Unit B-3: List of Subjects

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1 ES312 Energy Transer Fundamentals Unit B: First Law o Thermodynamics ROAD MAP... B-1: The Concept o Energy B-2: ork Interactions B-3: First Law o Thermodynamics B-4: Heat Transer Fundamentals Unit B-3: List o Subjects Control Mass Form o the First Law Energy Conversion Eiciencies Mechanical Eiciencies

2 PAGE 1 o 12 Control Mass orm o the First Law (1) Original Condition ork Transer in E i E i in CONTROL MASS FORM OF THE FIRST LA OF THERMODYNAMICS Let us consider an ideal riction-less piston-cylinder coniguration (ignore the weight o the piston) with a ixed amount o air contained within, without any leak (closed system): this is a simple example o Control Mass (CM) analysis Initially, the energy content within the system is: E i (this is purely internal energy as this is a stationary system without any kinetic or potential energy components within the system) ORK ENERGY TRANSFER IN Now, i the piston is pushed down by an externally applied orce, this can be modeled as an energy interaction due to work with the direction o in to the system: this is the work energy transer in in Ater this work energy interaction, the energy content within the system becomes: Ei in MOVING BOUNDARY ORK Note that this is a work due to a moving piston: this is commonly called, moving boundary or pdv work (meaning that the work due to the pressure change, associated with the volume change) in thermodynamics

3 PAGE 2 o 12 Control Mass orm o the First Law (2) Heat Transer in ork Transer out E i + in Q in out E i + in + Q in HEAT ENERGY TRANSFER IN In order to preserve the energy content within the system, a removable stopper-pin is plugged into the piston: the piston is locked into the ixed position Now, i the cylinder is heated by a burner, this can be modeled as an energy interaction due to heat with the direction o in to the system: this is the heat energy transer in Q in Ater this heat energy interaction, the energy content within the system becomes: Ei in Qin ORK ENERGY TRANSFER OUT Now, i the stopper-pin o the piston is removed, the higher pressure built inside the cylinder will push up the piston; this can be modeled as an energy interaction due to work with the direction o out o the system: this is the work energy transer out Ater this work energy interaction, the energy content within the system becomes: E Q i in in out out

4 PAGE 3 o 12 Control Mass orm o the First Law (3) Heat Transer out E i + in + Q in out Q out HEAT ENERGY TRANSFER OUT (Once again) in order to preserve the energy content within the system, a removable stopper-pin is plugged into the piston: the piston is locked into the ixed position Now, i the cylinder is cooled by an ice cold water bucket, this can be modeled as an energy interaction due to heat with the direction o out o the system: this is the heat energy transer out Q out Ater all possible energy transer (work and heat, in and out), the inal energy content o the system can be given by the simple expression as: Ei in Qin out Qout E THE FIRST LA OF THERMODYNAMICS: CONTROL MASS (CM) FORM ith multiple interactions o work/heat energy transer, the 1st law can be deined as: E Ei E Qin in Qout out or, in general: E Qin in Qout out This can also be expressed in the time rate o change (power) orm: de d Qin in d Qout out Qin in Qout out dt dt dt de In general (multiple interactions o work/heat energy transer): Qin in Qout out dt

PAGE 4 o 12 EXERCISE B-3-1 (Do-It-Yoursel) A rigid tank contains a hot luid that is cooled while being stirred by a paddle wheel. Initially, the internal energy o the luid is 800 kj.

5 PAGE 4 o 12 EXERCISE B-3-1 (Do-It-Yoursel) A rigid tank contains a hot luid that is cooled while being stirred by a paddle wheel. Initially, the internal energy o the luid is 800 kj. During the cooling process, the luid loses 500 kj o heat, and the paddle wheel does 100 kj o work on the luid. Determine the inal internal energy o the luid (in kj ). Neglect the energy stored in the paddle wheel. Solution A luid in a rigid tank loses heat while being stirred. The inal internal energy o the luid is to be determined. Assumptions (1) The tank is stationary and thus the kinetic and potential energy changes are zero. Thereore, internal energy is the only orm o the system s energy that may change during the process. (2) Energy stored in the paddle wheel is negligible. Take the contents o the tank as the system (shown in the igure). This is a closed system, since no mass crosses the boundary during the process. Also, the volume o the tank is constant (rigid tank). There is no moving boundary work. The heat is lost rom the system ( out ) and shat work is done on the system ( in ). Applying the energy balance on the system: E E Ei Qin in Qout out Since the energy stored within the system is only internal energy: U Ui sh,in Qout Thereore, U 800 kj 100 kj 500 kj => U 400 kj

PAGE 5 o 12 EXERCISE B-3-2 (Do-It-Yoursel) A room is initially at the outdoor temperature o 25 C. Now a large an that consumes 200 o electricity (when running) is turned on.

6 PAGE 5 o 12 EXERCISE B-3-2 (Do-It-Yoursel) A room is initially at the outdoor temperature o 25 C. Now a large an that consumes 200 o electricity (when running) is turned on. The heat transer rate between the room and the outdoor air is given as: ( ), where U = 6 /m 2 C is the overall heat transer coeicient, A = 30 m 2 is the exposed surace area o the room, and T i and T o are the indoor and outdoor air temperatures, respectively. Determine the indoor air temperature (in C ) when steady operating conditions are established. Solution A large an is turned on and kept on, in a room that loses heat to the outdoors. The indoor air temperature is to be determined when steady operation is reached. Assumptions (1) Heat transer through the loor is negligible. (2) There are no other energy interactions involved. The electricity consumed by the an is energy input or the room, and thus the room gains energy at a rate o 200 k. As a result, the room air temperature tends to rise. But, as the room air temperature rises, the rate o heat loss rom the room increases until the rate o heat loss equals the electric power consumption. At that point, the temperature o the room air, and thus the energy content o the room, remains constant. The conservation o energy or the room becomes: de Qin in Qout out dt de hen steady operating conditions are established: 0 dt => elect, in Q out 0 Thereore, elect, in Qout UATi To => /m 2 o C30 m 2 T 25 o i C o This gives: Ti 26.1 C

7 PAGE 6 o 12 ater is being heated in a closed pan on top o a range while being stirred by a paddle wheel. During the process, 30 kj o heat is transerred to the water, and 5 kj o heat is lost to the surrounding air. The paddle-wheel work is 500 Nm. Determine the inal energy o the system (in kj ) i its initial energy is 10 kj. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. e take the water in the pan as our system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system yields: E E Ei Qin in Qout out Since the energy stored within the system is only internal energy: U Ui Qin sh,in Qout Thereore, U 10 kj 30 kj 0.5 kj 5 kj => U 35.5 kj

PAGE 7 o 12 Energy Conversion Eiciencies Typical eiciencies o conventional and high-eiciency electric and natural gas water heaters A 15 compact luorescent lamp provides as much light as a 60

PERFORMANCE OF A SYSTEM Desired output Perormance o a system can be deined as: Perormance Required output Energy utilized by the system Eiciency o a system can be deined as: Eiciency Energy supplied

cooled to the room temperature Combustion eiciency can be deined as: Q Amount o heat released during combustion combustion = HV Heating value o the uel burned HEATING VALUE (HV) OF THE FUEL Lower

8 PAGE 7 o 12 Energy Conversion Eiciencies Typical eiciencies o conventional and high-eiciency electric and natural gas water heaters A 15 compact luorescent lamp provides as much light as a 60 incandescent lamp The eiciency o a cooking appliance represents the raction o the energy supplied to the appliance that is transerred to the ood The deinition o the heating value o gasoline PERFORMANCE OF A SYSTEM Desired output Perormance o a system can be deined as: Perormance Required output Energy utilized by the system Eiciency o a system can be deined as: Eiciency Energy supplied to the system EFFICIENCY OF COMBUSTION Heating Value (HV) o the uel: is the amount o heat released when a unit amount o uel at room temperature is completely burned and the combustion products are cooled to the room temperature Combustion eiciency can be deined as: Q Amount o heat released during combustion combustion = HV Heating value o the uel burned HEATING VALUE (HV) OF THE FUEL Lower Heating Value (LHV): when the water leaves as a vapor Higher Heating Value (HHV): when the water in the combustion gases is completely condensed and thus the heat o vaporization is also recovered For example, gasoline: LHV = 44,000 kj/kg / HHV = 47,300 kj/kg

9 PAGE 8 o 12 Mechanical Eiciencies Mechanical eiciency o a system Pump eiciency / Turbine eiciency Motor eiciency / Generator eiciency OTHER ENERGY CONVERSION FFICIENCIES Annual Fuel Utilization Systems (AFUE): accounts or the combustion eiciency as well as other losses such as heat losses to unheated areas and start-up and cool-down losses net,electric Eiciency o automotive engines can be deined as: overall = combustion thermal generator HHV m Lighting eicacy: the amount o light output in lumens per o electricity consumed Energy utilized Cooking appliance eiciency: cooking appliance = Energy supplied to appliance net MECHANICAL EFFICIENCY OF A SYSTEM Mechanical energy output Emech,out E Mechanical eiciency o a system: mech = 1 Mechanical energy input E E Pump eiciency: Note: pump Mechanical energy increase o the luid = Mechanical energy input pump,u = useul pumping power supplied to the luid E mech,in mech,luid shat,in pump, u pump mech,loss Mechanical energy output shat,out turbine Turbine eiciency: turbine = Mechanical energy decrease o the luid E turbine, e Note: turbine,e = useul turbine power extracted rom the luid Mechanical power output Motor eiciency: motor = Electric power input shat,out elect,in Electric power output Generator eiciency: generator = Mechanical power input elect,out shat,in mech,luid mech,in Combined eiciency: pump,u E pump-motor = pump motor elect,in mech,luid elect,in elect,out turbine-gen = turbine generator turbine, e E elect,out mech,luid

PAGE 9 o 12 EXERCISE B-3-3 (Do-It-Yoursel) The eiciency o cooking appliances aects the thermal heat gain rom them since an ineicient appliance consumes a greater amount o energy or the same task, and

10 PAGE 9 o 12 EXERCISE B-3-3 (Do-It-Yoursel) The eiciency o cooking appliances aects the thermal heat gain rom them since an ineicient appliance consumes a greater amount o energy or the same task, and the excess energy consumed shows up as heat in the living space. The eiciency o open burners is determined to be 73% or electric units and 38% or gas units. Consider a 2 k electric burner at a location where the unit costs o electricity and natural gas are $0.09/kh and $0.55/therm, respectively. Determine the rate o utilized energy by the electric burner (in k ) and the unit cost o utilized energy (in $/kh ) or both electric and gas burners. Solution The operation o electric and gas ranges is considered. The rate o energy consumption and the unit cost o utilized energy are to be determined. Unit Conversion 1 therm = 29.3 kh The eiciency o the electric heater is given to be 73 percent. Thereore, a burner that consumes 2 k o electrical energy will supply the useul power o: Qutilized Energy inputeiciency 2 k0.73 1,460 (1.46 k) The unit cost o utilized energy is inversely proportional to the eiciency, and is determined rom: Cost o energy input $0.09/kh Cost o utilized energy $0.123/kh Eiciency 0.73 The eiciency o a gas burner is 38 percent; the energy input to a gas burner that supplies utilized energy at the same rate (1.46 k) is: Q 1.46 k Eiciency 0.38 utilized Qinput,gas 3,840 (3.84 k) Note that 1 therm = 29.3 kh, the unit cost o utilized energy or a gas burner is determined to be: Cost o energy input $0.55/29.3 kh Cost o utilized energy $0.049/kh Eiciency 0.38

PAGE 10 o 12 Consider a 3 k hooded electric open burner in an area where the unit costs o electricity and natural gas are $0.07/kh and $1.2/therm, respectively.

11 PAGE 10 o 12 Consider a 3 k hooded electric open burner in an area where the unit costs o electricity and natural gas are $0.07/kh and $1.2/therm, respectively. The eiciency o open burners can be taken to be 73% or electric burners and 38% or gas burners. Determine (a) the utilized rate o energy (in k ) and the unit cost o utilized energy (in $/kh ) or electric burner. Also, (b) determine the equivalent energy input required or gas burner, in order to have the same utilized rate o energy as the electric burner (in k ) and the unit cost o utilized energy (in $/kh ) or gas burner. Properties Gas burner eiciency: gas 38% Electric burner eiciency: electric 73% The eiciency o the electric heater is given to be 73 percent. Thereore, a burner that consumes 3 k o electrical energy will supply the useul power o: Qutilized Energy inputeiciency 3 k0.73 2,190 (2.19 k) The unit cost o utilized energy is inversely proportional to the eiciency, and is determined rom: Cost o energy input $0.07/kh Cost o utilized energy $0.096/kh Eiciency 0.73 The eiciency o a gas burner is 38 percent; the energy input to a gas burner that supplies utilized energy at the same rate (2.19 k) is: Q 2.19 k Eiciency 0.38 utilized Qinput,gas 5,760 (5.76 k) Note that 1 therm = 29.3 kh, the unit cost o utilized energy or a gas burner is determined to be: Cost o energy input $1.20/29.3 kh Cost o utilized energy $0.108/kh Eiciency 0.38

12 PAGE 11 o 12 EXERCISE B-3-4 (Do-It-Yoursel) Solution A hydraulic turbine-generator is to generate electricity rom the water o a lake. The overall eiciency, the turbine eiciency, and the turbine shat power are to be determined. Assumptions (1) The elevation o the lake remains constant. (2) The mechanical energy o water at the turbine exit is negligible. Properties 3 The density o water can be taken to be: 1,000 kg/m Recall that the mechanical energy (and power) can be deine as: 2 p V emech low work kinetic energy potential energy gz 2 2 p V Emech memech m gz 2 The water in a large lake is to be used to generate electricity by the installation o a hydraulic turbine-generator at a location where the depth o the water is 50 m. ater is to be supplied at a rate o 5,000 kg/s. I the electric power output (rom the generator) is 1,862 k with the generator eiciency is 95%, determine (a) the combined eiciency o the turbine-generator, (b) the mechanical eiciency o the turbine, and (c) the shat power supplied by the turbine to the generator (in k ). (a) e take the bottom o the lake as the reerence level or convenience. Then kinetic and potential energies o water are zero, and the change in its mechanical energy per unit mass becomes: 2 emech emech,in emech,out gh 9.81 m/s 50 m 491 J/kg Emech,luid memech,in emech,out 5, 000 kg/s0.491 kj/kg 2, 455 k elect,out 1,862 k Overall turbine-generator eiciency is: turbine-gen 0.76 E 2,455 k mech,luid turbine-gen 0.76 (b) The turbine eiciency is: turbine-gen turbine generator => turbine 0.80 generator 0.95 (c) The shat power output is: shat,out turbine Emech,luid 0.802,455 k 1,964 k

13 PAGE 12 o 12 A geothermal pump is used to pump geothermal water (brine) whose density is 1,050 kg/m 3 at a rate o 0.3 m 3 /s rom a depth o 200 m. For a pump eiciency o 74%, determine the required power input to the pump (in k ). Disregard rictional losses in the pipes, and assume the geothermal water at 200 m depth to be exposed to the atmosphere. Assumptions (1) The pump operates steadily. (2) Frictional losses in the pipes are negligible. (3) The changes in kinetic energy are negligible. (4) The geothermal water is exposed to the atmosphere and its ree surace is at atmospheric pressure. Properties 3 The density o geothermal water is given as: 1,050 kg/m The elevation o geothermal water and thus its potential energy changes, but it experiences no changes in its velocity and pressure. Thereore, the change in the total mechanical energy o geothermal water is the change in its potential energy, which is g z (per unit mass) and mg z or a given mass low rate Emech memech mgz V gz 1, 050 kg/m 0.3 m /s9.81 m/s 200 m 618 k Then the required power input to the pump becomes: Emech 618 k pump,elect 835 k pump-motor 0.74

ENGG 3260: Thermodynamics. Home Assignment 2 (Chapter 2)

ENGG 3260: Thermodynamics. Home Assignment 2 (Chapter 2) ENGG 60: Thermodynamics Home Assignment (Chapter ) 1. A person gets into an elevator at the lobby level of a hotel together with his 0-kg suitcase, and gets out at the 10 th floor 5 m above. Determine