󰇗 = -193.8 kJ/s

ΔS󰇗 = mcn ln(T2/T1) = 4 (-0.2887) ln (500/333)

𝚫𝐒󰇗 = - 0.4694 kJ/s-K

b) For steady flow process P2 = 1184.9 kPa 𝑸󰇗 = -193.8 kJ/s

𝚫𝐒󰇗 = - 0.4694 kJ/s-K 󰇗 = m󰇗cp (T2 – T1) ΔH

󰇗 = 4 ( 0.8452) (500 - 333) ΔH 󰇗 = 𝟓𝟔𝟑. 𝟔 𝒌𝑱/𝒔 𝚫𝐇 󰇗 𝑊󰇗 nf = 𝑄󰇗 + ΔH

𝑊󰇗 nf = -193.8 – 563.6

𝑾󰇗 nf = -757.4 kJ/s ACTIVITY

Direction: Solve the following problems. Show your solution. 1. A reversible, nonflow, constant volume process decreases the internal energy by 316.5 kJ for 2.268 kgm of a gas for which R = 430 J/kgm-K and k = 1.35. For the process, determine (a) the work, (b) Q, and (c) ΔS. The initial temperature is 204.4oC. 2. Three pounds of a perfect gas with R = 38 ft. lbf/lbm.R and k = 1.667 have 300 Btu of heat added during a reversible nonflow constant pressure change of state. The initial temperature is 100oF. Determine the (a) final temperature, (b) ΔH, (c) W, (d) ΔU, and (e) ΔS. 3. Four pounds of air gain 0.491 Btu/oR of entropy during a nonflow isothermal process. If p1 = 120 psia and V2 = 42.5 ft3, find (a) V1 and T1, (b) W, (c) Q, and (d) ΔU. 4. One pound of an ideal gas undergoes an isentropic process from 95.3psig and a volume of 0.6 ft3 to a final volume of 3.6 ft3. If cp = 0.124 and cv = 0.093 Btu/lbm-R respectively. What are (a) T2, (b) p2, (c) ΔH, and (d) W. 5. The work required to compress a gas reversibly according to pV 1.30 = c is 67,790J, if there is no flow. Determine ΔU and Q if the gas is (a) air, (b) methane.

COMMON GAS CONSTANTS

REFERENCES 1. Yunus Cengel. (2019). Thermodynamics: An Engineering Approach - 9th Edition.

McGraw-Hill Education. 2. H. B. Sta. Maria and D. V. Gedaria. (2015). Thermodynamics - 2nd Edition. National 3. 4. 5. 6. 7. 8.

Bookstore, Inc. Y.A Cengel and M. A. Boles. (2015). Thermodynamics: An Engineering Approach - 8th Edition. McGraw-Hill Education. M. J. Moran, H. N. Shapiro, D. D Boettner & M. B. Bailey. (2011). Fundamentals of Engineering Thermodynamics - 7th Edition. John Wiley and Sons, Inc. R. T. Balmer. (2011). Modern Engineering Thermodynamics. Elsevier Inc. C. Borgnakke & R. E. Sonntag. (2009). Fundamentals of Thermodynamics - 7th Edition. John Wiley and Sons, Inc. R. K. Rajput. (2007). Engineering Thermodynamics - 3rd Edition. Laxmi Publications (P) Ltd. Boyle's law - Wikipedia, the free encyclopedia...