02 part3 work heat transfer first law

Work Transfer

S.Gunabalan Associate Professor Mechanical Engineering Department Bharathiyar College of Engineering & Technology Karaikal - 609 609. e-Mail : [email protected]

Energy interaction in a closed system

• A closed system and its surroundings can interact in two ways: – (a) by work transfer, – (b) by heat transfer.

These may be called energy interactions and these bring about changes in the properties of the system.

work

• work is the product of force and distance – The action of a force on a moving body is

identified as work – N.m or Joule [1 Nm = 1 Joule],

• Power is the rate at which work is done – J/s or watt

Work

• Work - basic modes of energy transfer

Work

• When work is done by a system, it is arbitrarily taken to be positive, and

• when work is done on a system, taken to be negative

Displacement Work

Quasi-static process : is an infinitesimal slow process, where each state of the system is pass through the equilibrium state. Work is a path function

pdV-Work

Point-path functions

Work is depends on the path of the process. work is a path function

Thermodynamic properties (P,V,T) are all point functions. For a given state, there is a definite value for each property. The change in a thermodynamic properties are depends only on the initial and final states of the system.

Cyclic Process

• the initial and final states of the system are the same.

• The change in any property is zero • cyclic integral o f a property is always zero

Thermodynamic Processes

• Constant pressure process (isobaric) • Pressure constant (p1= p2) • n = 0 • 푄 = ∆u + W (first law of thermodynamics)

• ∆u = 푚푐푣(푇2 − 푇1) • 푄 = 푚푐푝(푇2 − 푇1)

• W12 = ∫ 푝푑푉 = 푝(v2-v1)

Quasi-Static Processes

풑ퟏ풗ퟏ푻ퟏ

= 풑ퟐ풗ퟐ푻ퟐ


• Constant Volume process (isochoric process) • Volume constant (V1= V2) • n = ∞ • ∆u = 푚푐푣(푇2 − 푇1)

• W12 = ∫ 푝푑푉 = 푝(v1-v2) = 0

• 푄 = ∆u


풑ퟏ풗ퟏ푻ퟏ



• Constant Temperature process (isothermal process) Process in which pV = C • (T1= T2) • n = 1 • ∆u = 푚푐푣 푇2 − 푇1 = 0

• W12 = ∫ 푝푑푉 = P1V1 퐥퐧 푽ퟐ푽ퟏ

• 푄 = 푾ퟏퟐ


풑ퟏ풗ퟏ푻ퟏ


pdV-Work - Quasi-Static Processes

• Process in which pV = C • 푊12 = ∫ 푝푑푉

– PV = P1V1 = C

– P = P1V1

• 푊12 = ∫ P1V1 푑푉

• 푊12 = P1V1∫1 푑푉

• 푾ퟏퟐ = P1V1 퐥퐧 푽ퟐ푽ퟏ

OR 푾ퟏퟐ = P1V1 풍풏 푷ퟏ

푷ퟐ


Thermodynamic Processes Process in which pV n = C

Polytrophic process

푊12 = 푝푑푉

– 푝푉 = 푝1푉1

– P =

푊12 =푝1푉1

푉푑푉

푊12 = 푝1푉11

푉푑푉

푾ퟏퟐ = 푝1푉1 푉 푑푉


Process in which pV n = C

푾ퟏퟐ = 푝1푉1 푉 푑푉

푾ퟏퟐ = 푝1푉1푉−푛 + 1

푉2

푉1

푾ퟏퟐ = 푝1푉1푉2 − 푉1

−푛 + 1

푾ퟏퟐ =푝1푉1 푉2 − 푝1푉1 푉1

−푛 + 1

Polytrophic process

pdV-Work - Quasi-Static Processes Process in which pV n = C

푾ퟏퟐ =푝1푉1 푉2 − 푝1푉1 푉1

−푛 + 1

푾ퟏퟐ =풑ퟏ푽ퟏ풏푉2 − 푝1푉1

−푛 + 1

푝1푉1 = 푝2푉2 = c

푾ퟏퟐ =푝2푉2 푉2 − 푝1푉1

−푛 + 1

푾ퟏퟐ =푃2푉2 − 푃1푉1

−푛 + 1

푾ퟏퟐ =푷ퟏ푽ퟏ − 푷ퟐ푽ퟐ

풏 − ퟏ

Polytrophic process

Thermodynamic Processes Process in which pV n = C

푾ퟏퟐ = 푷ퟏ푽ퟏ 푷ퟐ푽ퟐ풏 ퟏ

= 푹푻ퟏ 푹푻ퟐ풏 ퟏ

풑ퟏ푽ퟏ풏 = 풑ퟐ푽ퟐ풏

∆퐮 = 풎풄풗 푻ퟐ − 푻ퟏ 푸 = ∆퐮 + 푾ퟏퟐ Also 푸 = 풄풗

휸 풏휸 ퟏ

(T2-T1)

푸 = 휸 풏휸 ퟏ

* 푾ퟏퟐ

푻ퟏ푻ퟐ

= 푽ퟐ푽ퟏ

풏 ퟏ

푻ퟐ푻ퟏ

= 풑ퟐ풑ퟏ

풏 ퟏ 풏

Polytrophic process


Process in which pV 휸 = C 푸 = ퟎ No heat is transferred to or from the system It require perfect thermal insulation

∆퐮 = 푾ퟏퟐ =푷ퟏ푽ퟏ − 푷ퟐ푽ퟐ

휸 − ퟏ

풑ퟏ푽ퟏ휸 = 풑ퟐ푽ퟐ휸

휸 -1= 푹풄풗

휸= 풄풑풄풗

Adiabatic process


Process in which pV = C

푾ퟏퟐ = 푹푻ퟏ 푽ퟐ/푽ퟏ

푸 = 풄풗 푻ퟐ− 푻ퟏ + 푾ퟏퟐ

Hyperbolic process

But not necessarily T = Constant

Free Expansion

• Expansion of gas against vacuum is called free expansion

• 푊 = 0 • 푞 = 0

Heat Transfer • Heat is defined as the form of energy that is

transferred across a boundary by virtue of a temperature difference.

• Energy transfer by virtue of temperature difference is called heat transfer (Q - Joules)

• Conduction : The transfer of heat between two bodies in direct contact.

• Radiation : Heat transfer between two bodies separated by empty space or gases through electromagnetic waves.

• Convection: The transfer of heat between a wall and a fluid system in motion.

Heat Transfer

• At constant pressure – 푄 = 푚푐푝푑푇

– 푄 = 푚 퐾푔 푐푝 푑푇(퐾)

– 푄 = KJ

• At constant Volume – 푄 = 푚푐푣푑푇 푚푐푣 – heat capacity

Heat Transfer is a boundary phenomenon, for isolated system Heat Transfer and work transfer is zero

Heat Transfer

• Heat flow into a system – positive • heat flow out o f a system - negative

Short Answer

• What is an indicator diagram ? – An indicator diagram is a trace made by a

recording pressure gauge.

• Define Quasi-static process ? – It is an infinitesimal slow process, where each

state of the system is pass through the equilibrium state.

Questions -1

A 280mm diameter cylinder fitted with a frictionless leak proof piston contains 0.02 Kg of steam at a pressure of 0.6MPa and a temperature of 200 oC As the piston moves slowly outwards through a distance of 305 mm, the steam undergoes a fully resisted expansion during which the steam pressure p and the steam volume V are related by 푝푉 = 퐶,푤ℎ푒푟푒푛푖푠푐표푛푠푡푎푛푡. The final pressure of the steam is 0.12MPa. Determine A) the value of n B) Work done by the steam

• (Apr-May 2012 Pondicherry university)

Given data – Cylinder size : 280 diameter (D) – Mass of steam 0.02 Kg (m) – pressure of 0.6MPa (p1) – temperature of 200 oC (T1) – As the piston moves a distance of 305 mm (h)

• 푝푉 = 퐶,푤ℎ푒푟푒푛푖푠푐표푛푠푡푎푛푡. • The final pressure of the steam is 0.12MPa. (p2)

– Determine • A) the value of n • B) Work done by the steam

Short Note • 푝푉 = 퐶

– 푛 log =log • P1,P2,t1,m Given

– Find V1 • 푝1푉1 = 푚푅푇1

푅 = 푅/M V2 – Final volume The conditions are

– Cylinder size : 280 diameter (D) – As the piston moves a distance of 305 mm (h)

푉2 = 휋푟 ℎ

Process in which pV n = C 푾ퟏퟐ =

푷ퟏ푽ퟏ − 푷ퟐ푽ퟐ풏 − ퟏ

Mass of steam 0.02 Kg (m) P1 = pressure of 0.6MPa (p1) temperature of 200 oC (T1) p2 = 0.12MPa

Ans

A) the value of n

푛 log =log

A) 푝푉 = 퐶

푝1푉1 = 푝2푉2

log 푝1 + 푛 log 푉1 = log 푝2 + 푛 log 푉2

푛 log 푉1 − 푛 log 푉2 = log 푝2 - log 푝1

푛 log =log

Concept poof

A) the value of n

푛 log =log . .

Here need v1 and v2

Find V1 – initial volume The initial conditions are

Mass of steam 0.02 Kg (m) pressure of 0.6MPa (p1) temperature of 200 oC (T1)

푝1푉1 = 푚푅푇1 0.6푀푃푎푥푉1 = 0.02퐾푔푥푅푠푡푒푎푚푥 200 + 273 퐾

푹 = 푴푹 푹 = 푹/M 푹− 푼풏풊풗풆풓풔풂풍푮풂풔풄풐풏풔풕풂풏풕푲푱 푲.풎풐풍푲⁄

푹 = ퟖ.ퟑퟏퟒퟒퟏ푲푱 푲.풎풐풍푲⁄

푴 − 푴풐풍풆풄풖풍풂풓풘풆풊품풉풕푲품 푲.풎풐풍

퐑 − 퐜퐡퐚퐫퐚퐜퐭퐞퐫퐢퐬퐭퐢퐜퐠퐚퐬퐜퐨퐧퐬퐭퐚퐧퐭 푲푱 풌품푲


푝1푉1 = 푚푅푇1 0.6푀푃푎푥푉1 = 0.02퐾푔푥푅푠푡푒푎푚푥 200 + 273 퐾

푹 = 푹/M

푹 = ퟖ.ퟑퟏퟒퟒퟏퟏퟖ.ퟎퟐ

= ?

R = 0.461 KJ/Kg K


Gas Molar Weight ( M)Kg/Kmol

Air 28.97

Nitrogen 28.01

Oxygen 32

Hydrogen 2.016

Helium 4.004

Carbon dioxide

44.01

Steam 18.02

푝1푉1 = 푚푅푇1

0.6푀푃푎푥푉1 = 0.02퐾푔푥0.461퐾퐽퐾푔퐾

푥 200 + 273 퐾

What is the unit of V1 ?

0.6풙ퟏퟎퟔ풙푵 풎ퟐ푥푉1

푚3

= 0.02퐾푔푥0.461ퟏퟎퟑ풙푵 풎ퟐ⁄

퐾푔퐾푥 200 + 273 퐾

V1 =?

V1 = 0.007268푚3


푴푷풂 = ퟏퟎퟔ풙푵 풎ퟐ 푀푃푎 = 106푥 푁 106푥푚푚2

푀푃푎 = 푁 푚푚2 KJ = ퟏퟎퟑ풙푵 풎ퟐ⁄

Exercise -1 Find V2 – initial volume The conditions are

– Cylinder size : 280 diameter (D) – As the piston moves a distance of 305 mm (h)

푉2 = 휋푟 ℎ 푉2 = 휋푥140 푥305 mm

푉2 = 18770920 mm3

푉2 =( )

m3

푉2 = 0.018770920 m3

A) the value of n 푝1푉1 = 푝2푉2

푛 log 푉1

푉2 = log

0.12푀푃푎0.6푀푃푎

Here need v1 and v2 Find 푉1 = 0.007268푚3

푉2 = 0.018770920 푚3

Substituting V1 and V2

푛 log .

. =log .

.

푛 = 1.6963


Process in which pV n = C 푾ퟏퟐ =

푷ퟏ푽ퟏ − 푷ퟐ푽ퟐ풏 − ퟏ

Process in which pV = Constant

푾ퟏퟐ = P1V1 퐥퐧 푽ퟐ푽ퟏ

OR 푾ퟏퟐ = P1V1 풍풏 푷ퟏ

푷ퟐ

Constant Volume process (isochoric process) dV = 0 푾ퟏퟐ = 0 Constant pressure process (isobaric) 푾ퟏퟐ = p(v2-v1)

– Determine B) Work done by the steam

• 푾ퟏퟐ = 푷ퟏ푽ퟏ 푷ퟐ푽ퟐ풏 ퟏ

• Values from given data

– P1 = 0. 6푀푃푎 – 푝2 = 0.12푀푃푎

• Calculated values – 푛 = 1.6963 – 푉1 = 0.007268푚3

– 푉2 = 0.018770920 푚3

푾ퟏퟐ

=0. 6푥106푃푎풙0.007268푚3 − 0.12푥106푃푎풙0.018770920 푚3

1.6963− ퟏ

푾ퟏퟐ = - 3735.67 Nm or J

– Determine The magnitude andsign of heat transfer

푸 = 풄풗휸 풏휸 ퟏ

(T2-T1)


* 푾ퟏퟐ

휸= 풄풑풄풗

푹 = 풄풑 − 풄풗

Ex - 1

A Cylinder containing 0.4 m3 of gas at 1 bar and 75 ℃. the gas is compressed to 0.15 m3, the final pressure is 4 bar. Take 훾 = 1.4, R = 0.2942 KJ/kg ℃ Find 1. Mass of the gas 2. n – (index of compression) 3. Work Done 4. increase in internal energy of gas 5. Heat transfer.

Ex – 1 - Note 1. Mass of the gas

• 푝1푉1 = 푚푅푇1

2. n – (index of compression) – 푝푉 = 퐶

» 푛 log =log

3. Work Done

푾ퟏퟐ = 푷ퟏ푽ퟏ 푷ퟐ푽ퟐ풏 ퟏ

4. increase in internal energy of gas 푢 = 푚퐶푣 푇2 − 푇1 C푣 = R = 0.2942 KJ/kg ℃ - so use Temperature in ℃ 5. Heat transfer. Q = u +W

푸 = 풄풗

휸 풏휸 ퟏ

(T2-T1)


* 푾ퟏퟐ

푻ퟏ푻ퟐ

= 푽ퟐ푽ퟏ

풏 ퟏ

푻ퟐ푻ퟏ

= 풑ퟐ풑ퟏ

풏 ퟏ 풏

Prob-2

• Air At 0.1 MPa at 25℃, initially occupies a volume of 0.016 m3. it is compressed reversibly and adiabatically to a pressure of 0.6MPa. Find the a) final temperature, b) Final Volume and c)Work Done. Take 휸 = 1.4 – for air – Final Temperature

• = 휸휸

– Final Volume • 풑ퟏ푽ퟏ휸 = 풑ퟐ푽ퟐ휸

∆퐮 = 푾ퟏퟐ =푷ퟏ푽ퟏ − 푷ퟐ푽ퟐ

휸 − ퟏ

Adiabatic process Q = 0

Prob-3

• 1.3 Kg of liquid having constant specific heat of 2.3 KJ/kgK, is stirred in a well insulated chamber, causing a temperature rise to 15 C.

• Calculate the work done and internal energy.

• well insulated – No heat transfer or loss Q = 0 Q = u + W Q = 0 W = -u U = m Cv (T2-T1)

Ex - 2

• 1.3 Kg of liquid having constant specific heat of 2.3 KJ/kgK, is stirred in a well insulated chamber, causing a temperature rise to 15 C. During the process 1.5 KJ of heat is transferred to the system

• Calculate the work done and internal energy.

• well insulated – heat transfer Q = 1.5KJ Q = u + W U = m Cv (T2-T1) W = Q-u

Reference • Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill, New

Delhi.