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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
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Objectives:
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.
state of the system changes, no heat is allowed to enter or leave the system, as seen in Figure 4.1.
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.
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
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.
Isochoric, and Isovolumic.
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.
Since the volume is constant and the derivative of a constant is zero so,
௦
ଶ
ଵ
Integrating the pressure,
௦
ଶ
ଵ
c. Heat, Q
The W = 0 for constant volume process so the resulting equation is,
௩
The formula for the change in enthalpy is ∆𝐻 = ∆𝑈 + 𝑊
then substitute it to the
equation for 𝑄 and 𝑊 = −∆𝑊
Therefore,
௩
d. Change in Enthalpy, ∆𝐻
e. Change in Internal Energy, ∆𝑈
௩
f. Change in Entropy, ∆𝑆
ௗொ
்
ଶ
ଵ
and 𝑑𝑄 = 𝑚𝑐
௩
Then the equation becomes,
ೡ
ௗ்
்
ଶ
ଵ
Integrate the variable T,
𝒗
𝒏
𝟐
𝟏
Figure 4.2: PV diagram for constant
volume process
(d) For 𝑄
Under constant volume process, 𝑄 = ∆𝑈 = −418.7 𝐵𝑡𝑢
(e) For ∆𝐻
(f) For ∆𝑆
௩
ଶ
ଵ
𝒏
Learning Activity 4.
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
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.
kPa, 278 K, and occuspies 0.06 cu. m. Find the heat (gain or loss) if the final temperature is 400 K.
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.
Integrating the volume,
ଶ
ଵ
௦
ଶ
ଵ
The work for steady flow systems under constant pressure process is zero,
௦
c. Heat, Q
The formula for the change in enthalpy is ∆𝐻 = ∆𝑈 + 𝑊
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
௩
f. Change in Entropy, ∆𝑆
ௗொ
்
ଶ
ଵ
and 𝑑𝑄 = 𝑚𝑐
Then the equation becomes,
ೡ
ௗ்
்
ଶ
ଵ
Integrate the variable T,
𝒑
𝒏
𝟐
𝟏
Figure 4.3: PV diagram for
constant pressure process
௩
௩
௩
௩
௩
(d) For ∆𝑆
ଶ
ଵ
𝒏
Learning Activity 4.
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.
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
o
F. Determine the specific heat for the system, Btu/lb-
o
process
o Working equations for constant temperature process:
a. Relationship between volume and temperature,
ଵ
ଵ
ଶ
ଶ
b. Work, W
-The equation for work under constant temperature
process can be derive by apply the formula for work
done by a system.
Since 𝑃
ଵ
ଵ
ଶ
ଶ
𝑃𝑉 = 𝑐 can be written as 𝑃 =
and we substitute it to our working equation for
work,
ଶ
ଵ
Rearranging the expression,
ଶ
ଵ
Integrate the volume,
ଶ
ଵ
Knowing that 𝑐 = 𝑃𝑉, substitute for c
ଶ
ଵ
ଵ
ଶ
ଵ
ଵ
ଶ
ଵ
ଶ
ଶ
ଶ
ଵ
௦
ଶ
ଵ
Since, 𝑃
ଵ
ଵ
ଶ
ଶ
𝑃𝑉 = 𝑐 can be written as 𝑉 =
and we substitute it to our working equation for
work,
௦
ଶ
ଵ
Rearranging the expression,
௦
ଶ
ଵ
Integrate the pressure,
௦
ଶ
ଵ
ଵ
ଶ
Knowing that 𝑐 = 𝑃𝑉, substitute for c
௦
భ
మ
మ
భ
Figure 4.4: PV diagram for constant
temperature process
d. ∆𝑈 and ∆𝐻
e. ∆𝑆
Solution
(a) For 𝑊
ଵ
ଶ
ଵ
ଶ
(b) For 𝑊
௦
௦
ଵ
ଶ
ଵ
ଶ
௦
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
ଵ
ଶ
Learning Activity 4.
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
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.
flow process (ΔP = 0, ΔK = 0), (b) ΔU, (c) Q, and (d) ΔS
k
process
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.
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,
ଵ
ଵ
ଶ
ଶ
் 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.
Since 𝑃
ଵ
ଵ
ଶ
ଶ
Figure 4.6: PV diagram for
constant entropy process
c. Heat, Q
system is totally insulated, 𝑄 = 0
d. Change in Enthalpy, ∆𝐻
e. Change in Internal Energy, ∆𝑈
௩
f. Change in Entropy, ∆𝑆
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
ଶ
, (b) ∆𝑈 and ∆𝐻, (c) Non-flow work, and (d) Steady-flow work.
Given:
ଵ
ଶ
ଵ
ଵ
ଷ
Using Table 3.1 to obtain the specific heats of helium,
ு
௧௨
ି °ோ
௩
Required:
a. 𝑉
ଶ
and 𝑡
ଶ
b. ∆𝑈 and ∆𝐻
c. 𝑊
d. 𝑊
௦
e. ∆𝑆
Solution
(a) For 𝑉
ଶ
and 𝑡
ଶ
process,
ଵ
ଵ
ଶ
ଶ
் మ
் భ
మ
భ
ೖషభ
ೖ
ଶ
ଷ
ଵ.
ଶ
ଵ.
ଶ
ଶ
ଷ
ଶ
ଶ
ଵ.ିଵ
ଵ.
ଶ
ଶ
ଶ
(b) For ∆𝑈 and ∆𝐻
ଶ
ଶ
ଶ
ଷ
௩
(c) For 𝑊
𝟐
𝟐
𝟏
𝟏
𝒇
𝟐
ଶ
ଶ
ଷ
ଶ
ଶ
ଶ
ଷ
𝒇
(d) For 𝑊
௦
௦
௦
௦
ଶ
ଵି
ଵ
ଵି
Simplify,
𝑷
𝟐
𝑽
𝟐
ି 𝑷
𝟏
𝑽
𝟏
𝟏ି 𝒏
௦
ଶ
ଵ
Since, 𝑃
ଵ
ଵ
ଶ
ଶ
= 𝑐 can be written as 𝑉 =
భ
భ
and we substitute it to our working equation
for work,
௦
ଵ
ଵ
ଶ
ଵ
Rearranging the expression,
௦
ଵ
න
ଵ
ଶ
ଵ
Integrate the pressure,
௦
ଵ
൦
ଶ
ିଵ
ଵ
ିଵ
Knowing that 𝑐 = 𝑃𝑉
, substitute for c
௦
ଵ
𝑉 ൦
ଶ
ଵି
ଵ
ଵି
Simplify,
௦
𝟐
𝟐
𝟏
𝟏
c. Heat, Q
Where: 𝐶
௩
ି
ଵି
d. Change in Enthalpy, ∆𝐻
e. Change in Internal Energy, ∆𝑈
௩
f. Change in Entropy, ∆𝑆
Apply the general equation for entropy, ∆𝑆 = ∫
ௗொ
்
ଶ
ଵ
and 𝑑𝑄 = 𝑚𝑐
Then the equation becomes,
ௗ்
்
ଶ
ଵ
Integrate the variable T,