LECTURE 5 cor 1

8/21/2019 LECTURE 5 cor 1

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LECTURE 5

(Fluid Mechanics and Thermodynamics)

1. Fluid Mechanics

1.1 PRIMARY THERMODYNAMIC PROPERTIES OF FLUIDS

Pressure (p) - is the (compression) stress at a point in a static fluid.

kPaapsiapatm 3.1017.14 Temperature (T) is a measure of the internal energy level of a fluid.

460 FR

273 CK

Density () - is its mass per unit volume.

V

m

Specific weight () - is its weight per unit volume.

V

mg

V

Wg

Specific gravity (SG) - is the ratio of a fluid density to a standard reference fluid, water (for liquids), and air (fo

gases).

air

gasgasSG

320.1 mkgair at 101.3 kPa and 21 C.

wat er

liquidliquidSG

31000 mkgwater

Example No. 1

What is the sea level (g = 32.2 ft/sec2) specific weight (lbf/ft3) of liquids with densities of 65.5 lbm/ft3?

A. 32.2

B. 42.2C. 65.5

D. 76.7

Solution:

cg

ftlbmindensitygg

3

2

3

2

232

565232

sec.

.sec.

lbfftlbm

ftlbmft

35.65 ftlbf (C)

1.2 VISCOSITY

Viscosity (1) - is that property of a real fluid by virtue of which it offers resistance to shear force.

Viscosity (2) is the fluid resistance to flow or the property of fluid to resist shear deformation.

Newtons law of viscosity states that the shear force to be applied for a deformation rate of ( dV/dy) over an area

A is given by,

dydVAF


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2

where F is the applied force in N, A is area in m2, dV/dy is the velocity gradient (or rate of deformation), 1/s

perpendicular to flow direction, here assumed linear, and is the proportionality constant defined as the

dynamic or absolute viscosity of the fluid.

Shear Stress = dydVAF Shear Strain = AFdydV

Viscosity Index is the rate at which viscosity changes with temperature.

Viscosimeter an instrument, consisting of standard orifice, used for measuring viscosity (in SSU and SSF).

Absolute (Dynamic) Viscosity is the viscosity determined by direct measurement of shear resistance (in Poise

or centiPoise.) Units are 1 reyn = 1 lb-sec/in2, 1 Poise = 1 dyne-sec/cm2 = 0.1 Pa-sec. 1 centiPoise (cP) = 0.01

Poise.

Kinematic Viscosity it the absolute viscosity of a fluid divided by the density (in Stoke or centiStoke.) Units are

ft2/s, m2/s, 1 stoke = 1 cm2/sec. = 0.0001 m2/sec. 1 centiStoke (cSt) = 0.01 Stoke.

Example No. 2

An oil has a kinematic viscosity of 1.25 x 10-4 m2/s and a specific gravity of 0.80. What is its dynamic (absolute)

viscosity in kg/(m-s)?A. 0.08

B. 0.10

C. 0.125

D. 1.0

Solution:

= 1.25 x 10-4 m2/s

SG = 0.80

= x

= SG xw x

= 0.80 x (1000 kg/m3) x (1.25 x 10-4 m2/s)

= 0.10 kg/(m-s)

1.3 REYNOLDS NUMBER

Reynolds number is a dimensionless number which is the ratio of the forces of inertia to viscous forces of the

fluids. It is the primary parameter correlating the viscous behaviour of all newtonian fluids.

v

DVDV

ForcesViscous

inertiaofForces

Re

where:

Re = Reynolds number, dimensionless

D = inside diameter, mV = velocity, m/s

= kinematic viscosity, m2/s

= absolute viscosity, Pa-sec

Example No. 3

Water is flowing in a pipe with radius of 12 inches and a velocity of 11 m/sec. The viscosity of water is 1.131 Pa

sec. What is the Reynolds Number?

A. 2964


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3

B. 5930

C. 58

D. 9189

Solution:

vDRe

59301311

113739

122

..

Re (B)

1.4 TYPES OF FLOW

Laminar Flow particles run parallel to each other. Laminar flow occurs if the Reynolds number is less than

2000.

Turbulent Flow particles run not in same direction. Turbulent flow occurs if the Reynolds number is greate

than 4000. Fully turbulent occurs at very high Reynolds number.

Transitional Flow also termed as critical flow in which this type of flow occurs if the Reynolds number is

between 2000 to 4000.

1.5 SURFACE TENSION AND CAPILLARY ACTION

Surface Tension is the membrane formed on the free surface of the fluid which is due to cohesive forces. Thereason why insects were able to sit on water is due to surface tension. The amount of surface tension decreases

as the temperature increases.

Capillary Action this is done through the behaviour of surface tension between the liquid and a vertical solid

surface.

1.6 COMPRESSIBILITY AND BULK MODULUS

Compressibility, - the measure of the change in volume of a substance when a pressure is exerted on thesubstance.

P

V

V

o

Where:V = change in volumeVo = original volume

P = change in pressure

Bulk modulus, EB - is defined as the ratio of the change in pressure to the rate of change of volume due to the

change in pressure. It is the inverse of compressibility.

1.7 HYDROSTATIC PRESSURE

Hydrostatic Pressure is the pressure of fluid exerted on the walls of the container.

Notes:

a. Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is

independent of the shape of the container. The pressure is the same at all points on a given horizontal planein the fluid. The pressure increases with depth in the fluid.

b. Any two points at the same elevation in a continuous mass of the same static fluid will be at the same

pressure.


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Pressure = Weight Density x Height

ghhp

Pressure Head,

gpph

where:

p = hydrostatic pressure (gage pressure)

h = height of liquid (pressure head)

= Density of liquid

Example No. 4

On a sea-level standard day, a pressure gage, moored below the surface of the ocean (SG = 1.025), reads an

absolute pressure of 1.4 MPa. How deep is the instrument?

A. 4 m

B. 129 m

C. 133 m

D. 140 m

Solution:

hSGhghpp watmabs

HmkNmkNmkN 322 81902513251011400 ... mh 129

Example No. 5

If the absolute pressure at the bottom of the ocean is 300 kPa, how deep is the water at this point?

A. 16.66 m

B. 19.66 m

C. 29.66 m

D. 39.66 m

Solution:

a tmga b s PPP

325.101300 gP


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5

kPaPg 675.198

mGS

Ph

g66.19

81.903.1

675.198

..

1.8 MANOMETER

Manometer is a device to measure pressure or mostly difference in pressure using a column of liquid to balance

the pressure.

BABA pppppppppp 332211 B AB A zzzzzzzzpp 3432321211

Example No. 6

In Fig. FE2.3, if the oil in region B has SG = 0.8 and the absolute pressure at point A is 1 atm, what is the absolute

pressure at point B?

A. 5.6 kPa

B. 10.9 kPa

C. 106.9 kPa

D. 112.2 kPa

Solution:

003.0

04.008.005.0

Bwo

wmww A

pmSG

mSGmSGp


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6

m

mmpp

w

AB

03080

04008056130501

..

....

mpp

w

AB56840.

mmkNmkNpB 56840819325101 32 ... kPapB 90106.

1.9 BUOYANCY

Buoyancy the tendency of a body to float when submerged in a fluid.

Two Archimedes Law of Buoyancy

a. A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces.

b. A floating body displaces its own weight in the fluid in which it floats.

BFW

VFB where:

FB = buoyant force


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W = weight of the body

V = volume of the body submerged or volume of the liquid displaced

= density of the liquid

Example No. 7

A stone weighs 105 lbs in air and 83 lb in water. Find the specific gravity of the stone.

A. 2.98

B. 0.35

C. 4.77 D. 2.21

Solution:

Volume (stone) = Volume (water displaced)

33525640

462

83105ft.

.

Specific Weight (stone) = 105 / 0.352564 = 297.82 lb/ft3

Specific gravity (stone) = 297.82 / 62.4 = 4.77 (c)

Example No. 8

An iceberg has a SG of 0.922. When floating in sea water (SG = 1.03), its exposed volume in % is nearest to,

A. 5.6B. 7.4

C. 8.9

D. 10.5

Solution:

su bm erged seaiceice VSGVSG

su bm erged ice VV 03.1922.0

icesu bm erged VV 895.0

%100895.0

%ice

iceice

V

VVosedexp

%5.10% osedexp

1.10 BASIC SCIENTIFIC LAWS USED IN THE ANALYSIS OF FLUID FLOW

a. Law of conservation of mass: This law when applied to a control volume states that the net mass flow

through the volume will equal the mass stored or removed from the volume. Under conditions of steady

flow this will mean that the mass leaving the control volume should be equal to the mass entering the

volume. The determination of flow velocity for a specified mass flow rate and flow area is based on the

continuity equation derived on the basis of this law.

b. Newtons laws of motion: These are basic to any force analysis under various conditions of flow. The

resultant force is calculated using the condition that it equals the rate of change of momentum. The reaction

on surfaces are calculated on the basis of these laws. Momentum equation for flow is derived based on

these laws.

c. Law of conservation of energy: Considering a control volume the law can be stated as the energy flow into

the volume will equal the energy flow out of the volume under steady conditions. This also leads to the

situation that the total energy of a fluid element in a steady flow field is conserved. This is the basis for the

derivation of Euler and Bernoulli equations for fluid flow.

d. Thermodynamic laws: are applied in the study of flow of compressible fluids.

1.11 CONTINUITY EQUATION


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Continuity Equation - This equation is used to calculate the area, or velocity in one dimensional varying area

flow, like flow in a nozzle or venturi.

2211 QQm

222111 VAVA

For incompressible flow. 21

21 QQQ

2211 VAVA

Example No. 9

Oil flow through a 30 tubes condenser with a velocity of 1.75 m/s. The internal diameter of tube is 20 mm an oil

density is 0.90 gm/mL. Find the volume flow in liters per second.

A. 16.5

B. 11.6

C. 15.6

D. 9.4

Solution:

AVQ

sec

104977.575.1020.0

4

342 m

Q

sec5.16301000sec

104977.53

34

Litubesm

LimQ

1.12 VELOCITY HEAD

Torricellis Theorem:

The velocity of a liquid which discharge under a head is equal to the velocity of a body which falls in the same

head.

g

Vh

2

2

where:

h = velocity head

V = velocity of the liquid

1.13 FRICTION HEAD LOSS IN PIPES

Darcy-Weisbach Equation

g

V

D

Lf

gD

fLVh f

22

22

where:

h f = velocity head, m or ft

V = velocity of the liquid, m/s or ft/sec

L = length of pipe, m or ftD = internal diameter, m or ft

f = coefficient of friction, friction factor (Darcy)

g = 9.81 m/s2 or 32.2 ft/sec2

Example No. 10

What is the expected head loss per mile of a closed circular pipe (17 in inside diameter, friction factor of 0.03

when 3300 gal/min of water flow under pressure?


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A. 38 ft

B. 3580 ft

C. 0.007 ft

D. 0.64 ft

Solution:

sec

min

60

1

12

12313300

3

3

in

ftgalingpmQ

ecsftQ 3

352437.

fpsft

ecsft

D

QV 66454

1217

35243744

2

3

2.

.

g

V

D

Lfh f

2

2

ftmileL 52801

2322

66454

1217

5280030

2

.

.. f h

fth f 7837.1.14 COEFFICIENT OF FRICTION, f

a. Coefficient of friction for laminar flow (Re < 2300).

Re

64f

b. Coefficient of friction for turbulent flow

Colebrook equation, turbulent flow only (Re > 2300)

7.3Re

51.2log2

21

D

ff

where, = nominal roughness of pipe or duct being used

A good approximate equation for the turbulent region of the Moody chart is given by Haalands equation:2

11.1

7.3Re

9.6log8.1

D

f

H. Blasius Equation (4000 < Re < 105)

4 Re

316.0f

Example No. 11

For flow of water at a Reynolds number of 1.03 x 106 through a 5-cm-diameter pipe of roughness height 0.5 mm

the approximate Moody friction factor is

A. 0.012

B. 0.018

C. 0.038


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D. 0.049

Solution:

Use Haaland approximation to Colebrook equation, Re > 2300

2111

73

9681

.

.Re

.log.

Df

2111

6 73

5050

10031

9681

.

.

.

.

.

log.f

0380.f

1.15 HYDRAULIC DIAMETER

For flow in non-circular ducts or ducts for which the flow does not fill the entire cross-section, we can define the

hydraulic diameter

P

ADh

4

where

A = cross-sectional area of actual flow,

P = wetted perimeter, i.e. the perimeter on which viscous shear acts

Example No. 12

Air at atmospheric pressure and with a mean velocity of 1.0 m/s, flows inside a square section duct of side b = 4

cm. If the air temperature is 350 K, determine the Reynolds Number. (at 350 K, viscosity = 20.76 x 10 -6 m2/s.)

A. Re = 1296.8

B. Re = 1926.8

C. Re = 1629.8

D. Re = 1962.8

Solution:

Hydraulic Diameter for square duct

mcmbb

bD 040444

2

.

81926

107620

01040

6.

.

..Re

DV(B)

Example No. 13

What is the hydraulic diameter of a rectangular air-ventilation duct whose cross section is 1 m by 25 cm?

A. 25 cm

B. 40 cm

C. 50 cm

D. 75 cm

Solution:

P

ADh

4

cmcm

cmcmDh 40

251002

251004

1.16 BERNOULLIS EQUATION


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Bernoullis equation is a general energy equation that is used for solving fluid flows. It relates elevation head

pressure head and velocity head. Some conditions of using Bernoullis equation: (1) No fluid friction, (2) fluid is

incompressible, and (3) negligible changes in thermal energy.

Bernoullis theorem Neglecting friction, the sum of the pressure head, velocity head and elevation head of a

point is equal to the sum of the pressure head, velocity head and elevation head of another point.

2

222

1

211

22Z

g

VpZ

g

Vp

where:

Pressure head =

p

Velocity head =g

V

2

2

Elevation head = Z

Example No. 14

Water flows through a horizontal pipe of cross sectional area of 20 cm2. At one section the cross sectional area is

4 cm2. The pressure difference between the two sections is 29.4 psi. How many cu. meters of water will flow out

of the pipe in 1 minute?

A. 1.208 m3

B. 0.0185 m3

C. 0.493 m3

D. 0.008 m3

Solution:

g

VVpp

2

21

2221

2211 VAVAQ

21 420 VV

12 5VV

21

21

22

2

pp

g

VV

kPapsi

kPapsipp 705.202

696.14

325.1014.2921

3

2

2

21

21

81.9

705.202

81.92

5

mkN

mkN

sm

VV

smV 11.41

smcmm

cmVAQ 11.4100

1

20

2

2

11

smQ33

1022.8

In 1 min.

Volume = (8.22 x 10-3)(60) = 0.4932 m3

Example No. 15


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A perfect venturi with throat diameter of 1.8 inches is placed horizontally in a pipe with a 5 in inside diameter

Eighty (80) lbm of water flow through the pipe each second. What is the difference between the pipe and

venture throat static pressure?

A. 29.9 psi

B. 34.8 psi

C. 5020 psi

D. 72.3 psi

Solution:

sftftlb

slbQ

3

3 282051

462

80.

.

fpsD

QV 40249

125

28205144

22

1

1.

.

fpsD

QV 549272

1281

28205144

22

2

2.

.

.

g

VVpp

2

2

1

2

221

2322

40249549272462144

22

21

...

. pp

psipp 823421

.

1.17 VENTURI, NOZZLE AND ORIFICE METERS

Venturi, Nozzle and Orifice meters are the three obstruction type meters commonly used for the measuremen

of flow through pipes.

Flow Rate:

ghAA

A

Q 212

12

2

This equation needs a modifying coefficient as viscous effects and boundary roughness as well as the velocity o

approach factor that depend on the diameter ratio have been neglected.

The coefficient is defined by,

dl theoreticaactual CQQ where Cd is the coefficient of discharge. Cd for venturi meters is in the range 0.95 to 0.98. Cd for flow nozzle is in

the range 0.7 to 0.9 depending on diameter ratio and Reynolds number to some extent. For orifice, The range

for coefficient of discharge is 0.6 to 0.65.

1.18 DISCHARGE MEASUREMENT USING ORIFICESActual flow rate:

ghACQ dact ual 20

where A0 is the area of orifice and Cd is the coefficient of discharge.

l theoretica

actual d

Q

QC

Coefficient of velocity Cv.


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13

gh

V

velocitylTheoretica

contractavenaatjetofvelocityA ctualC v

2

The value of Cv varies from 0.95 to 0.99 for different orifices depending on their shape and size.

Coefficient of contraction Cc.

0A

A

velocityofArea

contractavenaatjetofAreaC cc

The value of coefficient of contraction varies from 0.61 to 0.69 depending on the shape and size of the orifice.

Coefficient of discharge Cd .

vcd CCvelocitylTheoretica

velocityActual

arealTheoretica

areaActual

eargdischlTheoretica

eargdischActualC

Average value of Cd for orifices is 0.62.

Example No. 16

Calculate the discharge in liters per second through a 5 in diameter orifice under a head of 7.6 m of water.

Assume coefficient of discharge of 0.65.

A. 78B. 1547

C. 77

D. 100

Solution:

ghCACA VQ 2

msminminQ 6781920254054

650 22 ....

sLsmQ 1006010060 3

..

1.19 PERIPHERAL VELOCITY FACTOR

gh

DN

JetVelocityof

VelocityPeripheralC d

2

2. THERMODYNAMICS

2.1 MASS AND WEIGHT

Mass (1) a property of matter that constitutes one of the fundamental physical measurements or the amount

of matter a body contains. Units of mass are in lbm, slugs, or kg. Symbol m.

Mass (2) - is the absolute quantity of matter in it, an unchanging quantity for a particular mass when the speed

of the mass is small compared to the speed of light (no relativistic effect).

Weight (1) the force acting on a body in a gravitational field, equal to the product of its mass and the

gravitational acceleration of the field. Units of weight are in lbf or N. Symbol W. Formula W = mg. Where g =

9.81 m/s2 or 32.2 ft/s2.

Weight (2) - is the force exerted by a body when its mass is accelerated in a gravitational field.

2.2 VOLUME


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Volume the amount of space occupied by, or contained in a body and is measured by the number of cubes a

body contains. Units of volume are in ft3, gallons, liters, cm3, or m3. Symbol is V.

2.3 SPECIFIC VOLUME AND DENSITY

Density () of any substance is its mass (not weight) per unit volume. Units of density are lbm/ft3 or kg/m3

Symbol.

V

m

Specific Volume is the total volume of a substance divided by the total mass of a substance. Units of specific

volume are in ft3/lbm or m3/kg. Symbol is v.

m

Vv or

1v

Specific Weight () of any substance is the force of gravity on unit volume. Units of specific weight are ft3/lbfo

m3/N. Symbol is.

W

V or

gv

1

Specific Gravity (1) is a measure of the relative density of a substance as compared to the density of water at a

standard temperature. Symbol is SG.

Specific Gravity (2) a dimensionless parameter, it is defined as the ratio of the density (or specific weight) of a

substance to some standard density (or specific weight).

For Liquid substances:

stdatstdatSG

OH

l iquid

OH

l iquid

22

For Gaseous substances:

stdatstdatSG

air

gas

air

gas

Example No. 17

An iron block weighs 5 N and has a volume of 200 cm3. What is the density of the block?

A. 988 kg/cu. m.

B. 1255 kg/cu. m.

C. 2550 kg/cu. m.

D. 800 kg/cu. m.

Solution:

3

2

3

3

4.2548

81.9100

1200

5mkg

smcm

mcm

N

Vg

W

Example No. 18

Suppose two liquids of different densities, 1 = 1500 kg/m3 and2 = 500 kg/m3, were poured together inside a

100-L tank, filling it. If the resulting mixture density is 800 kg/m3, find the respective liquid amounts, in kg?

A. 35,45


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B. 53,54

C. 71,63

D. 89,72

Solution:

3

3

3

2

2

1

1 Vmmm

321 mmm

33321 Vmmm 80100800

21 .mm

12 80 mm

1005001500

21 . mm

100500

80

1500

11 .

mm

15080311

mm

kgm 451

kgm 3545802 Masses = 35 kg and 45 kg

2.4 STANDARD DENSITIES

Density of water

At approx. 4oC (39.2oF) pure water has it's highest density (weight or mass) = 1000 kg/m3 or 62.4 lb/ft3.

Density of air

At 70 F (21.1 C) and 14.696 psia (101.325 kPaa), dry air has a density of 0.074887 lbm/ft3 (1.2 kg/m3).

2.5 PRESSURE

Pressure is a measure of the force exerted per unit area on the boundaries of a substance (or system). It is

caused by the collisions of the molecules of the substance with the boundaries of the system. Units of pressure

are psi, kg/cm2, kN/m2 or kPa. Symbol is p. Formula p = F/A.

2.6 ATMOSPHERIC PRESSURE

Atmospheric pressure - is the force per unit area exerted against a surface by the weight of air above that

surface in the Earth's atmosphere.

The standard atmosphere (symbol: atm) - is a unit of pressure and is defined as being equal to 101.325 kPa.[1

The following units are equivalent, but only to the number of decimal places displayed: 760 mmHg (torr), 29.92

inHg, 14.696 psi.

Barometric pressure - is often also referred to as atmospheric pressure. Units is normally in Bar. 1 Bar = 100

kPaa.

Air pressure above sea level can be calculated as p = 101325(1 - 2.25577x10-5h)5.25588, where p = air pressure (Pa

and h = altitude above sea level (m).

Also

ohh

o epp

where:

p = atmospheric pressure, (measured in bars)


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16

h = height (altitude), km

p0 = is pressure at height h = 0 (surface pressure) = 1.0 Bar (Earth)

h0 = scale height = 7 km (Earth)

Or for every 1,000 feet, there is a corresponding pressure decrease of approximately 1 in Hg.

2.7 ABSOLUTE AND GAUGE PRESSURES

Absolute pressure, pabs - is measured relative to the absolute zero pressure - the pressure that would occur at

absolute vacuum. All calculation involving the gas laws requires pressure (and temperature) to be in absolute

units. It the sum of the gauge and atmospheric pressure.

Gauge pressure - the amount by which the total absolute pressure exceeds the ambient atmospheric pressure.

Formula pabs = patm + pg

Vacuum pressure (negative gauge pressure) - the amount by which the total absolute pressure is less than the

ambient atmospheric pressure.

Formula pabs = patm pv

Pressure gauge - is often used to measure the pressure difference between a system and the surrounding

atmosphere.

Example No. 19

A condenser vacuum gage reads 715 mm Hg when the barometer stands 757 mm Hg. State the absolute

pressure in the condenser in kN/m2.

A. 5.6 kN/m2

B. 5.9 kN/m2

C. 6.5 kN/m2

D. 5.2 kN/m2

Solution:

vacatma ppp


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Hgmmpa 42715757

2

2

65760

32510142 mkN

Hgmm

mkNHgmmpa .

.

Example No. 20

With a normal barometric pressure at sea level, atmospheric pressure at an elevation of 4000 ft is nearest to,

A. 26 in Hg

B. 27 in Hg

C. 28 in Hg

D. 29 in Hg

Solution:

A drop of 1 in Hg per 1000 ft

ftft

HginHginp 4000

1000

019229

..

Hginp 9225.

2.8 TEMPERATURE

Temperature is a measure of the molecular activity of a substance. It is a relative measure of how hot o

cold a substance is and can be used to predict the direction of heat transfer. It is an intensive property that is a

measure of the intensity of the stored molecular energy in a system.

Temperature Scales:

a. Fahrenheit (F) Scale 180 units from 32 F to 212 F.

b. Celsius (C) Scale or Centigrade Scale 100 units from 0 C to 100 C.

Relationship:

CF

5

932

9

532FC

Absolute zero - is the theoretical temperature at which entropy reaches its minimum value. The laws o

thermodynamics state that absolute zero cannot be reached using only thermodynamic means.

Absolute temperature - is the temperature measured relative to the absolute zero.

Absolute Temperature Scales:

a. Kelvin (K) Scale the absolute temperature scale that corresponds to the Celsius scale.

b. Rankine (R) Scale the absolute temperature scale that corresponds to the Fahrenheit scale.

Relationship:

460 FR

273 CK

* KK 15.273273

* RR 67.459460

Temperature Change:

RFKC

9

5

9

5


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KCRF

5

9

5

9

Ice point the temperature of a mixture of ice and air-saturated water at 1 atm = 0 C or 32 F.

Steam (boiling) point the temperature pure liquid water in contact with its vapour at 1 atm = 100 C or 212 F.

Triple point - The temperature and pressure at which a substance can exist in equilibrium in the liquid, solid, and

gaseous states. The triple point of pure water is at 0.01 degrees Celsius and 4.58 millimeters of mercury and is

used to calibrate thermometers.

2.9 ENERGY

Energy - is defined as the capacity of a system to perform work or produce heat.

Stored Energy otherwise known as possessed energy, it is the energy that is retrieved and stored within the

system; thus, dependent upon the mass flow.

Potential Energy (1) is defined as the energy of position. Symbol is P.E.

Potential Energy (2) energy due to the elevation and position of the system

cg

mgzPE

where m = mass (lbm, kg)

z = height above some reference level (ft, m)

g = acceleration due to gravity (ft/sec2, m/s2)

cg = gravitational constant.

=2

sec17.32 lbflbmft

= Nsmkg 21

Kinetic Energy (1) is the kinetic energy of motion. Symbol is K.E.

Kinetic Energy (2) energy or stored capacity for performing work; possessed by a moving body, by virtue of its

momentum.

cg

mvKE

2

2

where m = mass (lbm, kg)

v = velocity (ft/s, m/s)

g = acceleration due to gravity (ft/sec2, m/s2)

cg = gravitational constant.

=2

sec17.32 lbflbmft

= Nsmkg 2

1Joules Constant J 778 ft-lbf/ Btu

Internal Energy (1) is a microscopic forms of energy including those due to the rotation, vibration, translation

and interactions among the molecules of a substance.

Internal Energy (2) heat energy due to the movement of the molecules within the substance brought about its

temperature.


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19

Internal Energy (3) energy stored within a body or substance by virtue of the activity and configuration of its

molecules and the vibration of the atoms within the molecules.

Specific Internal Energy is the substance internal energy per unit mass. Unit is Btu/lbm or kJ/kg. Symbol is u.

P-V Energy is also called flow energy or flow work.

Specific P-V Energy is the substance P-V energy per unit mass.

Enthalpy (1) is the amount of energy possessed by a thermodynamic system for transfer between itself and itsenvironment. It is equal to PVUH .Enthalpy (2) the sum of the internal energy of a body and the product of pressure and specific volume. Unit is

Btu/lbm or kJ/kg. Symbol is h.

Specific Enthalpy is defines as Pvuh .

where u = specific internal energy

P = pressure

v = specific volume

Chemical Energy stored energy that is released or absorbed during chemical reactions.

Nuclear Energy energy due to the cohesive forces of the protons and neutrons within the atoms.

Example No. 21

A high velocity flow of gas at 800 ft/sec possesses kinetic energy nearest to which of the following?

A. 1.03 Btu/lb

B. 9.95 Btu/lb

C. 4.10 Btu/lb

D. 12.8 Btu/lb

Solution:

lbftVg

WKE

889937800

232

1

2

1

2

22 ..

lbBtuBtulbft

lblbftKE 7712

16778

889937.

.

.

(D)

2.10 HEAT AND WORK

Transitory Energy otherwise known as energy in transit or in motion; energy that loses its identity once it is

absorbed or rejected within the system; independent of mass flow stream.

Heat (1) - is the transfer of energy that occurs at the molecular level as a result of a temperature difference

Symbol Q. Unit is Btu, Btu/hr, kJ or kW.

Heat (2) energy in transition between a system and its surroundings because of a difference in temperature.

Q is positive (+) when heat is added to the body

Q is negative (-) when heat is rejected by the body

Heat transferred per unit mass =m

Qq


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20

Work is defined for mechanical system as the action of a force on an object through a distance. Symbol is W

Unit is ft-lb, kJ.

FdWWhere F = force (lbf, N)

d = displacement (ft, m )

Work is a process done by or on a system.

Power is the rate of doing work. Unit is hp or kW.

Time

WorkPower

2.11 SPECIFIC HEAT

Heat Capacity is the ratio of the heat (Q) added to or removed from a substance to the change in temperature

( T ).Specific Heat (1) is the heat capacity of a substance per unit mass. Unit is Btu/lbm-F or Btu/kg-C.

Specific Heat (2) is the heat required to raise the temperature of unit mass of a substance by a unit

temperature.

Specific heat at constant pressure is the change of enthalpy for a unit mass between two equilibrium states athe same pressure per degree change of temperature.

T

QC p

Tm

Qc p

T

qc p

Specific heat at constant volume is the change of internal energy for a unit mass per degree change o

temperature when the end states are equilibrium states of the same volume.

TQCv

Tm

Qcv

T

qcv

Specific heat ratio

u

h

c

ck

v

p

2.12 ENTROPY

Entropy (1) is a measure of inability to do work for a given heat transferred.

Entropy (2) is a measure of randomness of the molecules of a substance or measures the fraction of the tota

energy of a system that is not available for doing work.

Entropy (3) a property used to measure the state of disorder of a substance; a function of both heat and

temperature.

Entropy production is the increase in entropy.


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21

absT

QS

absT

qs

S = the change in entropy of a system during some process (Btu/R or kJ/K).

Q = the amount of heat transferred to or from the system during the process (Btu or kJ)

absT = the absolute temperature at which the heat was transferred (R or K).

s = the change in specific entropy of a system during some process (Btu/lbm-oR or kJ/kg.K).q = the amount of heat transferred to or from the system during the process (Btu/lbm or kJ/kg).

2.13 LAWS OF THERMODYNAMICS

a. ZEROTH LAW OF THERMODYNAMICS TEMPERATURE (THERMAL EQUILIBRIUM)

- states that when each of two systems is in equilibrium with a third, the first two systems must be in

equilibrium with each other. This shared property of equilibrium is the temperature.

- states that when two bodies have equality of temperature with a third body, they in turn have equalityof temperature with each other and the three bodies are said to be in thermal equilibrium. The third

body is usually a thermometer.

b. FIRST LAW OF THERMODYNAMICS LAW ON CONSERVATION OF ENERGY

- states that, because energy cannot be created or destroyed [setting aside the later ramifications of the

equivalence of mass and energy (Nuclear Energy)] the amount of heat transferred into a system plus

the amount of work done on the system must result in a corresponding increase of internal energy in

the system. Heat and work are mechanisms by which systems exchange energy with one another.

- states that during any cycle a system undergoes, the cyclic integral of heat is proportional to the cyclic

integral of work or for any system, total energy entering = total energy leaving.

c. SECOND LAW OF THERMODYMANICS ENTROPY

- states that the entropy that is, the disorder of an isolated system can never decrease. Thus, when an

isolated system achieved a configuration of maximum entropy, it can no longer undergo change: It has

reached equilibrium.

Significant statements:

- Clausius: It is impossible for a self-acting machine unaided by an external agency to move heat from

one body to another at a higher temperature.

- Kelvin-Planck: It is impossible to construct a heat engine which, while operating in a cycle produces

no effects except to do work and exchange heat with a single reservoir.- All spontaneous processes result in a more probable state

- The entropy of an isolated system never decreases.

- No actual or ideal heat engine operating in cycles can convert into work all the heat supplied to the

working substance.

- Caratheodory: In the vicinity of any particular state 2 of a system, there exist neighboring states 1

that are inaccessible via an adiabatic change from state 2.


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22

d. THIRD LAW OF THERMODYNAMICS - ABSOLUTE TEMPERATURE

- states that absolute zero cannot be attained by any procedure in a finite number of steps. Absolute zero

can be approached arbitrarily closely, but it can never be reached.

- The entropy of a substance of absolute temperature is zero.

2.14 CONSERVATION OF MASS

The law of conservation of mass states that the mass is indestructible.

Steady State is that circumstance in which there is no accumulation of mass or energy within the controvolume, and the properties at any point within the system are independent of time.

Continuity Equation of Steady Flow

2

22

1

11222111

v

A

v

AAAm

Usual English units are: fps, lb/ft3, ft3/lb, A ft2, m lb/sec.

Example No. 22

A fluid moves in a steady-flow manner between two sections in the same flow line. At section (1): A1 = 0.10 m2

V1 = 6 m/s, v1 = 0.33 m3/kg. At section (2): A2 = 0.20 m

2, 2 = 0.27 kg/m3. Calculate for the velocity of flow at

section (2).

A. 33.67

B. 37.63

C. 41.59

D. 45.55

Solution:

222111 VAVA

222

1

11 VAVA

2200270

330

6100V..

.

.

smV 67332

.

2.15 HELMHOLTZ FUNCTION

Helmholtz function - a thermodynamic property of a system equal to the difference between its internal energy

and the product of its temperature and its entropy. Symbol A. Formula A = u Ts.

2.16 GIBBS FUNCTION

Gibbs function - a thermodynamic property of a system equal to the difference between its enthalpy and the

product of its temperature and its entropy. It is usually measured in joules. Symbol G. Formula G = h Ts.

Example No. 23

Water is being heated by the exhaust gases from a gas turbine. The gases leave the gas turbine at 648 C and may

be cooled to 148 C. The water enters the heater at 93 C. The rate of gas flow is 25 kg/s and the water flow is


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23

31.5 kg/s. Assume that the mean specific heat of the gas and water are respectively 1.088 and 4.27 kJ/kg-C

What is the available energy removed from the hot gases in kw? Take available sink temperature as 311 K.

A. 8345.6

B. 4862.5

C. 6041.6

D. 6977.9

Solution:

gogg STQA

b

a poba pg

T

TmcTTTmcA ln

KTa 921273648

KTa 421273148

KTa 311

421

921ln088.125311421921088.125 gA

kWAg 9.977,61.622,6600,13

2.17 BOYLES LAW

Boyles Law (1) states that if the temperature of a given quantity of gas is held constant, the volume of gas

varies inversely with the absolute pressure during a quasistatic change of state.

Boyles Law (2) states that the volume of a gas varies inversely with its absolute pressure during change of

state if the temperature is held constant.

2211 VpVp

Example No. 24

An ideal gas is contained in a vessel of unknown volume at a pressure of 1 atmosphere. The gas is released andallowed to expand into a previously evacuated bulb whose volume is 0.500 liter. Once equilibrium has been

reached, the temperature remains the same while the pressure is recorded as 500 millimeters of mercury. What

is the unknown volume, V, of the first bulb?

A. 1.069 liter

B. 0.853 liter

C. 0.961 liter

D. 1.077 liter

Solution: 1 atm = 760 mm Hg

2211 VpVp LVV 50.0500760 11

LV 961.01

2.18 CHARLES LAW

Charles Law (1) the volume of a gas varies directly as the absolute temperature during a change of state if the

pressure of the gas is held constant.

2

2

1

1

T

V

T

V


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24

Charles Law (2) the pressure of a gas varies directly as the absolute temperature during a change of state if

the volume of the gas is held constant in.

2

2

1

1

T

p

T

p

Example No. 25

A closed vessel contains air at a pressure of 200 kN/m 2 gauge and a temperature of 32 C. The air is heated to 60

C with the atmospheric pressure of 750 mm Hg. What is the final gauge pressure?

A. 337.54

B. 127.54

C. 227.54

D. 427.54

Solution:

kPammHgPatm 100750

2

2

1

1

T

p

T

p

2736027332

1002002

p

kPagabskPap 54227543272

.. (C)

2.19 IDEAL GAS LAW

Ideal Gas Law - The equation of state of an ideal gas which is a good approximation to real gases at sufficiently

high temperatures and low pressures; that is, PV = RT, where P is the pressure, V is the volume per mole of gas

T is the temperature, and R is the gas constant.

mRTpV

2

22

1

11

T

Vp

T

Vp

where p = pressure, V = volume, m = mass, R = ideal gas constant, and T = absolute temperature.

Example No. 26

A volume of 400 cc of air is measured at a pressure of 740 mm Hg abs and a temperature of 18 C. What will be

the volume at 760 mm Hg abs and 0 C.

A. 376 cc

B. 326 cc

C. 356 cc

D. 366 cc

Solution:

1

11

2

22

T

Vp

T

Vp

27318

400740

2730

7602

V

ccV 43652

.

2.20 BASIC PROPERTIES OF IDEAL GAS


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25

KkgkJM

R 3143.8

or RlblbftM

R 1545

Rcc vp

kc

c

v

p

1

k

kRc p

1

k

Rcv

where:

R = gas constant

M = molecular weight

c p = specific heat at constant pressure

cv = specific heat at constant volume

k = specific heat ratio

Example No. 27

A 0.90 m3 tank contains 6.5 kg of an ideal gas. The gas has a molecular weight of 44 and is at 21 C. What is the

pressure of the gas?

A. 201.3 kPa

B. 301.3 kPa

C. 401.3 kPa

D. 501.3 kPa

Solution:

KkgJkmolkg

KkmolJ

M

RR

189

44

38314.

KT 29427321

kPaPaV

mRTp 340130040190

29418956 .,.

.

2.21 PROPERTIES OF AIR

airmoleairkgM 97.28

4.1kKkgkJRlblbftR 287.03.53

CkgkJCkgkcalRlbBtuc p 0.124.024.0

CkgkJCkgkcalRlbBtuc v 716.0171.0171.0

2.22 AVOGADROS LAW AND NUMBER

Avogadros Law states that equal volumes of all ideal gases at a particular pressure and temperature contains

the same number of molecules.

Avogadros number, NA is the number of molecules in a mole of any substance and is equal to 6.0225 x 102

gmole-1.

2.23 JOULES LAW


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26

Joules Law states that the change of internal energy of an ideal gas is a function of only the temperature

change.

2.24 DALTONS LAW OF PARTIAL PRESSURE

Daltons Law of Partial Pressure states that the total pressure pm of a mixture of gases is the sum of the

pressures that each gas would exert were it to occupy the vessel alone at the volume V m and temperature tm of

the mixture.

i

icbam ppppp

cbamcbam VVVVTTTT

mii pXp

1i

iX

Example No. 28

A mixture is formed at 689.48 kPaa, 37.8 C by bringing together these gases each volume before mixing

measured at 689.48 kPaa, 37.8 C: 3 mol CO2, 2 mol N2, 4.5 mol O2. Find the partial pressure of the CO2 afte

mixing.

A. 73 kPaa

B. 327 kPaa

C. 145 kPaa

D. 218 kPaa

Solution:

kPaapCO 73217486895423

3

2..

.

2.25 AMAGATS LAW OF PARTIAL VOLUMES

Amagat's Law of Partial Volumes - states that the volume Vm of a gas mixture is equal to the sum of volumes Vi o

the K component gases, if the temperature T and the pressure p remain the same.

i

icbam VVVVV

cbamcbam ppppTTTT

mii VXV

1i

iX

2.26 THERMODYNAMIC SYSTEMS AND PROCESSES

Thermodynamic Process is the path of the succession of states through which the system passes.

Cyclic Process or Cycle is a process where a system in a given initial state goes through a number of different

changes in state (going through various processes) and finally returns to its initial values.

Reversible Process is defined as a process that, once having taken place, can be reversed, and in so doing

leaves no change in either the system or surroundings.

Irreversible Process is a process that cannot return both the system and the surroundings to their origina

conditions. That is, the system and the surroundings would not return to their original conditions if the process

was reversed.

Adiabatic Process is one in which there is no heat transfer into or out of the system. The system can be

considered to be perfectly insulated.


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27

Isentropic Process is one in which the entropy of the fluid remains constant. This will be true if the process the

system goes through is reversible and adiabatic. An isentropic process can also be called a constant entropy

process.

Polytropic Process is a process when a gas undergoes a reversible process in which there is heat transfer, the

process frequently takes place in such a manner that a plot of the Log P (pressure) vs Log V (volume) is a straight

line. Or stated in equation form nPV = a constant.

Throttling Process is defined as a process in which there is no change in enthalpy from state one to state two

21 hh ; no work is done, 0W ; and the process is adiabatic, 0Q .Isobaric Process is an internally reversible (quasistatic, if nonflow) process of a pure substance during which

the pressure remains constant.

Isometric Process (Isochoric Process) a constant volume process that is internally reversible (quasi-static if

nonflow), involving a pure substance.

Isothermal Process is an internally reversible (quasistatic, if nonflow) constant temperature process of a pure

substance.

2.27 EQUATIONS FOR THERMODYNAMIC PROCESSES

Entropy = T

dQS

Non-Flow Equation

nfWUQ

Steady-Flow Equation, 0K 0P

sfWHQ

Steady-Flow Equation, 0K 0P

sf WPKHQ

where:

U = internal energy

H = enthalpyK = kinetic energyP = potential energy

nfW = non-flow work

sfW = steady flow work

Internal Energy

TmcU v

Enthalpy

TmcH p

Non-Flow Work

pdVWnfSteady Flow Work, 0K 0P

VdpWsfSteady-Flow Equation, 0K 0P

VdpWPK sf


28/44

28

Note:U

Hk

Example No. 29

In compressing 20 kg/min of air, its enthalpy was increased by 140 kJ/kg. If the power input is 50 kW, the heat

transfer in kW is,

A. 3.33

B. -3.33C. 6.67

D. -6.67

Solution:

WHQ

secminmin 60120140 kgkgkJH

kWH 6746.

kWW 50

kWQ 333506746 ..

Example No. 30

What is the temperature rise in water dropping at a height of 200 ft in F?

A. 0.257

B. 1.33

C. 2.25

D. 6.67

Solution:

J

WHtWc p

pJcHt

FlbBtuBtulbftft

t

0116778

200

..

Ft 2570.

Example No. 31

A non-flow system contains 1 lb of an ideal gas (cp = 0.24, cv = 0.17, both in Btu/lb-R). The gas temperature is

increased by 10 F while 5 Btu of work are done by the gas. What is the heat transfer in Btu?

A. -3.3

B. -2.6C. +6.7

D. +7.4

Solution:

WUQ BtuTmcU v 71101701 ..

BtuW 5BtuQ 76571 ..


29/44

29

2.28 ISOBARIC PROCESS

Isobaric Process is an internally reversible (quasistatic, if nonflow) process of a pure substance during which

the pressure remains constant.

p = Constant

1

2

1

2

T

T

V

V

12 TTmcH p

12 TTmcU v

Non-flow: nfWUQ

12 VVppdVWnf HQ

Entropy: T

dQS

2

1

1

2 lnlnV

Vmc

T

TmcS pp

Steady Flow: sfWHQ

0 VdpWsfHQ

Entropy: T

dQS

2

1

1

2 lnlnV

Vmc

T

TmcS pp

Specific Heat Ratio:

U

Q

U

Hk


30/44

30

2.29 ISOMETRIC PROCESS (ISOCHORIC PROCESS)

Isometric Process (Isochoric Process) a constant volume process that is internally reversible (quasi-static if

nonflow), involving a pure substance.

V = constant

1

2

1

2

T

T

p

p

12 TTmcH p

12 TTmcU v

Non-flow: nfWUQ

0 pdVWnf

UQ

Entropy: T

dQS

1

2

1

2 lnlnp

pmc

T

TmcS vv

Steady Flow: sfWHQ

12 ppVVdpWsf UQ

Specific Heat Ratio:

Q

H

U

H

k

Example No. 32

One kg of hydrogen are cooled from 450 C to 320 C in a constant volume process. The specific heat at constant

volume, cv, is 10.2 kJ/kg-K. How much heat is removed?

A. 1136 kJ

B. 1326 kJ

C. 1623 kJ


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31

D. 2136 kJ

Solution:

12

TTmcQ v kJQ 132632045021001 ..

2.30 ISOTHERMAL PROCESS

Isothermal Process is an internally reversible (quasistatic, if nonflow) constant temperature process of a pure

substance.

pV = constant, T = constant

12 TTmcH p

12 TTmcU v

Non-Flow Equation: nfWUQ

mRTCpVVpVp 2211

2

1

1

2

p

ppV

V

VpVWnf lnln

2

1

1

2 lnlnp

pmRT

V

VmRTWnf

nfWQ

Steady Flow: sfWHQ

2

1

1

2 lnlnp

ppV

V

VpVWs f

2

1

1

2 lnlnp

pmRT

V

VmRTWs f

sfWQ

Entropy:

1

2lnV

VmRS


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32

2

1lnp

pmRS

2

1

1

2 lnlnp

p

T

pV

V

V

T

pVS

Example No. 33

Five kmols of air initially at one atmosphere and 299 K are compressed isothermally to 8 atmospheres. How

much total heat is removed during the compression?A. 25,846.3 kJ

B. 28,922.9 kJ

C. 39,345.6 kJ

D. 44,680.9 kJ

Solution:

2

1

p

pnRTQ ln

atm

atmK

Kkmol

kJkmolQ

8

129931485 ln.

kJQ 2784625 ., (A)

2.31 ISENTROPIC PROCESS

Isentropic Process is a reversible adiabatic process with constant entropy.

kpV = constant, 0S , mRTpV , 0Q

k

V

V

p

p

2

1

1

2

,

1

2

1

1

2

k

V

V

T

T

,

Ork

k

p

p

T

T

1

1

2

1

2

12 TTmcH p 12 TTmcU v

Non-flow: nfWUQ

11

11

1

2

111

1

1

211

k k

nf p

p

k

Vp

V

V

k

VpW


33/44

33

11

11

11

1

1

21

1

2

11

1

1

21 k

k k k

nf p

p

k

mRT

V

V

k

mRT

V

V

k

mRTW

k

VpVp

k

TTmR

T

T

k

mRTWnf

111

1

112212

1

21

But,

1

kRcv

UTTmcW vnf 12

Steady Flow: sfWHQ

11

11

11

11 1

21

1

2

11

1

1

21

1

1

211

T

T

k

kmRT

V

V

k

kmRT

p

p

k

kmRT

p

p

k

VkpW

k k

k

k

k

s f

k

TTkmRWsf

1

12

But

1

k

kRc p

HTTmcW psf 12Note:

nf

sf

v

p

W

W

pdV

Vdp

U

H

c

ck

Example No. 34

Air is compressed in a diesel engine from an initial pressure of 13 psia and a temperature of 120 F to one-twelfthof its initial volume. Calculate the final temperature assuming the compression to be adiabatic.

A. 1110 F

B. 980 F

C. 987 F

D. 1560 F

Solution:1

2

1

1

2

k

V

V

T

T

141

1

12

12

1460120

460

.

V

Vt

Ft 1111072

.

Example No. 35

Nitrogen is expanded isentropically. The temperature change from 620 F to 60 F. If the pressure ratio, p1/p2 = 13

what is the isentropic index.


34/44

34

A. 1.26

B. 1.2

C. 1.4

D. 1.46

Solution:

k

k

p

p

T

T

1

2

1

2

1

kk 1

1346060

460620

kk 1

130772

.

131

0772 ln.lnk

k

28501

.

k

k

41.k

2.32 POLYTROPIC PROCESS

Polytropic Process is an internally reversible process during which pVn = C where n is any constant.

npV = constant, mRTpV

1

2

1

1

2

n

V

V

T

T

,

n

n

p

p

T

T

1

1

2

1

2

12 TTmcH p

12 TTmcU v

Non-flow Work: nfWUQ

11

11

1

2

111

1

1

211

nn

nf p

p

n

Vp

V

V

n

VpW


35/44

35

11

11

11

1

1

21

1

2

11

1

1

21 n

nnn

nf p

p

n

mRT

V

V

n

mRT

V

V

n

mRTW

n

VpVp

n

TTmR

T

T

n

mRTWnf

111

1

112212

1

21

But,

1

kRcv

11

112212

k

VpVp

k

TTmRU

1111

1

1

1

1

1

1 112211221122

nk

VpVpknVpVp

nkVpVp

nkQ

Steady Flow: sfWHQ

11

11

11

11 1

21

1

2

11

1

1

21

1

1

211

T

T

n

nmRT

V

V

n

nmRT

p

p

n

nmRT

p

p

n

VnpW

nn

n

n

n

s f

n

TTnmRWsf

1

12

But

1

k

kRc p

11

11221212

k

VpVpk

k

TTkmRTTmcH p

11221122111

1

1

1VpVp

n

n

k

kVpVp

nkQ

11

1122

nk

VpVpknQ

Entropy:

1

2ln1 T

T

n

knmcS v

Or

2

1lnV

VknmcS v

1

2

1

1

2 lnln1 p

p

n

knmc

p

p

n

knmcS v

n

n

v

Note:

Steady flow work:

VdpPKWsf


36/44

36

2.33 CURVES FOR DIFFERENT VALUES OF n.

2.34 TABLE OF IDEAL GAS FORMULAS

Ideal Gas Formulas

For constant mass systems undergoing internally reversible processes

Process

Isometric

CVIsobaric

CpIsothermal

CTIsentropic

CS

Polytropic

CpVn

p, V, T

relations 1

2

1

2

p

p

T

T

1

2

1

2

V

V

T

T

2211 VpVp

kkVpVp

21 21

1

2

1

1

2

k

V

V

T

T

kk

p

p

T

T1

1

2

1

2

nnVpVp

21 21

1

2

1

1

2

n

V

V

T

T

nn

p

p

T

T1

1

2

1

2

2

1

pdV 0 12 VVp 1

2

11 ln

V

VVp

k

VpVp

1

1122

n

VpVp

1

1122

2

1

Vdp 12 ppV 01

2

11 ln

V

VVp

k

VpVpk

1

1122

n

VpVpn

1

1122

12 UU dTcm v

12

TTmcv dTcm v

12 TTmcv

0 dTcm v

12 TTmcv

dTcm v

12 TTmcv

Q dTcm v

12 TTmcv

dTcm p

12 TTmcp

Tdsm

1

2

11 ln

V

VVp

0 dTcm n

12 TTmcn

n 0 1 k to

Specific heat, c vc pc 0

n

nkcc vn

1

Ck


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37

12 HH dTcm p

12 TTmcp

dTcm p

12 TTmcp 0 dTcm p

12 TTmcp

dTcm p

12 TTmcp

12 SS

TdTc

m v

1

2ln

T

Tmcv

TdTc

m p

1

2lnT

Tmcp

T

Q

1

2lnV

VmR

0 T

dTcm n

1

2lnT

Tmcn

1

2lnV

V

mRT

dTc

m

v

, 12

lnp

p

mRT

dTc

m

p

Example No. 36

After a series of state changes, the pressure and volume of 2.268 kg of nitrogen are each doubled. What isS? c=0.7442 kJ/kg-K, cp = 1.0414 kJ/kg-K.

A. 2.8 kJ/kg-K

B. 1.24 kJ/K

C. 2.8 kJ/K

D. 1.24 kJ/kg-K

Solution:

1

2

1

2

V

VmR

T

TmcS v lnln

1

2

1

2

p

pmc

V

VmcS v p lnln

KkJp

p

V

VS 8072

274420

2041412682

1

1

1

1 .ln.ln..

2.35 STAGNATION PROPERTIES

Stagnation Properties are those thermodynamic properties that a moving stream of compressible fluid would

have if it were brought to rest isentropically (no outside work, the kinetic energy brings about the compression).

Stagnation enthalpy

Jg

vhh

c2

2

0

Stagnation temperature

pc Jcg

vTT

2

2

0

TJcg

v

p

p

T

T

pc

k

k

21

21

00

2.36 MACH NUMBER

Mach Number is the ratio of the actual speed divided by the local speed of sound a in the fluid.

a

M 1M 1M

[SUBSONIC] [SUPERSONIC]

Acoustic speed:

21

kRTga c


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38

2.37 MOLLIER DIAGRAM

Mollier Diagram (h-s) is a chart on which enthalpy is the ordinate and entropy the abscissa.

2.38 TEMPERATURE-ENTROPY DIAGRAM


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39

2.39 ph-Chart

2.40 MIXTURES

Mixture substance made up of liquid and vapor portion or two-phase liquid-vapour system.

x= quality or dryness factor or vapour content

y = 1 x = moisture content or wetness

Properties of mixtures

fgf xvvv


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40

fgf xuuu

fgf xhhh

fgf xsss

2.41 PROCESSES INVOLVING PURE SUBSTANCES

a. Isobaric or constant pressure process: p1 = p2

b. Isothermal or constant temperature process: T1 = T2

Evaporation and condensation processes occur at constant pressure and constant temperature.c. Isometric or constant volume process: V1 = V2

For constant mass: v1 = v2

If the final state is a mixture: v1 = (vf+ xvfg)2

d. Isentropic or constant entropy process: s1 = s2

Isentropic process is reversible (no friction loss) and adiabatic (no heat loss, that is, completely insulated

system).

e. Throttling or isenthalpic (constant enthalpy) process: h1 = h2

If the final state is a mixture: h1 = (hf+ xhfg)2

If the initial state is a mixture, such as in steam calorimeter:

(hf+ xhfg)1 = h2

Example No. 37

After expanding 2.5 L of superheated steam at 2.5 MPaa and 400 C, its pressure was decreased to 0.01 MPaa. I

its dryness fraction is 90%, what is the final volume of the steam in L?

@ 2.5 MPaa and 400 C, v = 125.2 x 10-3 m3/kg

@ 0.01 MPaa. v f = 1.0102 x 10-3 m3/kg and vg = 14,674 x 10

-3 m3/kg.

A. 164

B. 264

C. 364

D. 464

Solution:

f g f vvxvv 2 3332 10010221106741490010010221 .,..v kgmv 332 1020613 ,

2

2

1

1

v

V

v

Vm

3

2

31020613102125

52

,.

. VL

LV 2642

Example No. 38Determine the heat transferred to the cooling fluid in a condenser operating under steady flow conditions with

steam entering with an enthalpy of 2300 kJ/kg and a velocity of 350 m/s. The condensate leaves with an

enthalpy of 160 kJ/kg and velocity of 70 m/s.

A. -2199 kJ/kg

B. -1922 kJ/kg

C. 2190 kJ/kg

D. 2910 kJ/kg


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41

Solution:

10002

2

1

2

2

12

VVhhQ

10002

350702300160

22 Q

2.42 THE CARNOT CYCLE

3421 TTTT and 3241 SSSS

411 SSTQ A

412322 SSTSSTQR

412411 SSTSSTQQW RA

411

412411

SST

SSTSST

Q

QQ

Q

We

A

RA

A

H

LH

T

TT

T

TTe

1

21

L

R

H

A

T

Q

T

Q

where e = Carnot cycle efficiency

T1 = TH = highest absolute temperature

T2 = TL = lowest absolute temperature

Example No. 39

Carnot engine receives 130 Btu of heat from a hot reservoir at 700 F and rejects 49 Btu of heat. Calculate the

temperature of the cold reservoir.

A. -21.9 F

B. -24.2 F

C. -20.8 FD. -22.7 F

Solution:

RTH 1160460700

H

LH

A

R A

A T

TT

Q

QQ

Q

We


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42

1160

1160

130

49130 LT

FRTL 772223437 ..

2.43 STIRLING AND ERICSSON CYCLE

Ideal Stirling Cycle is composed of two isothermal and two isometric processes, the regeneration occurring at

constant volume.

Ideal Ericsson Cycle consists of two isothermal and two isobaric processes, with the regeneration occurringduring constant pressure.

2.44 BASIC WORKING CYCLES FOR VARIOUS APPLICATIONS

Application Basic Working Cycle

Steam Power Plant Rankine Cycle

Gasoline Engine (Spark-Ignition) Otto Cycle

Diesel Engine (Combustion-Ignition) Diesel Cycle

Gas Turbine Brayton Cycle

Refrigeration System Refrigeration Cycle

Example No. 40

One kilogram of air at a pressure and temperature of 1 bar and 15 C initially, undergoes the following processes

in a cycle: isothermal compression to 2 bar; polytropic compression from 2 bar to 4 bar; isentropic expansion

from 4 bar to initial condition. What is the cycle work in kJ/kg?

A. 10.3

B. 13.0

C. 57.3

D. 70.3

Solution:

kPabarp 10011

KT 288273151

kPabarp 20022

kPabarp 40043 kgm 1

1-2, CTKTT 28812

kPabarp 20022

2-3, CpVn


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43

kPap 2002

KT 2882

kPap 4003

n

n

p

pTT

1

2

323

3-1, CpV

k

kPap 2003KT 2881

kPap 1001

k

k

p

pTT

1

1

313

Then

k

k

n

n

p

pT

p

pTT

1

1

31

1

2

323

k

k

n

n

p

p

p

p

1

1

3

1

2

3

4.1

14.11

100

400

200

400

n

n

333.2n

2

1

121

lnV

VmRTW

11

1

3

22

32

n

n

p

p

n

nmRTW

11

1

1

3113

k

k

p

p

k

kmRTW

kgm 1

1

2

12

1

121lnln

p

pmRT

V

VmRTW

kJW 3.57100

200ln288287.0121

11

1

3

2232

n

n

p

p

n

nmRTW


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1400

200

333.21

288287.01333.2 333.21333.2

32W

kJW 3.7032

11

1

1

3113

k

k

p

p

k

kmRTW

1100

400

4.11

288287.014.1 4.114.1

13W

kJW 6.14013

kJWWWWcyc 6.1403.703.57133221 kJWcyc 13

- End -

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LECTURE 5 cor 1