Thermodynamics

Chapter 4 | 173Note that the work is done by the system.(c) The work represented by the rectangular area (region I) is done againstthe piston and the atmosphere, and the work represented by the triangulararea (region II) is done against the spring. Thus,DiscussionW spring 1 1 kJ2 31320 2002 kPa4 10.05 m 3 2a1 kPa # b 3 kJm3This result could also be obtained fromW spring 1 2k 1x 2 2 x 2 12 1 1 kJ2 1150 kN>m2310.2 m2 2 0 2 4a b 3 kJ1 kN # m4–2 ENERGY BALANCE FOR CLOSED SYSTEMSEnergy balance for any system undergoing any kind of process wasexpressed as (see Chap. 2)or, in the rate form, asE in E out ¢E system 1kJ2⎫ ⎪⎬⎪⎭⎫⎪⎪⎬⎪⎪⎭Net energy transferby heat, work, and massRate of net energy transferby heat, work, and massChange in internal, kinetic,potential, etc., energiesE . in E . out dE system >dt1kW2Rate of change in internal,kinetic, potential, etc., energies(4–11)(4–12)For constant rates, the total quantities during a time interval t are related tothe quantities per unit time asQ Q # ¢t,W W # ¢t,and¢E 1dE>dt2 ¢t1kJ2The energy balance can be expressed on a per unit mass basis ase in e out ¢e system 1kJ>kg2(4–13)(4–14)which is obtained by dividing all the quantities in Eq. 4–11 by the mass mof the system. Energy balance can also be expressed in the differentialform asdE in dE out dE system orde in de out de system(4–15)For a closed system undergoing a cycle, the initial and final states are identical,and thus E system E 2 E 1 0. Then the energy balance for a cyclesimplifies to E in E out 0 or E in E out . Noting that a closed system doesnot involve any mass flow across its boundaries, the energy balance for acycle can be expressed in terms of heat and work interactions asW net,out Q net,in orW # net,out Q # net,in1for a cycle2(4–16)That is, the net work output during a cycle is equal to net heat input(Fig. 4–11).⎫⎪⎬⎪⎭⎫⎪⎪⎬⎪⎪⎭PINTERACTIVETUTORIALSEE TUTORIAL CH. 4, SEC. 2 ON THE DVD.Q net = W netFIGURE 4–11For a cycle E 0, thus Q W.V