CHAPTER. The First Law of Thermodynamics: Closed Systems

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1 CHAPTER 3 The First Law of Thermodynamics: Closed Systems

2 Closed system Energy can cross the boundary of a closed system in two forms: Heat and work FIGURE 3-1 Specifying the directions of heat and work. 3-1

3 Heat Energy transfer by virtue of temperature difference. If there is no heat transfer during a process, then the process is called adiabatic. 3-

4 30 kj heat Q=30 kj M= kg Δt=5 s Q =6 kw q=15 kj/kg Heat transfer per unit mass Heat transfer rate (kj/kg) (kj/s or kw) Sign Convention for Heat Heat in Q=+5 kj (+) ve if to the system (-)ve if from the system Q= - 5kJ Heat out 3-3

System boundary A spring is doing work on the surroundings Work An energy interaction which is not caused by a temperature difference between a system and its surroundings.

5 System boundary A spring is doing work on the surroundings Work An energy interaction which is not caused by a temperature difference between a system and its surroundings.(a rising piston, a rotating shaft,etc.) (kj/kg) W=5kJ The work done per unit time is called power and denoted (kj/s or kw) Sign Convention of Work W=-5kJ (+)ve if work done by a system (-)ve if work done on a system 3-4

6 Moving boundary work (P dv work): The expansion and compression work in a piston-cylinder device. FIGURE 3- A gas does a differential amount of work dw b as it forces the piston to move by a differential amount ds. 3-5

7 FIGURE 3-3 The area under the process curve on a P-V diagram represents the boundary work. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-6

8 FIGURE 3-4 The net work done during a cycle is the difference between the work done by the system and the work done on the system. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-7

9 First Law of Thermodynamics or the Conservation of energy Principle: Net energy transfer to (or from) the system = As heat and work Net increase (or decrease) İn the total energy of the system Q W = ΔU 3-9

10 (Remember from chapter 1: ΔE = ΔU + ΔKE + ΔPE ) Q W = ΔU + ΔKE + ΔPE where ΔU = m( - ) ΔKE =1/ m( - ) ΔPE =mg( - ) (kj) Most closed systems encountered in practice are stationary i.e. ΔPE =0 ΔKE =0 Q W = ΔU 3-10

11 Example: System boundary Q W U 0 0 KE PE Wpw = 18.5kJ 5 kg of steam Wpiston ΔU = m(u - u1 ) W=Wpw +Wpiston Q -(Wpw +Wpiston)= m(u - u1 ) Q = 80 kj u 1 u kj/kg kj/kg (+80kJ) ( ( -18.5kJ)+ Wpiston=(5kg)( )kJ/kg Wpiston =350kJ 3-1

12 Example: constant-pressure process, initially saturated water vapor. =? Q Wel = = -3.7 kj (-7.kJ)= 0.05 kg ( kj/kg) = kj/kg =300 kpa = =00ºC Table A Note: for constant pressure case: Q W other = ΔH and W = W boundary + W other

13 Example: constant volume Superheated steam T 1 v Q Q W U KE PE Q m( u ) u1 Note: for constant volume case: Q Wother U KE PE 3-14

14 Example: changing volume and pressure Evacuated system P 1 Q HO Partition V Q W U KE PE Q m( u ) u1 note: u1 uf@tsat u = uf +xufg 3-15

15 Specific Heats The energy required to raise the temperature of a unit of a substance by one degree. Cv : specific heat at constant volume Cp : specific heat at constant pressure Helium gas: V=constant m=1kg ΔT=1ºC Cv=3.13 kj/kgºc P=const m=1kg ΔT=1ºC Cp=5. kj/kgºc () (1) Cp > Cv Because at constant pressure, the energy required for expansion work must also be supplied to system. 3.13kJ 5.kJ 3-16

16 First law for constant volume: First law for constant pressure: 3-17

17 Ideal Gases: Pv RT Joule demonstrated that for ideal gases u=u(t) Cv=Cv(T) h u Pv Pv RT h u RT Since R is constant and u=u(t) h=h(t) Cp=Cp(T) and 3-18

18 Three ways of calculating u and h h h h 1 from tables h h 1 c c p p, avg ( T ) T dt Similarly: 3-19

19 differentiate dt c p dt =c v dt+rdt On a molar basis Specific heat ratio 3-0

20 3-1 Solids and Liquids: Can be approximated as incompressible: Again, specific heats depend on temperature only. The change in internal energy between states 1 and : c c c v p ) ( ) ( T T c u dt T c u u u avg

Example: insulated Q-W=ΔU or ΔU=0 m 50kg 80 o C Water 5ºC U U 0 mc( T T ) mc( T T ) m water iron 1 V v 5 C U iron water 0.5m 3 0.

21 Example: insulated Q-W=ΔU or ΔU=0 m 50kg 80 o C Water 5ºC U U 0 mc( T T ) mc( T T ) m water iron 1 V v 5 C U iron water 0.5m m 3 1 water 0 500kg kg 0.5 Specific heats are determined from table A kg(0.45 kj kg ) T C o 80 o C 500kg(4.18 kj kg o )( T C 5 o C) 0 T 5.6 o C 3-

Week 5. Energy Analysis of Closed Systems. GENESYS Laboratory

Week 5. Energy Analysis of Closed Systems Objectives 1. Examine the moving boundary work or PdV work commonly encountered in reciprocating devices such as automotive engines and compressors 2. Identify