Work Transfer

S.Gunabalan Associate Professor Mechanical Engineering Department Bharathiyar College of Engineering & Technology Karaikal - 609 609. e-Mail : gunabalans@yahoo.com

Part - 2

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

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

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

pdV-Work - Quasi-Static Processes

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)

Ex-4 A fluid is confined in a cylinder by a spring loaded frictionless piston so that the pressure in the fluid is a linear function of volume

푝 = 푎 + 푏푉 The internal energy of the fluid is given by the following equation

푢 = 34푎 + 3.15푝푉 Where 푢푖푛퐾퐽, 푝푖푛푘푃푎푎푛푑푉푖푛푚3. Find the fluid change from an initial state of 170kPa 0.03m3 to a final state of 400kPa, 0.06m3 with no work other than, that done on the piston. Find the direction and magnitude of the work and heat transfer.(Apr/May 2011)

Given Data 푝 = 푎 + 푏푉

푢 = 34푎 + 3.15푝푉 Where 푢푖푛퐾퐽,푝푖푛푘푃푎푎푛푑푉푖푛푚3. P1 = 170kPa V1 = 0.03m3 P2 = 400kPa, V2 = 0.06m3

work = ? heat transfer = ?

a, b – two constant need to be addressed 푝 = 푎 + 푏푉

Where 푢푖푛퐾퐽,푝푖푛푘푃푎푎푛푑푉푖푛푚3. P1 = 170kPa V1 = 0.03m3

푝 = 푎 + 푏푉 170 = 푎 + 0.03푏 −− −(1)

P2 = 400kPa, V2 = 0.06m3

400 = 푎 + 0.06푏 −− −(1) Find a,b

Work Done

푊12 = 푝푑푣

푝 = 푎 + 푏푉

푊12 = 푎 + 푏푉 푑푣

푊12 = 푎 푉2 − 푉1 + 푏(푉2

2− 푉2

1

2)

Heat transfer Q 푄 = ∆푢 + 푤12

푢 = 34푎 + 3.15푝푉

∆푢 = 푢2 - 푢1

∆푢 = (34푎 + 3.15푝푉)2 – (34푎 + 3.15푝푉)1 ∆푢 = 3.15[(푝푉)2 – (푝푉)1] Substitute and find Q

푄 = ∆푢 + 푤12

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

Delhi.