Thermodynamics

74 | ThermodynamicsMassinControlvolumeWQMassoutFIGURE 2–45The energy content of a controlvolume can be changed by mass flowas well as heat and work interactions.PQ net = W netFIGURE 2–46For a cycle E 0, thus Q W.VEnergy balance for any system undergoing any kind of process can beexpressed more compactly asor, 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 mass⎫⎪⎬⎪⎭⎫⎪⎪⎬⎪⎪⎭Change in internal, kinetic,potential, etc., energiesE . in E . out dE system >dt1kW2Rate of change in internal,kinetic, potential, etc., energies(2–35)(2–36)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(2–37)(2–38)which is obtained by dividing all the quantities in Eq. 2–35 by the mass m ofthe system. Energy balance can also be expressed in the differential form asdE in dE out dE system orde in de out de system(2–39)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(2–40)That is, the net work output during a cycle is equal to net heat input (Fig.2–46).EXAMPLE 2–10Cooling of a Hot Fluid in a TankU 1 = 800 kJU 2 = ?FluidQ out = 500 kJW sh, in = 100 kJFIGURE 2–47Schematic for Example 2–10.A rigid tank contains a hot fluid that is cooled while being stirred by a paddlewheel. Initially, the internal energy of the fluid is 800 kJ. During thecooling process, the fluid loses 500 kJ of heat, and the paddle wheel does100 kJ of work on the fluid. Determine the final internal energy of the fluid.Neglect the energy stored in the paddle wheel.Solution A fluid in a rigid tank looses heat while being stirred. The finalinternal energy of the fluid is to be determined.Assumptions 1 The tank is stationary and thus the kinetic and potentialenergy changes are zero, KE PE 0. Therefore, E U and internalenergy is the only form of the system’s energy that may change during thisprocess. 2 Energy stored in the paddle wheel is negligible.Analysis Take the contents of the tank as the system (Fig. 2–47). This is aclosed system since no mass crosses the boundary during the process. Weobserve that the volume of a rigid tank is constant, and thus there is nomoving boundary work. Also, heat is lost from the system and shaft work isdone on the system. Applying the energy balance on the system gives