BASIC THERMODYNAMICS

Course no. ME21202

Section 4
Dr. Sourav Mitra
RAC Lab(first floor)
Department of Mechanical Engineering

1
 First Law of
Thermodynamics

2
First Law of Thermodynamics:
Closed System

3
First Law of Thermodynamics: Statement
• The first formal statement of the first law of thermodynamics, by
Rudolf Clausius in 1850 was given for a thermodynamic system
undergoing a “cyclic” process:
• In all cases in which work is produced by the agency of heat,
a quantity of heat is consumed which is proportional to the
work done; and conversely, by the expenditure of an equal
quantity of work an equal quantity of heat is produced
p
4 3

Conversion of one form of energy transfer

(heat) to another form (work)
1 2
v
4
First Law of Thermodynamics: Corollary
p

2 1 2 1 2 A

∫ δ Q + ∫ δ Q = ∫ δW + ∫ δW
A B A B
1
1 2 1 2 B
2 1
v
∫ ( δ QA − δ WA ) =
1
− ∫ (δ QB − δ WB )
2

2 2

∫ (δ Q
1
A − δ WA ) = ∫ (δ Q
1
B − δ WB )

𝜹𝜹𝑸𝑸𝑨𝑨 − 𝜹𝜹𝑾𝑾𝑨𝑨 must represent some “state function” (path

invariant)
5
First Law of Thermodynamics: Corollary
𝜹𝜹𝐐𝐐 − 𝜹𝜹W must represent some “state p
function” (path invariant). Hence, it is A
2
property of the system.
1
This property is the total energy (E) B
possessed by the system
v
= δ Q − δW
dE
2 2 2

∫=
1
dE ∫ δ Q − ∫ δ W
1 1

⇒ E2 − E=
1 (Q 12, A )
− W12, A= (Q
12, B − W12, B )

Total energy change = heat transfer - work transfer

6
First Law for control mass (closed system)
= δ Q − δW
dE
δ Q − δ W = dEsystem = dU + d ( KE ) + d ( PE )
If we neglect change in KE and PE: δ Q − δW =
dU
Work transfer can be any of the modes discussed earlier: Boundary,
shaft/paddle, electrical, spring etc.

First law for transient processes will be given by: Q − W =

E
 δQ  δW E = dE
Q= W=
dt dt dt

If we neglect change in KE and PE: Q − W =

U 7
 Obtaining Internal Energy
• For real fluids use the tables to obtain the specific internal energy at
any given state similar to obtaining specific volume.

• For saturated mixture use the following formula (similar to vmix):

umix =(1 − x ) u f + xug =u f + xu fg

• The internal energy of the system U will depend on the specific internal
energy and the mass: U = Mu

In case of ideal gas: du = cv dT ⇒ dU =

mgas cv dT

8
 Example 1: Control mass (transient)
Water is in a piston/cylinder assembly with initial pressure P =
700 kPa, quality 50% with a volume of 0.5 m3. A heater is turned
on, heating the water in an isobaric process with 2.5 kW. How
long does it take to vaporize all the liquid? Ans: 1508 s

9
 Example 2: Control mass (transient)
Air enclosed inside a rigid insulated vessel of 1 m3 is initially at
30°C and 100 kPa. A fan is placed inside this vessel and it churns
the air inside consuming 30 W of electrical energy. Determine the
air temperature inside the vessel after the fan has operated for
30 mins. (Cv = 718 J/kg.K)
Ans: 95.4°C

10
 Example 3: Cycle
2 kg of an unknown ideal gas (with Cv =1 kJ/kg.K) is enclosed in a
piston cylinder assembly. It undergoes two processes A and B as
shown on the P-V chart. The work transfer for each process is
marked on the chart. P1 = 300 kPa, T1 = 100°C, P2 = 100 kPa.
Determine the heat transfer for process A and B?

Ans: For process A: -44 kJ

For process B: +26 kJ A

11
 Example 4: Control mass
Consider a system as shown in Fig.; the 400 L tank A contains Argon
(assume ideal gas with molar mass 40 g/mol) at 250 kPa and 30°C.
The cylinder B is initially empty and has a frictionless piston having
mass such that a gas pressure of 150 kPa floats it. As the valve is
opened, Argon flows into cylinder B and finally reaches a uniform
state of 150 kPa at 30°C in both tank A and cylinder B. What is the
heat transferred to Argon in the process? Ans: 40.05 kJ

12
 Example 5: Control mass
Consider a cylinder piston assembly as shown in Fig. The fluid is
saturated mixture of water. The initial pressure is 200 kPa, and
initial volume of the cylinder is 0.04 m3. Initially liquid phase
occupies 1/100th of the gas volume. The cylinder is heated in an
isothermal manner till it reaches a volume of 0.1 m3. Calculate the
heat transfer.
Ans: 158.64 kJ

Water
(saturated
mixture)

13
 Example 6: Control mass
Consider the process of free expansion of air (ideal gas with molar
mass 29 g). The cylinder volume is initially divided into two equal
halves by the piston (see Fig.) One side of the cylinder is
completely evacuated whereas the other side contains air at 100°C
and 100 kPa. As the pin is removed, piston moves rapidly towards
the top expanding air. Find the final pressure of air if the entire
cylinder is insulated. Ans: 50 kPa

14
 Example 7: Control mass
A piston held by a pin in an insulated cylinder, shown in Fig.
contains 2 kg of water at 100°C, with a quality of 98%. The piston
has a mass of 102 kg, with cross-sectional area of 100 cm2, and the
ambient pressure is 100 kPa. The pin is released, which allows the
piston to move. Determine the final temperature of water,
assuming the process to be adiabatic. (Will require iteration)
Ans: ~160°C, superheated

15
 Example 8: Control mass
A piston–cylinder device initially contains steam at 200 kPa,
200°C, and 0.5 m3. At this state, a linear spring is touching the
piston but exerts “no force” on it. Area of the piston is 0.1
m2. Heat is now slowly transferred to the steam, causing the
pressure and the volume to rise to 500 kPa and 0.6 m3,
respectively. Assuming gas as the system, what is the heat
transfer in this process?

Ans: 806.9 kJ

16
First Law of Thermodynamics:
Open System

17
Additional mode of energy transfer for
open system (control volume)
Energy transfer due to mass flow:
Mass flow in and out of the system: Additional mechanism
of energy transfer.
Energy lost
When mass enters a system, it carries
energy with it. Likewise, when some mass
leaves the system, the energy contained
within the system decreases because the
leaving mass takes out some energy with it.

E.g.: Cooling of a cup of coffee System

Energy lost as heat transfer

Energy lost due to mass of vapor exiting the system
In a “control volume” can there be any boundary work ?? 18
Additional mode of energy transfer for
open system
Flow work:
Mass flowing in/out of the system: This requires some work
to be done by/on the surroundings. This mode of work
transfer is called the flow work.

For the fluid element with area A

and length l entering the control
volume; the work done to push the
entire volume of fluid is given by:

19
 Additional mode of energy transfer for
open system
Flow work: It is the work done when fluid enters or leaves a
control volume.
W=flow F=.L PA=.L PV 1

The magnitude of flow work per

unit mass is given by: 2

PV
w=
flow = Pv
M

Specific flow work per unit mass is “state function”

Specific flow work at 1: P1v1
at 2: P2v2
20
Total modes of energy transfer for open system
1. Heat transfer
2. Work (shaft, electrical)
3. Energy transfer due to mass
flow
1. specific energy of fluid:
v 2
e =u + + gz
2
2. Specific flow work:
w flow = Pv

Specific flow work per unit mass is “state function”

21
 Specific Enthalpy
In an “open system” we can conclude that the total energy interaction
due to fluid entering/exiting the boundary of the system is given by:
v 2
e + Pv =u + + gz + Pv Looks familiar ??
2
Specific enthalpy of a system is defined as sum of specific internal
energy and the specific flow work:
h= u + Pv Intensive property

v 2
e + Pv =h + + gz
2
If we neglect the kinetic and potential energy terms, then the fluid flow
across a control volume (open system) produces an energy interaction
equal to its specific enthalpy
e + Pv =
h
22
First Law for control volume (open system)
m out
m in ECV hout
hin vout
vin zout
zin Q W
Control volume

dECV  vin2   vout

2
  
= m in  hin + + gzin  − m out  hout + + gzout  + Q − W
dt  2   2 

 vCV 2 
E=
CV mCV  uCV + + gzCV 
 2 
23
First Law for control volume : Transient
m out
hout
ECV
m in
hin

Q W
Control volume

Neglecting K.E and P.E.

dECV
= m in hin − m out hout + Q − W
dt
dECV d ( mCV uCV )
ECV = mCV uCV =
dt dt
24
 First Law for control volume: SSSF
Steady state: The specific energy of the C.V remains constant with time
 vCV 2 
E=
CV mCV  uCV + + gzCV 
 2

CONSTANT
 vCV 2 
Steady flow: No change in mass of the C.V.:=
ECV mCV  uCV +

+ gzCV 
CONSTANT  
2
m in m=
= out m

Steady State and Steady flow (SSSF):

dEC .V  vin2   vout

2
  
= m in  hin + + gzin  − m out  hout + + gzout  + Q − W
dt  2   2 
 vin2   vout
2
  
=0 m  hin + + gzin  − m  hout + + gzout  + Q − W
 2   2 
25
First Law for control volume : SSSF
m out
hout
m in ECV vout
hin zout
vin
zin Q W
Control volume

SSSF (Steady state steady flow process)

 vin2   vout
2

 hin + + gzin  −  hout + + gzout  + q − w =0
 2   2 
Q
Heat transfer per unit mass (kJ/kg) of fluid entering: q =
m
26
 First Law for control volume : SSSF
m
hout
ECV
m
hin

Q W
Control volume

Neglect change in K.E and P.E. hin − hout + q − w =0

Please note that here “w” includes all modes of work except “flow work”
Flow work already included in “Enthalpy” term

27
 Enthalpy of saturated mixture
In two-phase region: =
H H liq + H vap
=
Mh M liq hliq + M vap hvap

Similarly for specific enthalpy we can write: (1 x ) h f + xhg

h =−

Specific heat capacities

 ∂u 
Specific heat capacity at constant volume: cv =  
 ∂T v
 ∂h 
Specific heat capacity at constant pressure: cP =  
 ∂T  P
In case of ideal gas: ⇒ dh =
c p dT 28
 Examples of SSSF
1. Pump/Compressor/Turbine

pi > pe pi < pe

Assumption: Adiabatic (only work transfer)

0= m ( hi − he ) − W
29
Examples of SSSF
2. Heat Exchangers

Assumptions:
1. Each fluid stream is isobaric
2. No energy interaction of heat exchanger with
surroundings

=p1 p=
2, p3 p4

( m h + m fluid ,1h3 ) − ( m fluid ,2 h2 + m fluid ,1h4 ) =

fluid ,2 1 0
30
 Examples of SSSF
3. Nozzles/Diffusers

v1 Diffuser v2

Assumptions: 1
h1 + v12 =
1
h2 + v22
1. Adiabatic 2 2
2. No work transfer
3. K.E of fluid cannot be neglected
31
 Examples of SSSF
4. Throttling process

i e

pe < pi

i e

Assumptions:
1. Adiabatic hi = he
2. No work transfer
3. Exit pressure is lower than inlet pressure
32
 Example 1
Superheated steam at 15 bar and 350°C is allowed
to enter a steam turbine assembly. The exit happens
at saturated vapor condition at 10 kPa pressure.
Assume that the turbine is adiabatic, what is the
specific work generated by this turbine ? Ans: 563.4 kJ/kg

Sol: dE=cv
m in hin − m out hout + Q − Wshaft
dt
SSSF device with no heat interaction
dEcv
=m in hin − m out hout + Q − Wshaft
dt

 out + Wshaft =
mh  in
mh m in m=
= out m

33
 Example 2
Air at 10°C and 80 kPa enters the diffuser of a jet
engine steadily with a velocity of 200 m/s. The inlet
area of the diffuser is 0.4 m2. Air leaves the diffuser
with a velocity that is very small compared with the
inlet velocity. Determine:
(a) the mass flow rate of the air Ans: 78.8 kg/s, 30°C
(b) the temperature of the air leaving the diffuser

Sol: dEcv
= m in hin − m out hout + Q − W
dt  vin2   vout
2

 hin +  =  out
h + 
SSSF device with no heat and work interaction  2   2 
dEcv  vin2   vout
2
 
= m in  hin +  − m in  hout +  + Q − W
dt  2   2 

34
 Example 3
A hot-water stream at 80°C enters a mixing chamber with a mass flow
rate of 0.5 kg/s where it is mixed with a stream of cold water at 20°C. If
it is desired that the mixture leave the chamber at 42°C, determine the
mass flow rate of the cold-water stream. Assume all the streams are at
a pressure of 250 kPa. Ans: 0.865 kg/s

Sol: dEcv
= m 1h1 + m 2 h2 − m 3 h3 + Q − W
dt
SSSF device with no heat and work interaction
dEcv
= m 1h1 + m 2 h2 − m 3 h3 + Q − W
dt
Energy balance: m 1h1 + m 2 h2 =
m 3 h3 Mass balance: m 1 + m 2 =
m 3

35
 Example 4
In the condenser of a room air-conditioner, refrigerant-134a at 1 MPa and
90°C is to be cooled to 1 MPa and 30°C by air. The air enters at 100 kPa
and 27°C with a volume flow rate of 600 m3/min and leaves at 95 kPa and
60°C. Determine the mass flow rate of the refrigerant.
Ans: 1.67 kg/s
dEcv
= m 1h1 − m 2 h2 + m 3 h3 − m 4 h4 + Q − W
dt

SSSF device with no heat and work interaction

dEcv
= m 1h1 − m 2 h2 + m 3 h3 − m 4 h4 + Q − W
dt

Energy balance: m 1h1 + m 3 h3 = m 2 h2 + m 4 h4

Mass balance:
= m 1 m=
2 , m 3 m 4

36
 Example 5
Refrigerant R134a enters a throttling valve at 1017 kPa pressure and 30°C.
It leaves at 416 kPa pressure. Determine the refrigerant quality at exit of
throttle. Ans: 0.15

37
 Example 6
A 25-L tank, shown in Fig. that is initially evacuated is connected by a
valve to an air supply line flowing air at 20◦C, 800 kPa. The valve is
opened, and air flows into the tank until the pressure reaches 600 kPa.
Determine the final temperature and mass inside the tank, assuming the
process is adiabatic. Develop an expression for the relation between the
line temperature and the final temperature using constant specific heats.
Ans: 410 K, 0.1275 kg

d ( mu )cv
dEcv
= m in hin + Q − W ⇒ m in hin + Q − W
=
dt dt
t
d ( mu )cv t
 dmin 
⇒∫ ⇒ ( minu )cv ,t − ( mu )cv ,0 = min hin
0
dt ∫0  dt  hin
=

c pTin
⇒ Tcv =
cv

38
 Example 7
A 750-L rigid tank, shown in Fig., initially contains water at 250°C, which
is 50% liquid and 50% vapor by volume. A valve at the bottom of the tank
is opened, and liquid is slowly withdrawn. Heat transfer takes place such
that the temperature remains constant. Find the amount of heat transfer
required to reach the state where half of the initial mass is withdrawn.
Ans: 6750 kJ

39
 Example 8
A 200-L tank initially contains water at 100 kPa and a quality of 1%. Heat
is transferred to the water, thereby raising its pressure and temperature.
At a pressure of 2 MPa, a safety valve opens and saturated vapor at 2
MPa flows out. The process continues, maintaining 2 MPa inside until the
quality in the tank is 90%, then stops. Determine the total mass of water
that flowed out and the total heat transfer. Ans: 8.9 kg, 25.5 MJ

40
 Example 9
Pin = 15 bar
Superheated steam at a steady pressure of 15 bar and Tin = 350C
temperature of 350°C enters steam turbine. At the exit,
a tank of 50 m3 volume is kept which is initially empty.
Assume both turbine and tank to be thermally
insulated and steam doesn’t accumulate in the turbine.
Find the total work (in kJ) generated by this turbine and
the total mass that entered the tank when tank’s
pressure is 10 kPa and 50°C? Ans: 2367 kJ, 3.363 kg
V=50 m3

dEcv d ( mu )cv Initially empty

= m in hin + Q − W ⇒ m in hin + Q − Wshaft
= Pfinal = 10 kPa
dt dt
T final = 50C
d ( mu )cv  dm  d ( mu )cv  dm 
⇒ =  in  hin − Wshaft ⇒ =  in  hin − Wshaft
dt  dt  dt  dt 

⇒ ( mu )cv ,2 − ( mu=
)cv,1 min hin − Wshaft mcv ,2 − mcv ,1 = min
41