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Thermodynamics 1: Processes of Ideal Gas, Summaries of Thermodynamics

The different thermodynamic processes, equations for heat, work, and entropy, and how to solve problems for working substances undergoing thermodynamic processes. It also covers the processes of ideal gas, including adiabatic, reversible and irreversible, constant volume, and constant pressure processes. working equations for each process and explains how to calculate heat, work, and entropy for non-flow and steady-flow systems. It also includes PV diagrams for constant volume and constant pressure processes.

Typology: Summaries

2020/2021

Available from 06/25/2023

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UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE
THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 43
Week 4: Processes of Ideal Gas
Objectives:
1. Describe each of the thermodynamic process.
2. Obtain equations for heat, work, and entropy for each thermodynamic process.
3. Solve problems for working substance undergoing thermodynamic process.
XIV. Introduction
In solving mechanics problems, we isolate the body under consideration, analyze the external forces
acting on it, and then use Newton’s laws to predict its behavior. In thermodynamics, we take a similar
approach. We start by identifying the part of the universe we wish to study; it is also known as our system.
(We defined a system at the beginning of this chapter as anything whose properties are of interest to us; it
can be a single atom or the entire Earth.) Once our system is selected, we determine how the environment,
or surroundings, interact with the system. Finally, with the interaction understood, we study the thermal
behavior of the system with the help of the laws of thermodynamics.
The state of a system can change as a result of its interaction with the environment. The change in a
system can be fast or slow and large or small. The manner in which a state of a system can change from an
initial state to a final state is called a thermodynamic process.
XV. Processes of Ideal Gas
Adiabatic Process
- In an adiabatic process, the system is insulated from its environment so that although the
state of the system changes, no heat is allowed to enter or leave the system, as seen in Figure 4.1.
- When a system expands adiabatically, it must do work against the outside world, and
therefore its energy goes down, which is reflected in the lowering of the temperature of the
system. An adiabatic expansion leads to a lowering of temperature, and an adiabatic compression
leads to an increase of temperature.
Reversible and Irreversible Process
- A reversible process is a process in which the system and environment can be restored
to exactly the same initial states that they were in before the process occurred, if we go backward
along the path of the process. It is a process were any losses are neglected
- An irreversible process is what we encounter in reality almost all the time. The system
and its environment cannot be restored to their original states at the same time.
Figure
4
.1:
An insulated pisto n with a hot, compressed gas is released. The piston moves up, the volume expands, and
the pressure and temperature decrease. The internal energy goes into work. If the expansion occurs within a time frame in
w
hich negligible heat can enter the system, then the process is called adiabatic. Ideally, during an adiabatic process no heat
enters or exits the system.
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Week 4: Processes of Ideal Gas

Objectives:

  1. Describe each of the thermodynamic process.
  2. Obtain equations for heat, work, and entropy for each thermodynamic process.
  3. Solve problems for working substance undergoing thermodynamic process.

XIV. Introduction

In solving mechanics problems, we isolate the body under consideration, analyze the external forces

acting on it, and then use Newton’s laws to predict its behavior. In thermodynamics, we take a similar

approach. We start by identifying the part of the universe we wish to study; it is also known as our system.

(We defined a system at the beginning of this chapter as anything whose properties are of interest to us; it

can be a single atom or the entire Earth.) Once our system is selected, we determine how the environment,

or surroundings, interact with the system. Finally, with the interaction understood, we study the thermal

behavior of the system with the help of the laws of thermodynamics.

The state of a system can change as a result of its interaction with the environment. The change in a

system can be fast or slow and large or small. The manner in which a state of a system can change from an

initial state to a final state is called a thermodynamic process.

XV. Processes of Ideal Gas

 Adiabatic Process

  • In an adiabatic process, the system is insulated from its environment so that although the

state of the system changes, no heat is allowed to enter or leave the system, as seen in Figure 4.1.

  • When a system expands adiabatically, it must do work against the outside world, and

therefore its energy goes down, which is reflected in the lowering of the temperature of the

system. An adiabatic expansion leads to a lowering of temperature, and an adiabatic compression

leads to an increase of temperature.

 Reversible and Irreversible Process

  • A reversible process is a process in which the system and environment can be restored

to exactly the same initial states that they were in before the process occurred, if we go backward

along the path of the process. It is a process were any losses are neglected

  • An irreversible process is what we encounter in reality almost all the time. The system

and its environment cannot be restored to their original states at the same time.

Figure 4 .1: An insulated piston with a hot, compressed gas is released. The piston moves up, the volume expands, and

the pressure and temperature decrease. The internal energy goes into work. If the expansion occurs within a time frame in

which negligible heat can enter the system, then the process is called adiabatic. Ideally, during an adiabatic process no heat

enters or exits the system.

 Constant Volume Process

  • Another term for constant volume process are Isometric,

Isochoric, and Isovolumic.

  • Examples of systems under constant volume process are rigid

tank, rigid vessel, and indestructible tank.

o Working equations for constant volume process:

a. Relationship between pressure and temperature,

𝟏

𝟏

𝟐

𝟐

b. Work, W

-The equation for work under constant volume process

can be derive by apply the formula for work done by a system.

  • For non-flow systems

௡௙

Since the volume is constant and the derivative of a constant is zero so,

W = 0

  • For steady-flow systems

௦௙

Integrating the pressure,

௦௙

c. Heat, Q

  • For non-flow systems, apply the non-flow energy equation,

The W = 0 for constant volume process so the resulting equation is,

  • For steady-flow systems, applying the steady flow energy equation,

The formula for the change in enthalpy is ∆𝐻 = ∆𝑈 + 𝑊

then substitute it to the

equation for 𝑄 and 𝑊 = −∆𝑊

Therefore,

d. Change in Enthalpy, ∆𝐻

e. Change in Internal Energy, ∆𝑈

f. Change in Entropy, ∆𝑆

  • Apply the general equation for entropy, ∆𝑆 = ∫

ௗொ

and 𝑑𝑄 = 𝑚𝑐

Then the equation becomes,

௠௖

ௗ்

Integrate the variable T,

𝒗

𝒏

𝟐

𝟏

Figure 4.2: PV diagram for constant

volume process

൰ [(140 − 400)°𝑅]

  1. Solve for ∆𝑈

(d) For 𝑄

Under constant volume process, 𝑄 = ∆𝑈 = −418.7 𝐵𝑡𝑢

(e) For ∆𝐻

  1. The formula for ∆𝐻 is,

  1. Substitute the given values,

[(

]

  1. Solve for ∆𝐻

(f) For ∆𝑆

  1. Apply the equation for ∆𝑆 under constant volume process

  1. Substitute the given values,

𝒏

  1. Solve for ∆𝑆

Learning Activity 4.

  1. A reversible, nonflow, constant volume process decreases the internal energy by 316.6 kJ for 2.268 kg of

a gas for which R = 430 J/kg-K and k = 1.35. For the process, determine (a) the work, (b) the heat, and (c)

the change of entropy if the initial temperature is 204.

o

C.

  1. There are 1.36 kg of gas, for which R = 377 J/kg-K and k = 1.25, that undergo a nonflow constant volume

process from p 1

= 551.6 kPaa and t 1

o

C to p 2

= 1655 kPaa. During the process the gas internally stirred,

and there are also added 105.5 kJ of heat. Determine the (a) t 2

, (b) the work input, (c) Q, (d) ΔU, and (e) ΔS.

  1. A closed constant-volume system receives 10.5 kJ of paddle work. The system contains oxygen at 344

kPa, 278 K, and occuspies 0.06 cu. m. Find the heat (gain or loss) if the final temperature is 400 K.

 Constant Pressure Process

  • Another term for constant volume process is Isobaric

process.

o Working equations for constant pressure process:

a. Relationship between volume and temperature,

𝟏

𝟐

𝟏

𝟐

b. Work, W

-The equation for work under constant volume

process can be derive by apply the formula for work done

by a system.

  • For non-flow systems

௡௙

Integrating the volume,

௡௙

  • For steady-flow systems

௦௙

The work for steady flow systems under constant pressure process is zero,

௦௙

c. Heat, Q

  • For non-flow systems, apply the non-flow energy equation,

The formula for the change in enthalpy is ∆𝐻 = ∆𝑈 + 𝑊

  • For steady-flow systems, applying the steady flow energy equation,

The work for steady flow systems under constant pressure process is zero, 𝑊 =

So, 𝑄 = ∆𝐻 = 𝑚𝑐

d. Change in Enthalpy, ∆𝐻

e. Change in Internal Energy, ∆𝑈

f. Change in Entropy, ∆𝑆

  • Apply the general equation for entropy, ∆𝑆 = ∫

ௗொ

and 𝑑𝑄 = 𝑚𝑐

Then the equation becomes,

௠௖

ௗ்

Integrate the variable T,

𝒑

𝒏

𝟐

𝟏

Figure 4.3: PV diagram for

constant pressure process

  1. Substitute the given values,

  1. Solve for 𝑐

  1. The equation for ∆𝑈 under constant volume process is,

  1. Substitute the given values,

[(

]

  1. Solve for ∆𝑈

(d) For ∆𝑆

  1. Apply the equation for ∆𝑆 under constant pressure process

  1. Substitute the given values,

𝒏

  1. Solve for ∆𝑆

Learning Activity 4.

  1. A perfect gas has a value of R = 319.2 J/kg-K and k = 1.26. If 120 kJ are added to 2.27 kg of this gas at

constant pressure when the initial temperature is 32.

o

C, find (a) T 2

, (b) ΔH, (c) ΔU, and (d) work for a

nonflow process.

  1. Assume 5 lb of an ideal gas with R = 38.7 ft-lb/lb-

o

R and k = 1.668 have 300 Btu of heat added during a

reversible constant pressure change of state. The initial temperature is 80

o

F. Determine (a) final

temperature is 80

o

F. Determine (a) the final temperature, (b) ΔU, ΔH, and ΔS, (c) W

  1. For a constant pressure system whose mass is 80 lb, 1 hp-min is required to raise the temperature by

o

F. Determine the specific heat for the system, Btu/lb-

o

F

 Constant Temperature Process

  • Another term for constant volume process is Isothermal

process

o Working equations for constant temperature process:

a. Relationship between volume and temperature,

b. Work, W

-The equation for work under constant temperature

process can be derive by apply the formula for work

done by a system.

  • For non-flow systems

௡௙

Since 𝑃

𝑃𝑉 = 𝑐 can be written as 𝑃 =

and we substitute it to our working equation for

work,

௡௙

Rearranging the expression,

௡௙

Integrate the volume,

௡௙

Knowing that 𝑐 = 𝑃𝑉, substitute for c

௡௙

  • For steady-flow systems

௦௙

Since, 𝑃

𝑃𝑉 = 𝑐 can be written as 𝑉 =

and we substitute it to our working equation for

work,

௦௙

Rearranging the expression,

௦௙

Integrate the pressure,

௦௙

Knowing that 𝑐 = 𝑃𝑉, substitute for c

௦௙

Figure 4.4: PV diagram for constant

temperature process

d. ∆𝑈 and ∆𝐻

e. ∆𝑆

Solution

(a) For 𝑊

௡௙

  1. The equation for the non-flow work under isothermal process is,

௡௙

  1. Substitute the given values,

௡௙

  1. Solve for 𝑊

௡௙

௡௙

(b) For 𝑊

௦௙

  1. The equation for the non-flow work under isothermal process is,

௦௙

  1. Substitute the given values,

௦௙

  1. Solve for 𝑊

௦௙

௦௙

(c) For 𝑄

The heat and work are equal for isothermal process,

(d) For ∆𝑈 and ∆𝐻

∆𝑈 and ∆𝐻 are both functions of temperature difference and the temperature difference

under isothermal process is zero. Therefore,

(e) For ∆𝑆

  1. Apply the equation for ∆𝑆 under constant temperature process

  1. Substitute the given values,

  1. Solve for ∆𝑆

Learning Activity 4.

  1. During a reversible process there are abstracted 317 kJ/s from 1.134 kg/s of a certain gas while the

temperature remains constant at 26.

o

C. For this gas, c p

= 2.232 and c v

= 1.713 kJ/kg-K. The initial pressure

is 586 kPa. For both nonflow and steady flow (ΔP = 0, ΔK = 0) process, determine (a) V 1

, V

2

and p 2

, (b) the

work and Q, (c) ΔS and ΔH.

2 Air flows steadily through an engine at constant temperature, 400 K. Find the work per kilogram if the

exit pressure is one-third the inlet pressure and the inlet pressure is 207 kPa. Assume that the kinetic and

potential energy variation is negligible.

  1. Helium at 100 atm, 165 K expands isothermally to 1 atm. For 2 kg, find (a) W for nonlfow and for a steady

flow process (ΔP = 0, ΔK = 0), (b) ΔU, (c) Q, and (d) ΔS

 Constant Entropy Process (PV

k

= c)

  • Another term for constant entropy process is Isentropic

process

  • If a system has adiabatic walls, heat cannot enter or

leave the system, and the first cause of possible change

of entropy does not occur. A process, taking place

without heat entering or exiting, is called

an adiabatic process. If a process is quasi-static (also

known as reversible), there is no spontaneous increase

of entropy, hence, in short, an isentropic process is quasi-

static and adiabatic.

  • In actual practice, quasi-static (reversible) processes do not occur; the concept of

reversibility is an idealization, an unachievable limit, but useful for theoretical

considerations.

o Working equations for constant entropy process:

a. Relationship between pressure, volume and temperature,

் మ

் భ

௏ భ

௏ మ

௞ିଵ

் మ

் భ

௉ మ

௉ భ

ೖషభ

b. Work, W

-The equation for work under constant temperature process can be derive by

apply the formula for work done by a system.

  • For non-flow systems

௡௙

Since 𝑃

Figure 4.6: PV diagram for

constant entropy process

c. Heat, Q

  • Isentropic process is a reversible adiabatic process, when we say adiabatic the

system is totally insulated, 𝑄 = 0

d. Change in Enthalpy, ∆𝐻

e. Change in Internal Energy, ∆𝑈

f. Change in Entropy, ∆𝑆

  • Then entropy for isentropic process is constant so ∆𝑆 = 0

Example 4.

From a state defined by 300 psia, 100 cu ft and 240

o

F, helium undergoes isentropic process to 0.

psig. Find (a) 𝑉 ଶ

and 𝑡

, (b) ∆𝑈 and ∆𝐻, (c) Non-flow work, and (d) Steady-flow work.

Given:

Using Table 3.1 to obtain the specific heats of helium,

ு௘

஻௧௨

௟௕

ି °ோ

Required:

a. 𝑉

and 𝑡

b. ∆𝑈 and ∆𝐻

c. 𝑊

௡௙

d. 𝑊

௦௙

e. ∆𝑆

Solution

(a) For 𝑉

and 𝑡

  1. Apply the equations relating pressure, volume and temperature under isentropic

process,

் మ

் భ

௉ మ

௉ భ

ೖషభ

  1. Substitute the given values to determine 𝑉

ଵ.଺଺଻

[(

]

ଵ.଺଺଻

  1. Solve for 𝑉

  1. Substitute the given values to determine 𝑡

ଵ.଺଺଻ିଵ

ଵ.଺଺଻

  1. Solve for 𝑡

(b) For ∆𝑈 and ∆𝐻

  1. Apply the ideal gas equation of state to solve for the mass of air
  1. Substitute the given values,

[(

]

  1. Solve for m

  1. The formula for ∆𝐻 is,

  1. Substitute the given values,

[(

]

  1. Solve for ∆𝐻
  1. The formula for ∆𝑈 is,

  1. Substitute the given values,

[(

]

  1. Solve for ∆𝑈

(c) For 𝑊

௡௙

  1. The equation for the non-flow work under isentropic process is,

௡௙

𝟐

𝟐

𝟏

𝟏

  1. Substitute the given values,

௡௙

𝒇

𝟐

𝒇

  1. Solve for 𝑊

௡௙

௡௙

(d) For 𝑊

௦௙

  1. The equation for the steady-flow work under isentropic process is,

௦௙

௡௙

  1. Substitute the given values,

௦௙

  1. Solve for 𝑊

௦௙

௡௙

ଵି௡

ଵି௡

Simplify,

௡௙

𝑷

𝟐

𝑽

𝟐

ି 𝑷

𝟏

𝑽

𝟏

𝟏ି 𝒏

  • For steady-flow systems

௦௙

Since, 𝑃

= 𝑐 can be written as 𝑉 =

and we substitute it to our working equation

for work,

௦௙

Rearranging the expression,

௦௙

௡ න

Integrate the pressure,

௦௙

௡ିଵ

௡ିଵ

Knowing that 𝑐 = 𝑃𝑉

, substitute for c

௦௙

𝑉 ൦

ଵି௡

ଵି௡

Simplify,

௦௙

𝟐

𝟐

𝟏

𝟏

௡௙

c. Heat, Q

Where: 𝐶

௞ି௡

ଵି௡

d. Change in Enthalpy, ∆𝐻

e. Change in Internal Energy, ∆𝑈

f. Change in Entropy, ∆𝑆

Apply the general equation for entropy, ∆𝑆 = ∫

ௗொ

and 𝑑𝑄 = 𝑚𝑐

Then the equation becomes,

௠௖

ௗ்

Integrate the variable T,