Thermodynamics Second stage

3. The First Law of Thermodynamics

The first law can be stated as follows:

Whenever a system undergoes a cyclic change, the algebraic sum of work transfers is
equal to the algebraic sum of heat transfers.

∮ 𝛿𝑄 = ∮ 𝛿𝑊

where: 𝑄 = heat transfer

𝑊 = work transfer

The First Law of Thermodynamics for Non-Flow Processes

Non-flow process is the one in which there is no mass interaction across the system
boundaries during the occurrence of the process such as: heating and cooling of a fluid inside
a closed container, compression and expansion of a fluid in a piston-cylinder arrangement,
etc. For non-flow processes the first law can be written as:

𝑄 − 𝑊 = ∆𝑈 + ∆𝐾𝐸 + ∆𝑃𝐸 … … … … … … … (3.1)

For non- flow processes the kinetic and potential energies are very small and can be
neglected, so the energy equation becomes:

𝑄 − 𝑊 = 𝑈2 − 𝑈1 … … … … … … … (3.2)

𝑞 − 𝑤 = 𝑢2 − 𝑢1 … … … … … … … (3.3) per (kg)

where: state (1) refers to the initial state and state (2) refers to the final state.

For reversible processes:

2
𝑊 = ∫ 𝑃𝑑𝑉 … … … … … … … (3.4)
1

• For adiabatic processes (no heat transfer) 𝑄 = 0

• For constant volume processes 𝑊 = 0

• For constant temperature processes 𝛥𝑈 = 0

The following table contains the governing equations, displacement work equation
and heat interaction equation for different non-flow thermodynamic processes:

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Thermodynamics Second stage

Work
Process Governing equations 𝟐 Heat interaction
𝑾 = ∫ 𝑷𝒅𝑽
𝟏
𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Constant volume 𝑇1 𝑃1 𝑊=0 𝑄 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 )
(Isochoric) =
𝑇2 𝑃2
𝑃 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Constant pressure 𝑇1 𝑉1 𝑊 = 𝑃(𝑉2 − 𝑉1 ) 𝑄 = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 )
(Isobaric) =
𝑇2 𝑉2
𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉2 𝑉2
𝑊 = 𝑃1 𝑉1 ln ( ) 𝑄 = 𝑃1 𝑉1 ln ( )
Constant temperature 𝑃1 𝑉1 = 𝑃2 𝑉2 𝑉1 𝑉1
(Isothermal) 𝑉2 𝑃1 𝑉2 𝑉2
= 𝑊 = 𝑚𝑅𝑇ln ( ) 𝑄 = 𝑚𝑅𝑇ln ( )
𝑉1 𝑃2 𝑉1 𝑉1
𝑃2 𝑉1 𝛾
( )=( ) 𝑃1 𝑉1 − 𝑃2 𝑉2
𝑃1 𝑉2 𝑊=
𝑇2 𝑉1 𝛾−1 𝛾−1
Adiabatic ( )=( ) 𝑄=0
𝑇1 𝑉2
𝛾−1 𝑚𝑅(𝑇1 − 𝑇2 )
𝑇2 𝑃2 𝛾 𝑊=
( )=( ) 𝛾−1
𝑇1 𝑃1
𝑃2 𝑉1 𝑛
( )=( ) 𝑃1 𝑉1 − 𝑃2 𝑉2
𝑃1 𝑉2 𝑊= 𝛾−𝑛
𝑇2 𝑉1 𝑛−1 𝑛−1 𝑄=( )𝑊
Polytropic ( )=( ) 𝛾−1
𝑇1 𝑉2 𝑚𝑅(𝑇1 − 𝑇2 )
𝑛−1 𝑊=
𝑇2 𝑃2 𝑛 𝑛−1
( )=( )
𝑇1 𝑃1

Note:
• The heat transferred to the system is positive (+).

• The heat transferred from the system is negative (-).

• The work transferred from the system is positive (+).

• The work transferred to the system is negative (-).

Example (3.1): 1 kg of air enclosed in a rigid container is initially at 4.8 bar and 150°C.
The container is heated until the temperature is 200°C. Calculate the final pressure of
the air and the heat supplied during the process.

Solution:

Since we have a rigid vessel, then the processes takes place at constant volume. So (𝑉1 = 𝑉2 )
and 𝑊 = 0.

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Thermodynamics Second stage

𝑃1 𝑇1 4.8 (150 + 273)

= → = → 𝑃2 = 5.367 bar
𝑃2 𝑇2 𝑃2 (200 + 273)

𝑄 = 𝑈2 − 𝑈1 + 𝑊 = 𝑚. 𝐶𝑣 (𝑇2 − 𝑇1 ) + 𝑊 = 1 × 0.718(473 − 423) + 0

𝑸 = 𝟑𝟓. 𝟗 𝐤𝐉 Ans.

Example (3.2): In a piston-cylinder arrangement the steam at 1 MPa, 80% dryness

fraction and 0.05 m3 volume, is heated to increase its volume to 0.2 m3 while its pressure
remains constant. Determine the heat added.

Solution:

Initial states (1): 𝑃1 = 1 MPa, 𝑉1 = 0.05 m3 , 𝑥1 = 0.8

Final state (2): 𝑉2 = 0.2 m3 , 𝑃2 = 1 MPa

Work done during constant pressure process:

2
𝑊 = ∫ 𝑃𝑑𝑉 = 𝑃1 (𝑉2 − 𝑉1 )
1

𝑊 = 1000 × (0.2– 0.05)

𝑊 = 150 kJ

𝑉1
mass of gas (𝑚) =
𝑣1

From Saturated steam tables at 𝑃1 = 1 MPa:

𝑣𝑔 = 0.19444 m3 /kg

𝑢𝑓 = 761.68 kJ/kg

𝑢𝑓𝑔 = 1822 kJ/kg

so 𝑣1 = 𝑥1 . 𝑣𝑔 = 0.8 × 0.19444 = 0.15555 m3 /kg

0.05
Hence, 𝑚 = = 0.32144 kg
0.15555
𝑉2 0.2
𝑣2 = = = 0.6222 m3 /kg
𝑚 0.32144
Corresponding to this specific volume the final state is to be located in order to get the
internal energy. At 𝑃2 = 1 MPa

𝑣2 = 0.6222 > 𝑣𝑔 = 0.19444

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Thermodynamics Second stage

Hence, the state lies in the superheated region. From superheated steam tables at 𝑃2 = 1 MPa
and 𝑣2 = 0.6222 m3 /kg, we get the internal energy at the final state by interpolation:

0.6222 − 0.5871
𝑢2 = 4050.5 + ( ) (4255.1 − 4050.5) = 4205.27 kJ/kg
0.6335 − 0.5871
Internal energy at the initial state:

𝑢1 = 𝑢𝑓 + 𝑥1 . 𝑢𝑓𝑔 = 761.68 + 0.8 × 1822 = 2219.28 kJ/kg

From the first law of thermodynamics:

𝑄 − 𝑊 = 𝛥𝑈

𝑄 = (𝑈2 − 𝑈1 ) + 𝑊

= 𝑚(𝑢2 − 𝑢1 ) + 𝑊

= 0.32097 × (4205.27 − 2219.28) + 150

𝑸 = 𝟕𝟖𝟕. 𝟒𝟒 𝐤𝐉 Ans.

The First Law of Thermodynamics for Steady-Flow Processes

C1,P1,A1,
Flow processes can be classified into steady and U1
unsteady flow processes. For a steady-flow process, the
mass flow rate through the system must be constant
everywhere and the flow should not vary with time.
Q add
Consider one kilogram of a fluid has an internal energy
U, is moving with velocity C and at height Z above the
datum level
Z1

When the system is enclosed by a boundary it is W out

called a control volume. Assume that a steady flow of
heat 𝑄 is supplied to the system. So the total energy
C2,P2,A2,
entering the control volume is: U2
Z2

1
𝑈1 + 𝑃1 𝑉1 + . 𝑚. 𝐶12 + 𝑚. 𝑔. 𝑍1 + 𝑄 … … … … … … … (3.5)
2
where: 𝑃𝑉 is the flow energy,
1
. 𝑚. 𝐶 2 is the kinetic energy,
2

𝑚. 𝑔. 𝑍 is the potential energy.

In the same manner, the total energy leaving the control volume is:

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Thermodynamics Second stage

1
𝑈2 + 𝑃2 𝑉2 + . 𝑚. 𝐶22 + 𝑚. 𝑔. 𝑍2 + 𝑊 … … … … … … … (3.6)
2
Total energy entering the control volume = total energy leaving the control volume (energy
conservation statement). Thus the steady flow energy equation is:

1 1
𝑄 + 𝑈1 + 𝑃1 𝑉1 + . 𝑚. 𝐶12 + 𝑚. 𝑔. 𝑍1 = 𝑊 + 𝑈2 + 𝑃2 𝑉2 + . 𝑚. 𝐶22 + 𝑚. 𝑔. 𝑍2 … … … (3.7)
2 2
1 1
𝑞 + 𝑢1 + 𝑃1 𝑣1 + . 𝐶12 + 𝑔. 𝑍1 = 𝑤 + 𝑢2 + 𝑃2 𝑣2 + . 𝐶22 + 𝑔. 𝑍2 … … … (3.8) per (kg)
2 2
Since 𝐻 = 𝑈 + 𝑃𝑉, then equation (3.7) becomes:

1 1
𝑄 + 𝐻1 + . 𝑚. 𝐶12 + 𝑚. 𝑔. 𝑍1 = 𝑊 + 𝐻2 + . 𝑚. 𝐶22 + 𝑚. 𝑔. 𝑍2 … … … … … … … (3.9)
2 2
1 1
𝑞 + ℎ1 + . 𝐶12 + 𝑔. 𝑍1 = 𝑤 + ℎ2 + . 𝐶22 + 𝑔. 𝑍2 … … … … … … … (3.10) per (kg)
2 2
For steady-flow, the mass flow rate is constant at any section through the system 𝑚1° = 𝑚2° =
. . . . . 𝑚𝑛° (mass conservation statement).

For steady-flow processes:

2
𝑊 = − ∫ 𝑉𝑑𝑃 … … … … … … … (3.11)
1

• For adiabatic processes (no heat transfer) 𝑄 = 0

• For steady-flow constant pressure processes 𝑊 = 0

• For constant temperature processes ∆𝐻 = 0

Example (3.3): Steam is supplied to a fully loaded 1100 kW turbine at 15 bar with an
internal energy of 2935 kJ/kg, specific volume of 0.16 m3/kg and velocity of 110 m/s.
Exhaust takes place at 0.05 bar with an internal energy of 1885 kJ/kg, specific volume
equal to 26 m3/kg and velocity of 300 m/s. Heat loss from the steam in the turbine is 21
kJ/kg. Determine:

a) Shaft work output per kg.

b) Steam flow rate in kg/h.

Solution:

a) Applying the first law of thermodynamics and neglecting the potential energy:

1 1
𝑞 + 𝑢1 + 𝑃1 𝑣1 + . 𝐶12 = 𝑤 + 𝑢2 + 𝑃2 𝑣2 + . 𝐶22
2 2

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Thermodynamics Second stage

15 × 105 1 1102 0.05 × 105 1 3002

−21 + 2935 + × 0.16 + × = 𝑤 + 1885 + × 26 + ×
103 2 103 103 2 103
𝑤 = 1100 kJ/kg

1100
𝑃𝑜𝑤𝑒𝑟 = 𝑤 × 𝑚° → 𝑚° = = 1 kg/s
1100

𝒎° = 𝟑𝟔𝟎𝟎 𝐤𝐠/𝐡 Ans.

Reversible and Irreversible Processes

1. Reversible processes: which refer to the thermodynamic processes occurring in the
manner that states passed through are always in thermodynamic equilibrium and no
dissipative effects are present. Any reversible process occurring between states 1 and 2 upon
reversal, while occurring from 2 to 1 shall not leave any mark of process ever occurred as
states traced back are exactly similar to those in forward direction. Reversible processes are
thus very difficult to be realized and also called ideal processes. All thermodynamic
processes are attempted to reach close to the reversible process in order to give the best
performance.

2. Irreversible processes: which refer to the thermodynamic processes that do not fulfil the
conditions of a reversible process. Irreversibilities are the reasons causing processes to be
irreversible. Generally, the irreversibilities can be termed as internal irreversibilities and
external irreversibilities. Internal irreversibilities are there because of internal factors
whereas; external irreversibilities are caused by external factors at the system-surrounding
interface. Generic types of irreversibilities are due to: (friction, electrical resistance, inelastic
solid deformations, free expansion, heat transfer through a finite temperature difference, non-
equilibrium during the process, etc.).

Important Irreversible Thermodynamic Processes

1. Unrestricted or free expansion: consider the system shown in the figure. Tank A is filled
with a gas at pressure 𝑃𝐴 , tank B is initially completely evacuated. Valve V is initially closed.
When valve V is opened, the gas in A will expand rapidly to fill both tanks at a lower
pressure. This process is highly irreversible since, the fluid is eddying continuously. The
process is a non-flow process. Applying the first law of thermodynamics:

𝑞 = (𝑢2 − 𝑢1 ) + 𝑤

If the system is thermally insulated 𝑞 = 0

Also, no work is done on or by the fluid 𝑤 = 0

Thus, 𝑢2 − 𝑢1 = 0 → 𝑢2 = 𝑢2 → 𝐶𝑣 𝑇2 = 𝐶𝑣 𝑇1

Finally, 𝑇2 = 𝑇1 this means that the process is isothermal.

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Thermodynamics Second stage

2. Throttling: in this process the velocities before and after throttling are either equal or
negligibly small.

Also no heat is exchanged with the surroundings 𝑞 = 0

Moreover, no work done on or by the fluid 𝑤 = 0

This process is a flow process so applying the first law of

thermodynamics:

ℎ1 = ℎ2

So the process is considered isenthalpic.

This process is highly irreversible due to eddy generated at the restriction.

For a perfect gas, 𝐶𝑝 𝑇1 = 𝐶𝑝 𝑇2 → 𝑇1 = 𝑇2

So the process is also isothermal.

3. Adiabatic mixing: this process is a steady flow process and can be assumed adiabatic.

𝑚3° = 𝑚1° + 𝑚2°

𝑞=0

No work is done on or by the fluid.

𝑤=0

Applying the first law of thermodynamics for flow processes and neglecting the change in
kinetic energy:

ℎ1 + ℎ2 = ℎ3 … … … … … … … (3.12)

If the same fluid flows in both streams:

𝑚1° × ℎ1 + 𝑚2° × ℎ2 = 𝑚3° × ℎ3 … … … … … … … (3.13)

𝑚1° × 𝐶𝑝 × 𝑇1 + 𝑚2° × 𝐶𝑝 × 𝑇2 = 𝑚3° × 𝐶𝑝 × 𝑇3 … … … … … … … (3.14)

Since 𝑚3° = 𝑚1° + 𝑚2° , then the above equation becomes:

𝑚1° × 𝑇1 + 𝑚2° × 𝑇2
𝑇3 = … … … … … … … (3.15)
𝑚1° + 𝑚2°

Example (3.4): A superheated steam at 6 bar and 300°C is mixed in steady adiabatic
flow with wet steam at 6 bar and dryness fraction of 0.9. Calculate the mass of the wet
steam required per one kilogram of superheated steam to produce dry saturated steam
at 6 bar.

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Thermodynamics Second stage

Solution:

𝑚1° × ℎ1 + 𝑚2° × ℎ2 = 𝑚3° × ℎ3

Since 𝑚3° = 𝑚1° + 𝑚2° , then:

𝑚1° × ℎ1 + 𝑚2° × ℎ2 = (𝑚1° + 𝑚2° ) × ℎ3

From superheated steam tables at 6 bar and 300°C, (ℎ1 = 3061.6 kJ/kg)

ℎ2 = ℎ𝑓 + 𝑥. ℎ𝑓𝑔

From saturated steam tables at 6 bar (ℎ𝑓 = 670.56 kJ/kg, ℎ𝑓𝑔 = 2086.3 kJ/kg) so:

ℎ2 = 670.56 + 0.9 × 2086.3 = 2548.23 kJ/kg

From saturated steam tables at 6 bar (ℎ3 = ℎ𝑔 = 2756.8 kJ/kg)

Substituting ℎ1 , ℎ2 and ℎ3 we get:

1 × 3061.6 + 𝑚2° × 2548.23 = (1 + 𝑚2° ) × 2756.8

𝒎°𝟐 = 𝟏. 𝟒𝟔𝟏 𝐤𝐠 Ans.

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 Thermodynamics Second stage

Exercises
p245 Rajput Problem (3.1): 0.05 kg of carbon dioxide (molecular weight 44), occupying a volume of 0.03
m3 at 1.025 bar, is compressed reversibly until the pressure is 6.15 bar. Calculate the final
temperature, the work done on the gas and the heat flow to or from the cylinder walls when:

a) The process is according to a law 𝑃𝑉 1.4 = 𝐶.

b) The process is isothermal.
c) The process takes place in a perfectly thermally insulated cylinder.

Assume carbon dioxide to be a perfect gas and take γ=1.3.

Ans. (540.06 K, -5.138 kJ, 1.713 kJ, 323.68 K, -5.51 kJ, -5.51 kJ, 324 K, -5.25 kJ, 0 kJ)

Problem (3.2): The gases in the cylinder of an internal combustion engine have a specific
internal energy of 800 kJ/kg and a specific volume of 0.06 m3/kg at the beginning of
expansion. The expansion of the gases may be assumed to take place according to a
reversible law 𝑃𝑣 1.5 = 𝐶, from 55 bar to 1.4 bar. The specific internal energy after expansion
is 230 kJ/kg. Calculate the heat rejected to the cylinder cooling water per kilogram of gases
during the expansion stroke.

Ans. (-104 kJ/kg)

Problem (3.3): A mass of gas at an initial pressure of 28 bar and with an internal energy of
1500 kJ, is contained in a well-insulated cylinder of volume 0.06 m3. The gas is allowed to
expand behind a piston until its internal energy is 1400 kJ, the law of expansion is 𝑃𝑉 2 = 𝐶.
Calculate:

a) The work done.

b) The final volume.
c) The final pressure.

Ans. (100 kJ, 0.148 m3, 4.59 bar)

Problem (3.4): 0.2 m3 of air at 4 bar and 130°C is contained in a system. A reversible
adiabatic expansion takes place till the pressure falls to 1.02 bar. The gas is then heated at
constant pressure till the enthalpy increases by 72.5 kJ. Calculate the work done.

Example 4.23 from Rajput Ans. (85.55 kJ)

Problem (3.5): 0.1 m3 of an ideal gas at 300 K and 1 bar is compressed adiabatically to 8 bar.
It is then cooled at constant volume and further expanded isothermally so as to reach the
condition from where it started. Calculate:

a) Pressure at the end of constant volume cooling.

b) Change in internal energy during constant volume process.
c) Net work done and heat transferred during the cycle.

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Thermodynamics Second stage

Assume Cp = 14.3 kJ/kg.K and Cv = 10.2 kJ/kg.K.

Ans. (4.4 bar, -20.27 kJ, -5.45 kJ)

Problem (3.6): 0.15 m3 of an ideal gas at a pressure of 15 bar and 550 K is expanded
isothermally to 4 times the initial volume. It is then cooled to 290 K at constant volume and
then compressed back polytropically to its initial state. Calculate the net work done and heat
transferred during the cycle.

Ans. (81.03 kJ, 81.03 kJ)

Problem (3.7): A steam turbine receives a steam flow of 1.35 kg/s and the power output is
500 kW. The heat loss from the casing is negligible. Calculate:

a) The change of specific enthalpy across the turbine when the velocities at entrance and exit
and the difference in elevation are negligible.
b) The change of specific enthalpy across the turbine when the velocity at the entrance is 60
m/s, the velocity at exit is 360 m/s and the inlet pipe is 3 m above the exhaust pipe.

Ans. (370 kJ/kg, 433 kJ/kg)

Problem (3.8): A turbine under steady-flow conditions receives steam at the following state:
pressure 13.8 bar, specific volume 0.143 m3/kg, specific internal energy 2590 kJ/kg and
velocity 30 m/s. the state of the steam leaving the turbine is as follows: pressure 0.35 bar,
specific volume 4.37 m3/kg, specific internal energy 2360 kJ/kg and velocity 90 m/s. Heat is
rejected to the surroundings at the rate of 0.25 kW and the rate of steam flow through the
turbine is 0.38 kg/s. Calculate the power developed by the turbine.

Ans. (102.7 kW)

Problem (3.9): The pressure in a steam main is 12 bar. A sample of steam is drawn off and
passed through a throttling calorimeter, the pressure and temperature at exit from the
calorimeter being 1 bar and 140°C respectively. Calculate the dryness fraction of the steam in
the main, stating any assumptions made in the throttling process.

Ans. (0.986)

Problem (3.10): 229 kg/h of air at 40°C enters a mixing chamber where it mixes with 540
kg/h of air at 15°C. Calculate the temperature of the air leaving the chamber, assuming steady
flow conditions. Assume that the heat loss is negligible.

Ans. (22.4°C)

23