Heat and Work in Thermodynamics

CHAPTER 3
HEAT ,WORK & 1ST LAW OF
THERMODYNAMICS
By: Natnael Mesfin (Msc.)

Introduction
• The forms of energy already discussed, in chp.1, which
constitute the total energy of a system, can be contained
or stored in a system, and thus can be viewed as the static
forms of energy.
• The forms of energy not stored in a system can be viewed
as the dynamic forms of energy or as energy interactions.
• The dynamic forms of energy are recognized at the
system boundary as they cross it, and they represent the
energy gained or lost by a system during a process.

cont.
• The only two forms of energy interactions associated
with a closed system are heat transfer and work.
• An energy interaction is heat transfer if its driving
force is a temperature difference.
 Mechanical energy
 Many engineering systems are designed to transport a
fluid from one location to another at a specified flow
rate, velocity, and elevation difference, and the system
may generate mechanical work in a turbine or it may
consume mechanical work in a pump or fan during this
process.

Cont.
• Mechanical energy can be defined as the form of energy
that can be converted to mechanical work completely
and directly by an ideal mechanical device such as an
ideal turbine.
• Kinetic and potential energies are the familiar forms of
mechanical energy.
• A pump transfers mechanical energy to a fluid by raising
its pressure, and a turbine extracts mechanical energy
from a fluid by dropping its pressure.
• Therefore, the pressure of a flowing fluid is also
associated with its mechanical energy.

Cont.
• Note that; pressure itself is not a form of energy but a
pressure force acting on a fluid through a distance
produces work, called flow work, in the amount of P/ρ
per unit mass.
• The mechanical energy of a flowing fluid can be
expressed on a unit mass basis as:
Where; P/ρ is the flow energy, is the kinetic energy,
and gz is the potential energy of the fluid, all per unit mass.
It can also be expressed in rate form as

Energy Transfer by Heat
Energy can cross the boundary of a closed system in two
distinct forms: heat and work .
• It is important to distinguish between these two forms of
energy.
• Heat is defined as the form of energy that is transferred
between two systems (or a system and its surroundings)
by virtue of a temperature difference.

….
• Heat is energy in transition. It is recognized only as it
crosses the boundary of a system.
 a system exchanges no heat with its surroundings.
A process during which
there is no heat transfer is
called an adiabatic process

Cont.
 There are two ways a process can
be adiabatic when:
 The system is well insulated so that
only a negligible amount of heat
can pass through the boundary
 Both the system and the
surroundings are at the same
temperature and therefore there is
no driving force (temperature
difference) for heat transfer.
 Temperature difference is the
driving force for heat transfer.
The larger the temperature
difference, the higher is the
rate of heat transfer.

Cont.
• The amount of heat transferred during the process
between two states (states 1 and 2) is denoted by Q1-2 or
just Q.
• Heat transfer per unit mass of a system is denoted q and
is determined from:
Sometimes it is desirable to know the rate of heat
transfer(the amount of heat transferred per unit time)

Cont.
 Heat is transferred by three mechanisms:
 conduction,
 convection, and
 radiation

Energy Transfer by Work
• Work, like heat, is an energy interaction between a system
and its surroundings.
• As mentioned earlier, energy can cross the boundary of a
closed system in the form of heat or work.
• Therefore, if the energy crossing the boundary of a closed
system is not heat, it must be work.
• More specifically, work is the energy transfer associated
with a force acting through a distance.

Cont.
• A rising piston, a rotating shaft, and an electric wire
crossing the system boundaries are all associated with
work interaction
• The work done during a process between states 1&2 is
denoted by W12 or simply W.
• The work done per unit mass of a system is denoted by
w and is expressed as;
The rate at which work is done by, or upon, the system is
known as power. The unit of power is J/s or watt.

Cont.
Heat and work are directional quantities, and thus the
complete description of a heat or work interaction
requires the specification of both the magnitude and
direction.
• One way of doing that is to adopt a sign convention. The
generally accepted formal sign convention for heat and
work interactions is as follows:
 Heat transfer to a system and work done by a system are
positive;
 Heat transfer from a system and work done on a system
are negative.

Boundary Work
• One form of mechanical work frequently encountered in
practice is associated with the expansion or compression
of a gas in a piston–cylinder device.
• During this process, part of the boundary (the inner face
of the piston) moves back and forth. Therefore, the
expansion and compression work is often called moving
boundary work, or simply boundary work.

Cont.
• Consider the gas enclosed in the piston–cylinder device
shown in Fig. below. The initial pressure of the gas is P,
the total volume is V, and the cross-sectional area of the
piston is A.
• If the piston is allowed to move a distance ds in a quasi-
equilibrium manner, the differential work done during
this process is:

Cont.
• The total boundary work done during the entire
process as the piston moves is obtained by adding
all the differential works from the initial state to
the final state
The total area A under the process curve 1-2 on fig.
below is obtained by adding these differential areas:
The above Equation reveals that the area under the process curve on
a P-V diagram is equal, in magnitude, to the work done during a
quasi-equilibrium expansion or compression process of a closed
system.

…
 The area under the process curve on a P-V diagram
represents the boundary work.

 Constant volume ((Isochoric process)
• If the volume is held constant, dV = 0, and the
boundary work equation becomes:
EXAMPLE
1) A rigid tank contains air at 500
kPa and 150°C. As a result of heat
transfer to the surroundings, the
temperature and pressure inside the
tank drop to 65°C and 400 kPa,
respectively. Determine the
boundary work done during this
process.
Solution : Since a rigid tank has dV = 0 Therefore, there is no
boundary work done during this process. i.e the Wb during a constant-
volume process is always zero.

 Constant pressure (Isobaric)
• If the pressure is held constant, the boundary work
equation becomes:
 Constant temperature, ideal gas (isothermal)
If the temperature of an ideal gas system is held
constant, or PV=C then the equation of state provides
the pressure-volume relation
Then, the boundary work is

 Polytropic Process
During actual expansion and compression processes of
gases, pressure and volume are often related by PVn = C,
where n and C are constants. A process of this kind is called
a polytropic process.
The pressure for a polytropic process can be expressed as
Substituting this relation into Wb equation we obtain;
since

For the special case of n = 1 the boundary work becomes
when the ideal gas equation of state is introduced into the
general equation of work the following expressions are
obtained, respectively

Example
2) A piston–cylinder device initially contains 0.4m3of air
at 100kPa and 80°C.The air is now compressed to 0.1m3in
such a way that the temperature inside the cylinder remains
constant. Determine the work done during this process
Solution
The (-) sign indicates that this work is
done on the system

Examples
3) Air undergoes a polytropic compression in a piston-
cylinder assembly from P1=1bar,T1=22oC to P2=5bars.
Employing the ideal gas model, determine the work per
unit mass, in kJ/kg, if n=1.3
Solution
Assumptions:
 The air is a closed system.
 The air behaves as an ideal gas.
 The compression is polytropic with n=1.3.

Schematic and Given Data:
The work can be evaluated in this case from the expression

The temperature at the final state,T2, is required. This can be
evaluated from;
The work is then;

Exercise
1)Suppose that 10 moles of Oxygen gas are allowed to
expand isothermally @ temperature of 300 K from an
initial volume of 10 liters to a final volume of 30 liters.
How much work does the gas do on the piston?

2. Sketch a P-V diagram showing the following processes
in a cycle?
Process 1-2: isobaric work output of 10.5 kJ from an initial volume of
0.028 m3and pressure 1.4 bar
Process 2-3:isothermal compression and
process 3-1: isochoric heat transfer to its original volume of 0.028
m3and pressure 1.4 bar.
Calculate
(a) the maximum volume in the cycle, in m3
(b) the isothermal work, in kJ, (c) the net work, in kJ, and (d) the
heat transfer during isobaric expansion, in kJ.

Chapter 4:
First Law of Thermodynamics

…
• So far, we have considered various forms of energy such
as heat Q, work W, and total energy E individually, and
no attempt is made to relate them to each other during a
process.
• The first law of thermodynamics, also known as the
conservation of energy principle, provides a sound basis
for studying the relationships among the various forms of
energy and energy interactions.

The first law or the conservation of energy relation
with the help of some familiar examples;
• Consider some processes that involve heat transfer but
no work interactions.
The increase in the energy of
a potato in an oven is equal
to the amount of heat
transferred to it.
As a result of heat transfer to
the potato, the energy of the
potato will increase. If we
disregard any mass transfer
(moisture loss from the
potato), the increase in the
total energy of the potato
becomes equal to the amount
of heat transfer.

Cont.
• If 15 kJ of heat is transferred to the water from the
heating element and 3 kJ of it is lost from the water to
the surrounding air, the increase in energy of the water
will be equal to the net heat transfer to water, which is
12 kJ.
In the absence of any work interactions, the energy change of a
system is equal to the net heat transfer.

Energy Balance
• The net change (increase or decrease) in the total energy
of the system during a process is equal to the difference
between the total energy entering and the total energy
leaving the system during that process.
or
This relation is often referred to as the energy balance and
is applicable to any kind of system undergoing any kind of
process.

The successful use of this relation to solve engineering
problems depends on understanding the various forms of
energy and recognizing the forms of energy transfer.
Energy Change of a System, ∆Esystem
 The determination of the energy change of a system
during a process involves the evaluation of the energy of
the system at the beginning and at the end of the process,
and taking their difference. That is,
 Energy change = Energy at final state - Energy at
initial state
 The change in the total energy of a system during a process is the
sum of the changes in its internal, kinetic, and potential energies
and can be expressed as:

.
• Where
Most systems encountered in practice are stationary,
that is, they do not involve any changes in their velocity
or elevation during a process
Thus, for stationary systems the, ∆KE=∆PE=0), and the
total energy change relation reduces to ∆E = ∆U

Mechanisms of Energy Transfer
• Energy can be transferred to or from a system in three
forms: heat, work and mass flow.
 Mass Flow,(m ): Mass flow in and out of the system
serves as an additional mechanism of energy transfer.
Noting that energy can be transferred in the forms of heat,
work, and mass, and that the net transfer of a quantity is
given by;

 The energy transport with mass Emass is zero for systems
that involve no mass flow across their boundaries (i.e.
closed systems).
Energy Balance For Closed Systems
 The heat transfer Q is zero for adiabatic systems, the
work transfer W is zero for systems that involve no
work interactions, and
 Energy balance for any system undergoing any kind
of process can be expressed as
The rate form,

..
For a closed system undergoing a cycle, the initial and
final states are identical, and thus:
∆Esystem = E2 - E1 = 0.
Then the energy balance for a cycle simplifies to;
Ein - Eout = 0 or Ein = Eout.
Noting that a closed system does not involve any mass
flow across its boundaries, the energy balance for a cycle
can be expressed in terms of heat and work interactions
as:

Cont.
For a cycle ∆E =0
Thus, Q = W.
 Various forms of the first-law
relation for closed systems.

Examples
1.A rigid tank contains a hot fluid that is cooled while
being stirred by a paddle wheel. Initially, the internal
energy of the fluid is 800kJ. During the cooling process,
the fluid loses 500kJ of heat, and the paddle wheel does
100 kJ of work on the fluid. Determine the final internal
energy of the fluid.
Solution
Assumptions:
Tank is stationary and thus KE=PE=0.
• Analysis:
Q-W=U2-U1

Cont.
2. A closed system of mass 2kg undergoes an adiabatic
process. The work done on the system is 30kJ. The
velocity of the system changes from 3m/s to 15m/s.
During the process, the elevation of the system increases
45 meters. Determine the change in internal energy of
the system.
Solution….?exercise
3.Steam @ 1100kPa and 92 % quality is heated in a
rigid container until the pressure is 2000kPa. For a mass
of 0.05kg, calculate the amount of heat supply (in kJ) ?

Solution (3)
u1= uf1+xufg
=779.78+0.92(1805.7)
=2441.02kJ/kg
For a rigid container,
v2=v1=0.1634 m3/kg
State2
@ P=2000kpa
by linear interpolation we have

Specific Heats (Heating Capacity)
The specific heat is defined as the energy required to
raise the temperature of a unit mass of a substance by one
degree
In thermodynamics, we are interested in two kinds of
specific heats: specific heat at constant volume cv and
specific heat at constant pressure cp.
 The specific heat at constant volume cv can be viewed as
the energy required to raise the temperature of the unit
mass of a substance by one degree as the volume is
maintained constant.
o The energy required to do the same as the pressure is
maintained constant is the specific heat at constant
pressure cp.

Cont.
If we add heat to a system, there are two general
destinations for the energy:
 It will “heat up” the system (i.e., raise T).
It can make the system do work on the
surroundings.
Q = ∆U + W

Cont.
 The specific heat at constant pressure cp is always greater
than cv because at constant pressure the system is allowed
to expand and the energy for this expansion work must
also be supplied to the system.
 Consider a fixed mass in a stationary closed system
undergoing a constant-volume process.
 The conservation of energy principle ein - eout = esystem for
this process can be expressed in the differential form as:
 The left-hand side of this equation represents the net
amount of energy transferred to the system.

From the definition of Cv this energy must be equal to
Cv.dT, where dT is the differential change in temperature.
Thus;
or
Similarly, an expression for the specific heat at constant
pressure cp can be obtained

Cont.
Consider the two systems shown below. In Case I, the
gas is heated at constant volume; in Case II, the gas is
heated at constant pressure.
Compare QI, the amount of heat needed to raise the
temperature 1ºC in system I to QII, the amount of
heat needed to raise the temperature 1ºC in system II.

Cont.
In Case I, Wby = 0, because the volume does not
change.
In Case II, Wby > 0, because the gas is expanding.
Both cases have the same ∆U, because the
temperature rise is the same.
→ more heat is required in Case II
→ Cp > Cv

Internal Energy, Enthalpy, And Specific Heats Of Ideal
Gases
 Using the definition of enthalpy (h = u + pv) and
writing the differential of enthalpy, the relationship
between the specific heats for ideal gases is:
 The specific heat ratio, k is defined as;

.
 Joule found experimentally that the internal energy
of an ideal gas is a function of temperature only
Fig. Schematic of the experimental apparatus used by Joule.

.
For ideal gases u, h, Cv, and Cp are functions of
temperature alone. The Δu and Δh of ideal gases can be
expressed as
The change in internal energy or
enthalpy for an ideal gas during a
process from state 1 to state 2 is
determined by integrating these
equations

Reading Assignment
 INTERNAL ENERGY, ENTHALPY, AND
SPECIFIC HEATS OF SOLIDS AND LIQUIDS

.
The First Law of Thermodynamics
for
Open System (Control Volume)

Conservation of Mass
 Conservation of mass is one of the most fundamental
principles in nature.
 For control volume, mass can cross the boundaries. So the
amount of mass entering and leaving the control volume must be
considered.
Mass and Volume Flow Rates
 Mass flow through a cross-
sectional area per unit time
is called the mass flow
rate. The dot over the mass
symbol indicates a time
rate of change. It is
expressed as;

Principle of Conservation of Mass
• The conservation of mass principle for a control volume can be
expressed as
Control volume involves two main processes(steady and unsteady)
 steady-flow process, a process during which a fluid flows
through a control volume steadily.
 That is, the fluid properties can change from point to point
within the control volume, but at any point, they remain constant
during the entire process. (Remember, steady means no change
with time.)
 During a steady-flow process, no intensive or extensive
properties within the control volume change with time.

.These observations greatly simplify the analysis.
 The mass balance for a general steady-flow system is
given as
The mass balance for a single-stream (one-inlet and one-
outlet) steady-flow system was given as
 During a steady-flow process, the total energy content
of a control volume remains constant ECV = constant
,and thus the change in the total energy of the
control volume is zero (∆ECV = 0).
Multiple inlets
and exits
Single
stream

Cont.
 Thus, the volume V, the mass m, and the total energy
content E of the control volume remain constant .
 As a result, the boundary work is zero for steady-flow
systems (since VCV = constant), and the total mass or
energy entering the control volume must be equal to
the total mass or energy leaving it.
 (since mCV = constant and ECV = constant).

 Under steady-flow conditions,
the mass and energy contents
of a control volume remain
constant.
 Under steady-flow conditions,
the fluid properties at an inlet
or exit remain constant (do not
change with time).

Energy Balance for Steady-Flow Systems
Total Energy of a Flowing Fluid
 The total energy consists of three parts for a non-flowing fluid and
four parts for a flowing fluid.

Cont.
The fluid entering or leaving a control volume
possesses an additional form of energy - the flow
energy Pv, Then the total energy of a flowing fluid on a
unit-mass basis (denoted by ϴ) becomes
 Therefore, the amount of energy entering a control
volume in all forms (by heat, work, and mass) must
be equal to the amount of energy leaving it.

Then the rate form of the general energy balance reduces for a steady-
flow process to
Noting that energy can be transferred by heat, work, and mass only,
the energy balance for a general steady-flow system can also be
written more explicitly as

.
 The first-law or energy balance relation in that case for a general
steady-flow system becomes
 For single-stream devices, the steady-flow energy balance
equation becomes:
Dividing the above equation by m gives the energy balance on a
unit-mass basis as

Cont.
 Where q = 𝑄/ṁ and
w = Ẇ/ṁ - are the heat transfer and work
done per unit mass of the working fluid, respectively.
 When the fluid experiences negligible changes in its
kinetic and potential energies (that is, ∆ke ≈ 0, ∆pe ≈ 0),
the energy balance equation is reduced further to

Some Steady-flow Engineering Devices
Many engineering devices operate essentially under the
same conditions for long periods of time.
The components of a steam power plant (turbines,
compressors, heat exchangers, and pumps), for example,
operate non-stop for months before the system is shut
down for maintenance.
Therefore, these devices can be conveniently analyzed as
steady-flow devices.

.
Figure : Some of Steady-flow Engineering Devices

Nozzles and Diffusers
 Nozzle - device that increases the velocity fluid at the expense
of pressure.
 Diffuser - device that increases pressure of a fluid by slowing
it down.
Commonly utilized in jet engines, rockets, space-craft and even
garden hoses.
 The rate of heat transfer between the fluid flowing through a
nozzle or a diffuser and the surroundings is usually very small
( = 0) since the fluid has high velocities, and thus it does not spend
enough time in the device for any significant heat transfer to take
place.
 Nozzles and diffusers typically involve no work (Ẇ = 0) and any
change in potential energy is negligible (∆pe = 0).

Therefore, the kinetic energy changes must be accounted for in
analyzing the flow through these devices (∆ke ≠ 0).

Energy balance (nozzle & diffuser):

Turbines and Compressors
 In steam, gas, or hydroelectric power plants, the device that drives
the electric generator is the turbine. As the fluid passes through
the turbine, work is done against the blades, which are attached to
the shaft. As a result, the shaft rotates, and the turbine produces
work.
 Compressors, as well as pumps and fans, are devices used to
increase the pressure of a fluid. Work is supplied to these devices
from an external source through a rotating shaft. Therefore,
compressors involve work inputs. Even though these three devices
function similarly, they do differ in the tasks they perform.
 A fan increases the pressure of a gas slightly and is mainly used to
mobilize a gas. A compressor is capable of compressing the gas to
very high pressures. Pumps work very much like compressors
except that they handle liquids instead of gases.

Cont.
Fig: A modern land-based gas turbine used for electric power
production.

Note that turbines produce power output whereas compressors,
pumps, and fans require power input.
Turbine - a work producing device through the expansion of a fluid.
Compressor (as well as pump and fan) - device used to increase
pressure of a fluid and involves work input.
Q = 0 (well insulated),
ΔPE = 0, ΔKE = 0 (very small compare to ΔH).

 Energy balance For Turbine

Energy balance for compressor
Compressor
Inlet
Exit
Win

For pump:
In case of pump the following assumption can be made:

Throttling Valves
• Throttling valves are any kind of flow-restricting devices that cause
a significant pressure drop in the fluid. Some familiar examples are
ordinary adjustable valves, capillary tubes, and porous plugs.
 Throttling valves are usually
small devices, and the flow
through them may be assumed to
be adiabatic (q ≈ 0) since there is
neither sufficient time nor large
enough area for any effective heat
transfer to take place. Also, there
is no work done (w ≈ 0), and,
(∆pe ≈ 0)&(∆ke ≈ 0). if any, is
very small.

Mixing Chamber Mixing two or more fluids is a
common engineering process.
Mixing
Chamber
The mixing chamber does not have
to be a distinct “chamber.”
An ordinary T-elbow, or a Y-elbow
in a shower, for example, serves as
the mixing chamber for the cold-
and hot-water streams as shown in
the figure.

Energy Balance for Mixing Chamber
Mixing
Chamber
Mixing
Chamber
e
i
m m
 

 
Material Balance
3
2
1 m
m
m





Energy balance
i e e
i
m h m h
 

 
3
3
2
2
1
1 h
m
h
m
h
m






Heat Exchanger
As the name implies, heat exchangers are devices where
two moving fluid streams exchange heat without mixing.
Heat exchangers are widely used in various industries, and
they come in various designs.
 Heat exchangers typically involve
no work interactions (w=0) and
negligible kinetic and potential
energy changes (∆ke ≈ 0, ∆pe ≈ 0)
for each fluid stream.

.
Mass Balance
with equal inlet and outlet 1 2 3 4
,
m m m m
 
 The heat transfer rate associated with heat exchangers depends on
how the control volume is selected. Heat exchangers are intended
for heat transfer between two fluids within the device, and the outer
shell is usually well insulated to prevent any heat loss to the
surrounding medium.

.
• Mass Balance
– Considering one single stream
– with one inlet and one outlet flow rates
 Energy balance
– One inlet
– One outlet
– Plus heat transfer
1 2
m m

2 2 1 1
Q m h m h
 
1 1 3 3 2 2 4 4
m h m h m h m h
  
Single stream
Energy balance
For two inlets & two outlets

Example and Exercise
1.Steam at 0.4 Mpa, 300ºC, enters an adiabatic nozzle with a low
velocity and leaves at 0.2Mpa with a quality of 90%. Find the
exit velocity?
 solution
Simplified energy balance

.
2.Air at 10°C and 80 kPa enters the diffuser of a jet engine steadily
with a velocity of 200m/s. the inlet area of the diffuser is 0.4 m2.
The 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 and (b) the temperature of
the air leaving the diffuser?
Solution
(a)

3. Air enters a 28cm radius pipe steadily at 200kpa and 20°C with a
velocity of 5 m/s. Air is heated as it flows, and leaves the pipe at
180kpa and 40°C. Determine
a. The volume flow rate of air at the inlet,
b.The mass flow rate of air, and
c. the velocity and volume flow rate at the exit.
 Use the gas constant for air 0.287kJ/kg.k

(c) Noting that mass flow rate is constant, the volume flow rate
and the velocity at the exit of the pipe are determined from

4.The power output of an adiabatic steam turbine is 5 MW, and the
inlet and the exit conditions of the steam are as indicated in the figure.
a) Compare the magnitude of h,
ke, and pe.
b) Determine the work done per
unit mass of the steam flowing
through the turbine.
c) Calculate the mass flow rate of
the steam.

5.Air at 100 kPa and 280 K is compressed steadily to 600 kPa and
400 K. The mass flow rate of the air is 0.02 kg/s, and
a heat loss of 16 kJ/kg occurs during the process .Assuming the
changes in kinetic and potential energies are negligible, determine
the necessary power input to the compressor.
solution

6.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 250kPa
Noting that T < Tsat @ 250 kPa =
127.41°C, the water in all three
streams exists as a compressed
liquid,
which can be approximated as a
saturated liquid at the given
temperature.
Solution
h1 ≅ hf @ 80°C = 335.02 kJ/kg
h2 ≅ hf@ 20°C = 83.915 kJ/kg
h3 ≅ hf @ 42°C = 175.90 kJ/kg

 The mass and energy balances for this steady-flow system can be
expressed in the rate form as;

Exercise
1.Steam at 5 Mpa and 400°C enters a nozzle steadily with a velocity
of 80 m/s, and it leaves at 2 Mpa and 300°C. The inlet area of the
nozzle is 50 cm2, and heat is being lost at a rate of 120 kJ/s.
Determine;
a) the mass flow rate of the steam,
b) the exit velocity of the steam, and
c) the exit area of the nozzle.

Exercise
2.Steam enters a turbine at steady state with a mass flow rate of 4600
kg/h. The turbine develops a power output of 1000kW. At the inlet
the pressure is 0.05 Mpa, the temperature is 400 °C, and the velocity
is 10m/s. At the exit, the pressure is 10kPa, the quality is 0.9, and the
velocity is 50m/s. Calculate the rate of heat transfer between the
turbine and surroundings, in kW.

Assignment(2)
1.Write short note on Unsteady (Transient) flow process?

Heat and Work in Thermodynamics

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