Thermodynamics

orE # in E # out dE system >dt 0⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭⎫⎪⎪⎬⎪⎪⎭⎫⎬⎭⎫⎬⎭Rate of net energy transferby heat, work, and massRate of change in internal, kinetic,potential, etc., energies(5–33)Energy balance: E . in E . out 1kW2(5–34)Rate of net energy transfer inby heat, work, and massRate of net energy transfer outby heat, work, and mass¡ 0 (steady)Noting that energy can be transferred by heat, work, and mass only, theenergy balance in Eq. 5–34 for a general steady-flow system can also bewritten more explicitly asChapter 5 | 231Q # in W # in ainm # u Q # out W # out aoutm # u(5–35)orQ # in W # in ainm # a h V 22 gz b Q# out W # out aoutm # a h V 22 gz b123123for each inlet(5–36)since the energy of a flowing fluid per unit mass is u h ke pe h V 2 /2 gz. The energy balance relation for steady-flow systems first appearedin 1859 in a German thermodynamics book written by Gustav Zeuner.Consider, for example, an ordinary electric hot-water heater under steadyoperation, as shown in Fig. 5–20. A cold-water stream with a mass flow rateṁ is continuously flowing into the water heater, and a hot-water stream ofthe same mass flow rate is continuously flowing out of it. The water heater(the control volume) is losing heat to the surrounding air at a rate of Q . out ,and the electric heating element is supplying electrical work (heating) to thewater at a rate of Ẇ in . On the basis of the conservation of energy principle,we can say that the water stream experiences an increase in its total energyas it flows through the water heater that is equal to the electric energy suppliedto the water minus the heat losses.The energy balance relation just given is intuitive in nature and is easy touse when the magnitudes and directions of heat and work transfers areknown. When performing a general analytical study or solving a problemthat involves an unknown heat or work interaction, however, we need toassume a direction for the heat or work interactions. In such cases, it is commonpractice to assume heat to be transferred into the system (heat input) at arate of Q . , and work produced by the system (work output) at a rate of Ẇ, andthen solve the problem. The first-law or energy balance relation in that casefor a general steady-flow system becomesQ # W # aoutm # a h V 22 gz b a m # a h V 2in 2 gz b123 123for each exitfor each inletfor each exit(5–37)Obtaining a negative quantity for Q . or W . simply means that the assumeddirection is wrong and should be reversed. For single-stream devices, thesteady-flow energy balance equation becomesm ˙ 2 = m˙ 1HotwateroutHeatlossQ˙outCV(Hot-water tank)ElectricheatingelementẆ inṁ 1ColdwaterinFIGURE 5–20A water heater in steady operation.Q # W # m # c h 2 h 1 V 2 2 2 V 1 g 1z 2 z 1 2d2(5–38)