3) Chapter 2 4 Energy Transfer Analysis

CDB 1053 Introduction to Engineering Thermodynamics

1

Chapter 2 & 4: Energy Transfer & Analysis of Closed Sysyems

By the end of this lecture, you are expected to:

• Understand the concept of energy & its forms.

• Understand mechanism of energy transfer &

concept of electrical & mechanical work.

• Apply the 1st Law of Thermodynamics &

energy balance of a system to closed systems

2

Learning Outcome

CDB 1053 Into to Eng Thermodynamics

3

Contents


2 - 3 Heat transfer

2 - 4 Energy transfer by work

2 - 5 Mechanical forms of work

2 - 6 Conservation of mass principle

2 - 7 Flow work & the energy of a flowing fluid

4CDB 1053 Into to Eng Thermodynamics

2-3 Heat transfer

The first law of thermodynamics

the conservation of energy principle.

Closed System

Energy can cross the boundaries of a closed system in the form of heat or work.

Energy


Heat

E.g. when a body is left in a medium that is at a differenttemperature, energy transfer takes place between the body &surrounding medium until thermal equilibrium is established(reach the same T).

Direction of energy transfer: Higher T to Lower T. Heat addition & heat rejection.

Energy is recognized as heat transfer only as it crosses the system boundary.


Adiabatic process

How ? (1) System is well insulated

(2) Both system & surroundings

are at same T

Mechanisms of heat transfer:

(a) Conduction

transfer of energy from more energetic particles

of a substance to the adjacent less energetic ones

as a result of interaction between particles.

(b) Convectiontransfer of energy between a solid surface & adjacent fluid that is in motion & involves the combined effects of conduction & fluid motion.

(c) Radiationtransfer of energy due to the emission of electromagnetic waves(or photons)


The net heat transferred to a system is defined as

Unit of heat kJ(or Btu)Heat transfer per unit mass of a system (kJ/kg)

indicate amount of heat transfer per unit timeunit kJ/s = kW

2-4 Energy transfer by workIf energy crossing the boundary of a closed system is not heat, it must be work.Work is the energy transfer associated with a

force acting through a distance.e.g. rotating shaft, rising piston, electricwire

Work done per unit mass of a system is denoted by w & expressed as

(kJ/kg)

outinnet QQQ

m

Qq

Q

m

Ww

Work done per unit mass, power (W) unit kJ/s or kW Heat & work requires the specification of both magnitude &

direction. Formal sign convention:

heat transfer to a system heat transfer from a systemwork done by a system work done on a system

use subscripts in & out to indicatedirection

e.g a work input = Win

a heat loss = Qout

can assume a direction for interaction +ve assumed direction is right -ve direction of interaction is the opposite


2-5 Mechanical forms of work

The work done is proportional to the force applied (F) & the distance traveled (s).

W = Fs (kJ)

2 requirements for a work interaction between a system & its surrounding to exist:(1) the boundary must move(2) there must be a force acting on the boundary

Some common forms of mechanical work are :

W F ds Fds

cos


(1) Moving boundary work (boundary work/ P dV work) associated with the expansion & compression of a gas in a piston-

cylinder devise. Piston is allowed to move a distance ds in a quasi-equilib. manner, the differential work done during process,

(valid for quasi-equilibrium process)P = absolute pressuredV = volume change (+ve expansion)

(-ve compression) To calculate the boundary work, the process by which the

system changed states must be known. I.e. functional relationship between P & V during process.

P = f(V) should be available: equation of the process path on a P-V diagram.


The boundary work = area under the process curve plotted on pressure-volume diagram

Total boundary work from initial

state to final state:

(kJ)

Note:P is the absolute pressure and is always positive.

When dV is positive, Wb is positive.When dV is negative, Wb is negative

2

1

PdVWb



each process gives a different value for the boundary work

The boundary work done during a process depends on the path followed as well as the end states.


Some typical processes:

(a) Boundary work during a constant-volume process

If the volume is held constant, dV = 0, boundary work equation becomes

01

2 PdVWb

P

V

1

2

P-V diagram for V = Constant


(b) Boundary work for a constant-pressure process

12

2

1

2

1VVPdVPPdVW ob

P

V

2 1

P-V diagram for P = constant

For the constant pressure process shown above, is the boundary work positive or negative and why?


(c) Isothermal compression of an ideal gas

If the temperature of an ideal gas system is held constant, then the equation of state provides the temperature-volume relation:

Then, the boundary work is

Note: The above equation is the result of applying the ideal gas assumption for the equation of state. For real gas undergoing an isothermal process, the integral in the boundary work equation would be done numerically.

V

mRTP

dVV

mRTPdVWb

2

1

2

1

1

2lnV

VmRT


(d) Polytropic process

During expansion & compression processes of gases, P & V are often related by: PVn = C where C are constants

The value of n - to + depending on process

Some of the common values are given below:

Process Exponent nConstant pressure 0Constant volume

Isothermal & ideal gas 1 Adiabatic & ideal gas k = CP/CV

Cp = specific heat at constant pressure

Cv = specific heat at constant volume


Schematic & P-V diagram for a polytropic process:

How to determine the boundary work for polytropic process?

2

1PdVWb

nCVP

dVV

Cn

2

1


For an ideal gas (PV = mRT), this equation can also be written as

For special case of n = 1 the boundary work becomes

2

1

1dV

VC

n

1

1

1

1

2

n

VVC

nn

n

VPVP

1

1122

since C = P1V1n =P2V2

n

1,

1

12

n

n

TTmRWb

1

2lnV

VmRTWb


(2) Shaft work

Energy transmission through rotating shaft

For a specified constant torque, the work done during n revolutions:

A force F acting through a moment arm r generates a torque T of

T = Fr F = T/r

This force acts through a distance s, related to radius r by

s = (2r)n

Shaft work is determined from Wsh = Fs (kJ)

The power transmitted through the shaft is the shaft work per unit time, which can be expressed as

(kW)

where n is the no. of revolutions per unit time.

TnWsh 2


(3) Spring work

Work done, when the length of a spring changes by a differential amount dx under the influence of a force F is

For linear elastic spring, the displacement x is proportional to the force applied, that is

F = kx (kN)

where k = spring constant & unit kN/m

Substitution & integrate

(kJ)

where x1 & x2 are the initial & final displacements of the spring, measured from the undisturbed position of the spring.

FdxWspring


Problem 3-15, Cengel

A frictionless piston-cylinder device initiallycontains 200L of saturated liquid refrigerant-134a. The piston is free to move, and its mass issuch that it maintains a pressure of 800 kPa onthe refrigerant. The refrigerant is now heateduntil its temperature rises to 50oC. Calculatethe work done during this process.



A mass of 1.2 kg of air at 150 kPa and12oC is contained in a gas-tight,frictionless piston-cylinder device.The air is now compressed to a finalpressure of 600 kPa. During theprocess, heat is transferred from theair such that the temperature insidethe cylinder remains constant.Calculate the work done during thisprocess.



Nitrogen at an initial state of 300 K,150 kPa and 0.2 m3 is compressedslowly in an isothermal process to afinal pressure of 800 kPa.Determine the work done during thisprocess.



A gas is compressed from an initial volume of 0.42 m3 to afinal volume of 0.12 m3. During the quasi-equilibriumprocess, the pressure changes with volume according tothe relation P = aV + b, where a = -1200 kPa/m3 and b =600 kPa. Calculate the work done during this process

a) by plotting the process on a P-V diagram and finding the area under the process curve, and

b) by performing the necessary integrations.



During some actual expansion andcompression processes in piston-cylinderdevices, the gases have been observedto satisfy the relationship PVn=C, wheren and C are constants. Calculate thework done when a gas expands from astate of 150 kPa and 0.03 m3 to a finalvolume of 0.2 m3 for the case of n = 1.3.


2-6 Conservation of mass principle

One of the most fundamental principles in nature.

Mass, like energy is a conserved property, & it cannot be created or destroyed. Mass (m) & energy (E) can be converted to each other:

Mass of a system will change when its energy changes.

For closed system mass of the system remains constant during process

For control volumes mass can cross boundaries

(open system)

2mcE


Mass flow rate

The amount of mass flowing through a cross section area per unit time is called mass flow rate,

where is the mean fluid velocity normal to A, m/s

is the density of fluid, kg/m3

A is cross sectional area normal to flow direction, m2

Actual & mean velocity profiles for flow in a pipe (the mass flow rate is the same for both cases)

m

AVm m

mV


Volume flow rates,

The volume of fluid flowing through a cross section per unit time

(m3/s)

The mass & volume flow rates are related by

(kg/s)

Example:

Air at 100 kPa, 50 C flows through a pipe with a volume flow rate of 40 m3/min. Find the mass flow rate through the pipe.

Assume air to be an ideal gas,

V

AVV m

VVm


Conservation of mass principle can be expressed as: net mass transfer to or from a system during a process is equal to the net change (increase or decrease) in the total mass of the system during that process.

kJ

kPam

kPa

K

kgK

kJ

P

RTv

3

100

)27350(287.0

kg

m3

9270.0

s

kg

skgm

m

v

Vm

719.0

60

min1

/9270.0

min/403

3

Total mass entering the

system

Total mass leaving the

system

Net change in mass within the system

- =


or

where rate of change of mass within the system boundaries.

Mass balance for control volume can be expressed more explicitly as

where i = inlet; e = exit; 1 = initial stage; 2 = final state of

control volume Conservation of mass equation is often referred to as continuity

equation in fluid mechanics.

)/(

)(

skgmmm

kgmmm

systemoutin

systemoutin

systemm

systemei

systemei

mmm

and

mmmm

)( 12


Mass balance for steady-flow processes

Total amount of mass contained within a control volume does not change with time (mCV = constant)

Conservation of mass requires that the total amount of mass entering a control volume equal the total amount of mass leaving it.

For a general steady-flow system with

multiple inlets & exists:

Total mass entering = Total mass leaving

CV per unit time CV per unit time

Steady flow: (kg/s) ei mm


For single-stream steady-flow systems:

E.g. nozzles, diffusers, turbines, compressors & pumps

21 mm 222111 AvAv


2-7 Flow work & energy of a flowing fluid

Control volume involve mass flow across their boundaries, & some work is required to push the mass into or out of the control volume.

Flow work (flow energy) necessary for maintaining continuous flow through a control volume.

Force applied,

Work flow (kJ) is

Work flow per unit mass,

(kJ/kg)

PAF

PVPALFLWflow

Pvw flow


Total energy of a flowing fluid

Total energy of a simple compressible system (internal,kinetic & potential energy):

on a unit-mass basis: (kJ/kg)

For flowing fluid, an additional form of energy Flow energy (Pv)

Total energy of a flowing fluid on a unit-mass basis, ( )

But Pv + u = h (enthalpy), this reduces to

(kJ/kg)

gzv

upekeue 2

2

)( pekeuPvePv


Energy transport by mass

amount of energy transport:

(kJ)

rate of energy transport:

(kW)

)2

(2

gzv

hmmEmass

)2

(2

gzv

hmmEmass

3) Chapter 2 4 Energy Transfer Analysis

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