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Chapter 9
THE FIRST LAW
APPLIED TO
OPEN SYSTEMS
When the infinity of space and universe was first revealed he was terrified; he felt lost and humbled - he was no longer the central point of all but an insignificant, infinitely small grain of dust. − Nicholas Berdyaev (Man and Machine)
So far, we have learned to apply the first law of thermodynamics to solve
various kinds of thermodynamic problems associated with closed systems.
In this chapter, we will learn to apply the first law of thermodynamics to
open systems. An open system, like a closed system, allows heat and work
to enter and/or leave the system. In addition, an open system allows mass
to enter and/or leave the system. Study of open systems is important,
since a large number of engineering applications involve open systems.
190 Chapter 9
9.1 Example of an Open System
Consider a container to which water at room temperature is supplied
from one end, and hot water is drawn from the other end, as shown in
Figure 9.1. A heating coil is placed within the container to heat the water
flowing through the container. A stirrer is used to stir the water, and thus
does stirring work on the water.
Choose the open system to be the water present within the container.
We may consider the dashed line of Figure 9.1 as the boundary of the sys-
tem. It is however not correct. The boundary of the system should enclose
only the water in the system, and should exclude the stirrer and the heating
coil. The stirrer and the heating coil are part of the surroundings.
stirrer
heater
^ water
water inlet^
^ ^ water outlet
Figure 9.1 An example of an open system.
An open system is therefore best represented by a block diagram, as
shown in Figure 9.2. The system comprises only the water in the container,
marked by the boundary of the system, given by the dashed line of Figure
9.2, but excludes the heater and the stirrer.
Let the rate at which heat enters the system from the heater be Q˙in,
and the rate at which work enters the system from the stirrer be W˙in. Let
the rates at which mass enters and leaves the system be m˙i and m˙e, re-
spectively. Let the mass of the system at time t be ms, and the rate of
increase in mass within the system can be given by dms/dt. The energy of
192 Chapter 9
Teacher: Please, go ahead.
Student: What do you mean by the rate of increase in energy within the sys- tem?
Teacher: Consider the system shown in Figure 9.1 or in Figure 9.2. Let us suppose the temperature of the water in the tank increases with time, which means the internal energy of water increases with time. The rate at which the energy of the system increases is what we refer to as the rate of increase in energy within the system, and it is denoted by dEs/dt. Note that the unit of dEs/dt may be kJ/s, or kJ/hour, or kJ/day, or kJ/year.
Student: Thanks, Teacher.
9.2 Mass Balance for an Open System
The law of conservation of mass applied to the open system of Figure
9.2 yields the following:
rate of increase in mass within the system = rate at which mass entering the system − rate at which mass leaving the system,
which is equivalent to
dms
dt
= ˙mi − m˙e (9.1)
where (dms/dt) represents the rate of increase in mass within the system,
and m˙i and m˙e represent the respective rates at which mass entering and
leaving the system.
Note that if m˙i is larger than m˙e then dms/dt will be positive denoting
that mass accumulates within the system. If m˙e is larger than m˙i then
dms/dt will be negative denoting that the mass within the system is de-
creased. If m˙i is the same as m˙e then dms/dt is zero so that the mass of
the system remains a constant with time.
The First Law applied to Open Systems 193
Integrating (9.1) over a time interval Δt (= tf − to), we get
∫ msf
mso
dms =
∫ tf
to
m ˙i dt −
∫ tf
to
m ˙e dt
which can be written as
msf − mso = mi − me (9.2)
where mso is the mass of the system at the initial time to, msf is the mass
of the system at the final time tf , mi is the total mass entering the system
during the time interval Δt, and me is the total mass leaving the system
during the time interval Δt.
Equations (9.1) and (9.2) are applicable for open systems with one inlet
and one exit. If an open system has many inlets and exits for the mass to
flow then (9.1) must be expanded to
dms
dt
= ˙mi 1 + ˙mi 2 + ˙mi 3 + · · · − m˙e 1 − m˙e 2 − m˙e 3 − · · ·
and (9.2) must be expanded to
msf − mso = mi 1 + mi 2 + mi 3 + · · · − me 1 − me 2 − me 3 − · · ·
where the subscripts i 1 , i 2 , i 3 , ··· stand for inlet 1, inlet 2, inlet 3, etc., and
e 1 ,^ e 2 ,^ e 3 ,^ ··· stand for exit 1, exit 2, exit 3, etc.
9.3 Energy Balance for an Open System
The energy of a closed system is changed only by heat and work in-
teractions between the system and its surroundings. In an open system,
however, an additional mechanism is seen to change the total energy of the
system: that is, the flow of matter in and out of the system. When matter
enters a system, it brings in some energy with it. Likewise, when matter
leaves the system, it takes out some energy with it.
The First Law applied to Open Systems 195
The dE (^) s /dt term
For simple compressible systems, the total energy of the system, Es,
is only the internal energy of the system, Us. If the system as a whole
undergoes a change in its elevation then we will add in the gravitational
potential energy term. If the velocity of the whole system changes then
we will add in the translational kinetic energy term. As beginners in ther-
modynamics, we will not consider systems that could possess any forms
of energy other than the internal energy, the gravitational potential energy
and the translational kinetic energy.
The Q˙in term
If there is a temperature difference between the system and the sur-
roundings, then heat will be transferred across the system boundary. If the
system is adiabatic, there will be no heat transfer across the boundary, and
Q^ ˙in = 0.
The W˙ (^) in term
Three of the work forms that we often come across with open systems
are discussed below.
Boundary Work ( W˙boundary):
It is a work that results when the system boundary moves. We have
learned about this work in detail in Chapter 7.
Shaft Work ( W˙shaft):
It is a form of work that is realized when a shaft is given a motion by
the fluid flowing through an open system, or when the motion of a
shaft is used to do work on the fluid flowing through an open system.
Flow Work ( W˙flow):
Work associated with a fluid mass flowing into or out of an open
system is called flow work, and is relevant only to an open system.
Let us consider an opening through which mass flows into a system
as shown in Figure 9.3. The dashed box shows the volume dV of a
very small mass dm that is being pushed into the system in time dt.
196 Chapter 9
P
dl
A
system
Figure 9.3 Schematic for flow work.
dm
The fluid mass dm is taken to be sufficiently small for the properties
within the fluid mass to be considered uniformly distributed. The
pressure force acting on the fluid mass dm to push it into the system
is given by the pressure P multiplied by the cross-sectional area A.
The distance across which the fluid mass is pushed is given by dl.
The work done in time dt to push the fluid mass dm into the system
across the inlet is given by
dWf low = (P A) (dl) = P dV = P v dm
since A dl = dV and since dV = v dm, where v is the specific
volume. Dividing the above expression by dt, we get the following:
dWf low
dt
P v dm
dt
which can be written as
W^ ˙f low = ˙m P v (9.5)
The above equation gives the rate at which flow work is done to
maintain a mass flow rate m˙ at the inlet concerned. Flow work
enters the system when matter flows into the system, and leaves the
system when matter flows out of the system.
198 Chapter 9
Substituting from (9.7) for the energy entering and leaving the system
with the flowing fluid in (9.6), we get
dEs
dt
= Q˙in + ( W˙boundary)in + ( W˙shaf t)in + ( ˙m P v)i − ( ˙m P v)e
m u ˙ + ˙m
c^2
+ ˙m gz
i
mu ˙ + ˙m
c^2
+ ˙m gz
e
which can be regrouped as
dEs
dt
= Q˙in + ( W˙boundary)in + ( W˙shaf t)in
m P v ˙ + ˙m u + ˙m
c^2
+ ˙m gz
i
m P v ˙ + ˙mu + ˙m
c^2
+ ˙m gz
e
Using h = u + P v from (4.2) in the above equation, we get
dEs
dt
= Q˙in + ( W˙boundary)in + ( W˙shaf t)in
m h ˙ + ˙m
c^2
+ ˙m gz
i
m h ˙ + ˙m
c^2
+ ˙m gz
e
Equation (9.8) uses enthalpy which includes both the internal energy
and the energy associated with pushing the fluid into or out of the open
system. It is therefore the energy of a fluid stream flowing into or out of an
open system is represented by enthalpy, kinetic energy and potential energy,
and no reference will be made to flow work.
Equation (9.8) is the first law of thermodynamics applicable to open
systems at any given time, and each term in this equation is a rate term
The First Law applied to Open Systems 199
having the unit of energy per time. Equation (9.8) is, however, inadequate
to completely describe the open systems. We need to use the mass balance
given by (9.1), along with (9.8), to solve problems involving open systems.
Integrating (9.8) over the time interval Δt (= tf − to), we get
∫ Esf
Eso
dEs =
∫ tf
to
Q^ ˙indt +
∫ tf
to
( W˙boundary)indt +
∫ tf
to
( W˙shaf t)indt
∫ tf
to
m h ˙ + ˙m
c^2
+ ˙m gz
i
dt
∫ tf
to
m h ˙ + ˙m
c^2
+ ˙m gz
e
dt
which can be rewritten as
Esf − Eso = Qin + (Wboundary)in + (Wshaf t)in
∫ tf
to
m h ˙ + ˙m
c^2
+ ˙m gz
i
dt
∫ tf
to
m h ˙ + ˙m
c^2
+ ˙m gz
e
dt (9.9)
where Eso is the energy of the system at the initial time to, Esf is the energy
of the system at the final time tf , and Qin, (Wboundary)in and (Wshaf t)in
are the respective amounts of net heat, net boundary work and net shaft
work entering the system during the time interval Δt.
Equation (9.9) is the first law of thermodynamics applicable to open
systems over a chosen time interval, such as tf − to, and each term in this
equation takes the unit of energy. Equation (9.9) is used together with the
mass balance given by (9.2) to solve problems involving open systems.
Equations (9.8) and (9.9) are applicable for open systems with not
more than one inlet and one exit. If an open system has many inlets
and exits then we must include
m h ˙ + ˙m c
2
2 + ˙m gz
-term in (9.8), and
∫ tf
to
m h ˙ + ˙m c
2
2 + ˙m gz
dt-term in (9.9), at the respective inlets and exits
with appropriate signs; positive sign for inlet and negative sign for outlet.
The First Law applied to Open Systems 201
entire charging period becomes
Esf − Eso =
∫ (^) tf
to
m h ˙ + ˙m
c^2 2
i
dt
Since the system is stationary, the kinetic energy of the system does not change. The potential energy of the system changes with nitrogen entering the system. Let us however ignore the change in potential energy, and take the change in total energy Es of the system be the change in internal energy Us alone. Also, by neglecting the kinetic and potential energies at the inlet, the above equation is reduced to
Usf − Uso =
∫ (^) tf
to
( ˙m h)i dt (9.10)
where Usf and Uso are the internal energies of the nitrogen in the tank at time tf and to, respectively. Pressure and temperature in the nitrogen supply line remain constant through- out the charging process, and therefore enthalpy at the inlet of the system also remains constant at hi. Using this information, (9.10) can be simplified to
Usf − Uso = hi
∫ (^) tf
to
m ˙i dt = hi mi
where mi is the total mass of nitrogen entering the tank during the entire charging process. Replacing the internal energies of the above equation by specific internal energies, we get msf usf − mso uso = hi mi (9.11)
where msf and mso are the mass of the nitrogen in the tank at time tf and to, respectively, and usf and uso are the specific internal energies of the nitrogen in the tank at time tf and to, respectively, which are assumed to be uniform throughout the system. The mass balance given by (9.2), when applied to the given system, reduces to msf − mso = mi (9.12) Combining (9.11) and (9.12) to eliminate mi, we get
msf usf − msouso = hi (msf − mso) (9.13)
In classical thermodynamics, the absolute values of u and h are not known. Only the changes in u and h are known. Therefore, (9.13) is transformed to a
202 Chapter 9
convenient form using the following standard procedure. Replacing the h-term in (9.13) using h = u + P v, we get
msf usf − msouso = (u + P v)i (msf − mso) which can be rearranged to give
msf (usf − ui) − mso (uso − ui) = (P v)i (msf − mso) (9.14)
Since nitrogen is assumed to behave as an ideal gas, the differences in spe- cific internal energies of (9.14) can be expressed in terms of the differences in temperatures using Δu =
Cv dT. Taking Cv as a constant and using the ideal gas equation of state P v = RT , (9.14) can be rewritten as
msf Cv (Tsf − Ti) − mso Cv (Tso − Ti) = R Ti (msf − mso)
which can be rearranged, using the ideal gas relationship Cp = Cv + R and the definition of γ = Cp/Cv, to give the following:
msf (Cv Tsf − Cv Ti − R Ti) = mso (Cv Tso − Cv Ti − R Ti) msf (Cv Tsf − Cp Ti) = mso (Cv Tso − Cp Ti) msf (Tsf − (Cp/Cv) Ti) = mso (Tso − (Cp/Cv) Ti) msf (Tsf − γ Ti) = mso (Tso − γ Ti) (9.15)
Since the masses msf and mso are unknown, they can be expressed using the ideal gas equation of state by
msf =
Psf Vsf R Tsf and mso = Pso Vso R Tso
Since the tank is rigid, its volume remains a constant throughout. Therefore
Vsf = Vso (9.17)
Using (9.16) and (9.17), we can rewrite (9.15) as follows: Psf Tsf (Tsf − γ Ti) =
Pso Tso (Tso − γ Ti) (9.18)
Substituting the numerical values known from the problem statement in (9.18), we determine the value of Tsf as follows:
6 bar Tsf
× (Tsf − 1. 4 × 350 K) = 1 bar 300 K
× (300 K − 1. 4 × 350 K)
6 ×
1 − 1. 4 ×
350 K
Tsf
1 − 1. 4 ×
Tsf = 443 K
204 Chapter 9
Example 9.
Air at 15 bar and 500 K is supplied to a rigid
tank of volume 1 m^3 containing air initially at 1 bar and 300 K, until the
air in the tank reaches 10 bar and 450 K. Determine the mass of air fed to
the tank and the heat interaction between the tank and the surroundings.
Assume that air behaves as an ideal gas, and that R and γ for air are 287
J/kg · K and 1.4, respectively.
Solution to Example 9.
Initially the air in the tank is at 1 bar and 300 K, and therefore the initial mass of air in the tank is
mso = (10^5 Pa) × (1 m^3 ) (287 J/kg · K × (300 K)
= 1. 16 kg (9.19)
Finally the air in the tank is at 10 bar and 450 K, and therefore the final mass of air in the tank is
msf = (10^6 Pa) × (1 m^3 ) (287 J/kg · K × (450 K)
= 7. 74 kg (9.20)
Therefore, the mass of air added to the tank is 6.58 kg.
To determine the heat interaction between the system and the surroundings, let us use the following procedure. The problem to be solved here is similar to the problem of Example 9.1, except for the heat interaction between the system and the surroundings. Therefore, (9.9) applied to the given system becomes
Esf − Eso = Qin +
∫ (^) tf
to
m h ˙ + ˙m c^2 2
i
dt
Neglecting the potential and kinetic energy changes of the system and of the inlet stream, the above equation is reduced to
Usf − Uso = Qin +
∫ (^) tf
to
( ˙m h)i dt
which can be rewritten, in terms of the specific internal energies, as
msf usf − mso uso = Qin + hi mi (9.21)
The First Law applied to Open Systems 205
where hi is the specific enthalpy of the incoming nitrogen taken as a constant, and mi is the mass of nitrogen entering the tank during the entire charging process, given by
∫ (^) tf to m˙i^ dt. Mass balance (9.2) applied to the system yields
msf − mso = mi (9.22)
Combining (9.21) and (9.22) to eliminate mi and rearranging the resulting equation, we get
Qin = msf usf − mso uso − hi (msf − mso) (9.23)
Replacing the specific enthalpy hi in (9.23) by (ui + RTi) applicable for an ideal gas, we get
Qin = msf (usf − ui − RTi) − mso(uso − ui − RTi) = msf [Cv(Tsf − Ti) − RTi] − mso[Cv(Tso − Ti) − RTi] = msf (CvTsf − CpTi) − mso(CvTso − CpTi) = msf Cv(Tsf − γTi) − msoCv(Tso − γTi) (9.24)
in which mso and msf are known from (9.19) and (9.20), Tso = 300 K, Tsf = 450 K and Ti = 500 K. We can calculate Cv using R/(γ − 1) as 717.5 J/kg · K. Substituting these numerical values in (9.24), we get Qin = −1055 kJ, which is to say that about 1055 kJ of heat is lost to the surroundings during the entire charging process.
Example 9. A tank contains 0.25 kg of nitrogen at 300 kPa and 300 K. It is discharged until its pressure becomes 100 kPa. If the tank is well insulated, what will be the final temperature in the tank? Determine the amount of nitrogen discharged from the tank. Assume ideal gas behaviour, and γ for nitrogen to be 1.4.
Solution to Example 9.
Figure 9.5 shows discharging of nitrogen from a tank. Initially the tank is filled with mso = 0.25 kg of nitrogen at Pso = 300 kPa and Tso = 300 K. It is
The First Law applied to Open Systems 207
where me is the mass of nitrogen leaving the tank during the entire discharging process, given by
∫ (^) tf to m˙e^ dt. Rewriting (9.27) in terms of specific internal energies, we get
msf usf − msouso = −
hso + hsf 2
me (9.28)
Applying (9.2) to the discharging process, we get
msf − mso = −me (9.29)
Combining (9.28) and (9.29) to eliminate me, we get
msf usf − mso uso =
hso + hsf 2
(msf − mso) (9.30)
which can be rearranged to give
msf (2 usf − hso − hsf ) = mso (2 uso − hso − hsf )
Using u = h - P v, the above equation can be written as
msf (hsf − hso − 2 Psf vsf ) = mso (hso − hsf − 2 Psovso)
Now, transform the differences in specific enthalpies into differences in tem- peratures using Δh =
Cp dT , with the assumption that nitrogen behaves as an ideal gas. Taking Cp as a constant, and using the ideal gas equation of state P v = RT in the above equation, we get
msf [Cp (Tsf − Tso) − 2 R Tsf ] = mso [Cp (Tso − Tsf ) − 2 R Tso]
Substituting mso = 0.25 kg, Tso = 300 K, γ= 1.4, R = (8.314/28) kJ/kg · K, and therefore Cp = γ R/(γ − 1) = 1.039 kJ/kg · K, we get
msf (0. 445 Tsf − 311 .7) = 33. 4 − 0. 260 Tsf (9.31)
We need to determine Tsf from (9.31), but we do not know msf. Therefore, we need to look for another independent equation containing Tsf and msf. Applying the ideal gas equation of state at the final state, we get
msf =
Psf Vsf R Tsf
of which we know Psf = 100 kPa, but Vsf is unknown.
208 Chapter 9
Since the tank is rigid, its volume remains a constant throughout. Therefore
Vsf = Vso (9.33)
Applying the ideal gas equation of state at the initial state, we get
mso =
Pso Vso R Tso
Combining (9.32), (9.33) and (9.34), we get
msf =
Psf R Tsf
×
mso R Tso Pso
Psf Tso mso Pso Tsf
which gives msf = 25/Tsf (9.35) Eliminating msf from (9.31) using (9.35), we get
- 260 T (^) sf^2 − 22. 3 Tsf − 7792 .5 = 0
which gives Tsf = 221. 2 K Substituting the value of Tsf in (9.35), we get
msf = 0. 113 kg
The amount of nitrogen discharged from the tank is therefore given by
mso − msf = 0. 25 kg − 0. 113 kg = 0. 137 kg
9.5 Summary
- Mass balance for an open system is given by
dms dt
= ˙mi − m˙e (9.1)
where the subscript (^) s stands for system, (^) i for inlet, and (^) e for exit.
