Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University
Examine the moving boundary work or P d work commonly encountered in reciprocating devices such as automotive engines and compressors. Identify the first law of thermodynamics for closed systems. Develop the energy balance applied to closed systems. Define the specific heat at constant volume and the specific heat at constant pressure. Relate the specific heats to the calculation lation of the changes in internal energy and enthalpy of ideal gases. Describe incompressible substances and determine the changes in their internal energy and enthalpy. Solve energy balance problems for closed systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances.
4 MOING BOUNDARY WORK Boundary work (W b ): work associated with a moving boundary W b is one form of mechanical work frequently encountered in practice is associated with the expansion or compression of a fluid in a piston cylinder device. Quasi-equilibrium process: a process during which the system remains nearly in equilibrium at all times. It is also called a quasi-static process, is closely approximated by real engines, especially when the piston moves at low velocities 4
δw b F ds PAds P d (4-) W b P d (kj) (4-) A da P d (4-) Area under the process curve on ap P- diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system. 5
Work is a path function (i.e., it depends on the path followed as well as the end states). The cycle shown in Fig. 4 5 produces a net work output (W net ) because the work done by the system during the expansion process (area under path A) is greater than the work done on the system during the compression part of the cycle (area under path B), and the difference between these two is the net work done during the cycle (the colored area). 6
W b for a Constant-olume Process W b P d 0. 0 W b for a Constant-Pressure Process (Isobaric Process) W b P d P( ) P m(v v ) 7
Isothermal Compression or expansion of an Ideal Gas C C P or C mrt P o C ln P C ln d C P d W b P P mrt P P o P 8
Polytropic Process: thermodynamics process that obeys: During actual expansion and compression processes of gases, pressure and volume are often related by P n C, where n and C are constants. -n n P C (4-8) P P W b -n+ -n+ (4-9) -n P P P d C d C n+ n n C For Gas and Liquid For an ideal gas (P mrt),eq. (4-9) can also be written as W b P P mr(t T ) n n n (4-0) For the special case of n the boundary work becomes For an ideal gas this - W b P d C d P ln result is equivalent to the isothermal process 9
Example: A piston cylinder device contains 50 kg of water at 50 kpa and 5 C. The cross-sectional area of the piston is 0. m. Heat is now transferred to the water, causing part of it to evaporate and expand. When the volume reaches 0. m, the piston reaches a linear spring whose spring constant is 00 kn/m. More heat is transferred to the water until the piston rises 0 cm more. Determine (a) the final pressure and temperature and (b) the work done during this process. Also, show the process on a P- diagram. 0
Solution: (a) T P Fs k x ( 00 )( 0. ) P P + P + 50+ 450 A A 0. o 5 C v v 0 0. 0000 m /kg f@5 C 50kPa kpa mv ( 50)(. ). m 0 0000 0 05 + xap. + ( 0. )( 0. ) 0. 0 m 0. m 0. v 0. 0044 m 50 m /kg
At 450 kpa, v f 0.00088 m /kg and v g 0.49 m /kg. Noting that v f < v < v g, the final state is a saturated mixture and thus the final temperature is T Tsat@ kpa 47. 9 450 o C (b) The pressure remains constant during process - and changes linearly (a straight line) during process -. Then the boundary work during this process is simply the total area under the process curve, P + P Area P( ) + ( ) 50+ 450 50( 0. 0. 05) + ( 0. 0. ) W b 44. 5 kj
Example: A piston cylinder device contains 0.5 kg of air initially at MPa and 50 C C. Theairisfirst is expanded isothermally to 500 kpa, then compressed polytropically p ywith a polytropic p exponent of. to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle.
Solution: R air 0.87 kj/kg.k (Table A-). For the isothermal expansion process: mrt ( 0. 5 )( 0. 87 )( 50 + 7 ) 04 0. m P 000 mrt ( 0. 5 )( 0. 87 )( 50 + 7 ) 0564 0. m P 500 0. 0564 W b, P ln( ) ( 000)( 0. 04)ln( ) 7. 8 0. 04 kj 4
For the polytropic compression process: P n n.. P ( 500)( 0. 0564 ) 000( ) 0. 069 500 m P P ( 000)( 0. 069) ( 500)( 0. 0564) W b, 4. 86 kj n. For the constant pressure compression process: W b, P ( ) 000 ( 0. 04 0. 069 ) 6. 97 The net work for the cycle is the sum of the works for each process W + W + W 78. + ( 4. 86) + ( 6. 97) 4. kj Wnet b, b, b, 65 kj 5
4 ENERGY BALANCE FOR CLOSED SYSTEMS Ein Eout ΔE system (kj) (4-) Net energy transfer Change in internal, kinetic, by heat, work, and mass potential etc.. energies Ein Eout (Qin Qout ) + (Win Wout ) ΔEsystem ΔU + ΔKE+ ΔPE net heat input: net work output: Q W net, in Q net,in W or, in the rate form,as E in Eout Q net, out de system net,out in W /dt Q out ΔE out W system in (kw) (4-) Work: moving boundary, electr., shaft, etc ) Rate of net energy transfer Rate of change in internal, by heat, work, and mass kinetic, potential etc.. energies 6
For constant rates, the total quantities during a time interval Δt are related to the quantities per unit time as Q Q Δt,W W Δt, and ΔE (de/dt) Δt (kj) (4-) The energy balance (Eq.4-) can be expressed on a per unit mass basis as e e Δe (kj/kg) (4-4) in out system For a closed system undergoing a cycle, the initial and final states are identical: ΔE E W in system E net,out out Q E 0 net,in E or or E in 0 E W net,out out Qnet,in (4-6) 7
Taking heat transfer to the system and work done by the system to be positive quantities, the energy balance for a closed system can also be expressed as Q W ΔE ΔUU + ΔKE + ΔPE Q: net heat input W: net work output Work (W) is the sum of boundary and other forms of work (such as electrical and shaft, etc ). W W other + W b For a constant-pressure process: W b + ΔU ΔH 8
Example: A piston cylinder device contains steam initially at MPa, 450 C, and.5 m. Steam is allowed to cool at constant pressure until it first starts condensing. Show the process on a T-v diagram with respect to saturation lines and determine (a) the mass of the steam, (b) the final temperature, and (c) the amount of heat transfer. 9
Solution: Ein Eout ΔE system (kj) Q W ΔU m(u u ) ( since KE PE out b, out 0 Q m(h h ) Q out since ΔU + W b ΔH during a constant pressure quasi-equilibrium ilib i process. The properties of water are (Tables A-4 through A-6) ) P T MPa 450 o C v h 0. 045 m /kg 7. kj/kg m. 5 7. v 0. 045 565 kg 0
(b) P MPa 9 0 T Tsat@ Mpa 799. C Sat.vapor h h 777. kj/kg g@ Mpa (c) Substituting, the energy balance gives Q out m(h h ) 7. 565 ( 7. 777. ) 4495 kj
: A piston cylinder device initially contains 0.8 m of saturated water vapor at 50 kpa. At this state, the piston is resting on a set of stops, and the mass of the piston is such that a pressure of 00 kpa is required to move it. Heat is now slowly transferred to the steam until the volume doubles. Show the process on a P-v diagram with respect to saturation lines and determine (a) the final temperature, re (b) the work done during this process, and (c) the total heat transfer.
Q Solution: W ΔU Ein Eout ΔE system (kj) m(u u ) ( since KE PE in b, out 0 Q + in m(u u ) W b,out, stationary system) The properties of steam are (Tables A-4 through A-6) P 50 kpa v vg@ 50 kpa 0. 787 m /kg Sat. vapor u u 56. kj/kg. m 0 8. kg v 0. 787. 6 v. 475 m /kg m. g@ 50 kpa 8 P 00 kpa v. 475 T 66 0 C u 4. 4 kj/kg
(b) The work done during process - is zero (since const) and the work done during the constant pressure process - is: (c) () W b, out P d P( ) 00 (. 6 0. 8 ) 40 Q in m(u u ) + W b, out. ( 44. 568. ) + 40 kj kj 4
4 SPECIFIC HEATS Specific heat: a property that will enable us to compare the energy storage capabilities of various substances. The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way Two kinds of specific heats: a- Specific heat at constant volume (c v ): can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. u c v ( ) (4-9) T 5
b- Specific heat at constant volume (c P ): can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant. h c P ( ) P (4-0) T c P > c v because at constant pressure the system is allowed to expand and the energy for this expansion work must also be supplied to the system. c P, c v units: kj/kg C or kj/kg K. ΔT( C) Δ T(K) 6
c P, c v units: kj/kg C or kj/kg K. ΔT( C) Δ T(K) The specific heats are sometimes given on a molar basis. They are then denoted by c and c and have the unit C P v kj/kmol or kj/kmol K. The unit kj/kmol is very convenient in the thermodynamic analysis of chemical reactions. 7
4 4 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES State equation for ideal gas: Pv RT The internal energy for ideal gas is a function of the temperature only: u u (T) (4-) The enthalpy of an ideal gas is also a function of temperature only: h h (T) (4-) h u+pv h u + RT Pv RT Since u and h depend only on temperature for an ideal gas, the specific heats c P and c v cp also v depend, at most, on temperature only. 8
The differential changes in the internal energy and enthalpy of an ideal gas can be expressed as: du cv(t ) dt (4-) dh cp(t ) dt (4-4) The change in internal energy or enthalpy for an ideal gas during a process from state to state is determined by u u u cv(t ) dt ( kj / kg ) Δ (4-5) v Δu u u Δ u Δu M h h h cp(t ) dt ( kj / kg ) Δ h Δh M Δu cv (T ) dt ( kj / kmol ) M: molar mass Δ (4-6) h h c (T ) dt ( kj / kmol ) Δh P At low pressures, all real gases approach ideal-gas behavior, and therefore their specific heats depend on temperature t only 9
The specific heats of real gases at low pressures are called ideal-gas specific heats, or zero-pressure specific heats, and are often denoted c p0 and c v0 The use of ideal-gas specific heat data is limited to low pressures, but these data can also be used at moderately high pressures with reasonable accuracy as long as the gas does not tdeviate from ideal-gas bh behavior significantly. ifi By performing the integrations in Eqs. 4 5 and 4 6 then u and h for a number of ideal gases have been tabulated in kj/kg for air (Table A 7) and in kj/kmol for other gases Table A-8 to A-5 The variation of specific heats with temperature is smooth and may be approximated as linear over small temperature intervals 0
Therefore the specific heat functions in Eqs. 4 5 and 4 6 can be replaced by the constant average specific heat values. u u cv, avg (T T ) kj / kg (4-7) h h c (T T p, avg ) kj / kg (4-8) The specific heat values for some common gases are listed as a function of temperature in Table A b.
The average specific heats c p,avg and c v, avg evaluated from Table A- at the average temperature (T +T )/. If the final temperature T is not known, the specific heats may be evaluated at T or at the anticipated average temperature. Then T can be determined by using these specific heat values. The value of T can be refined, if necessary, by evaluating the specific heats at the new average temperature. Another way of determining the average specific heats is to evaluate them at T and T and then take their average. Usually both methods give reasonably good results, and one is not necessarily better than the other.
The presence of the constant-volume specific heat c v in an equation should not lead one to believe that this equation is valid for a constant-volume process only. To summarize, there are three ways to determine the internal energy and enthalpy changes of ideal gases () Δu u u Table A-7, A-8 to A-5 For Air In mole for different gases () ΔuΔ u cv (T ) dt Table A-c () Δ u c v, avg Δ T Table A-a(at 00K or near 00 K), A-b
Specific Heat Relations of Ideal Gases A special relationship between c p and c v for ideal gases can be obtained by differentiating the relation: h u + RT dh du + R dt replacing dh by c p dt and du by c v dt and dividing the resulting expression by dt, we obtain: cp cv + R ( kj / kg.k ) (4-9) When the specific heats are given on a molar basis, R in the above equation should be replaced by the universal gas constant R u c Specific heat ratio k, defined as: cp k c (4-) cv Ru ( kj / kmol.k ) (4-0) P + v 4
Example: A piston cylinder device contains 0.8 kg of nitrogen initially iti at 00 kpaand 7 C. The nitrogen is now compressed slowly in a polytropic process during which P. constant until the volume is reduced by one-half. Determine the work done and the heat transfer for this process. 5
Solution: R 0.968 kpa.m /kg.k (Table A-). c v 0.744 kj/kg.k (Table A-b) @ T avg (69+00)/ 5 K or you can use (Table A-a) to get c v 0.74 kj/kg.k Ein Eout ΔE system (kj) W Q ΔU m(u u ) m c (T T ) b, in out v The final pressure and temperature of nitrogen are:.. / / P P P ( ) P ( 00 ) P P P 46. T T 0. 5 00 T T P 00 64. kpa 69. K 6
The boundary work for this polytropic process: P P W b, in P d n mr(t T n ( 0. 8 )( 0. 968 )( 69. 00 ) 54. 8 kj. ) Q W b,in m cv(t T ) 54. 8 ( 0. 8 )( 0. 744 )( 69. out. 6 kj 00 ) 7
A piston cylinder device, with a set of stops on the top, initially contains kg of air at 00 kpa and 7 C. Heat is now transferred to the air, and the piston rises until it hits the stops, at which point the volume is twice the initial volume. More heat is transferred until the pressure inside the cylinder also doubles. Determine the work done and the amount of heat transfer for this process. Also, show the process on a P-v diagram. 8
Solution: R 0.87 kpa.m /kg.k (Table A-). Ein Eout ΔE system (kj) Q in W b,out ΔU m(u u ) Q m(u u ) + W in b,out The initial and the final volumes and the final temperature of air are: mrt 0. 87 00 P 00. 9. 58 m P P P T T T T P. 9 m 400 00 00 00 K 9
No work is done during process - since. The pressure remains constant during process - and the work done during this process is W b P d P ( ) 00 (. 58. 9 ) 58 The initial and final internal energies of air are (Table A-7) u u u u @ 00 K @00K 4. 07 9. kj/kg kj/kg Q in ( 9. 4. 07 ) + 58 46 Alternative solution: c v 0.800 kj/kg.k @T avg (00+00)/750 K (Table A-b) Q kj m(u u ) + Wb,out mcv(t T ) + W in kj 0. 800 ( 00 00) + 58 48 b, out kj 40
4 5 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Liquids and solids can be approximated as incompressible substances (substance whose specific volume (or density) is constant) The constant-volume and constant-pressure specific heats are identical for incompressible substances c c c (4-) p v 4
Internal Energy Changes Specific heats of incompressible substances depend on temperature only du cv dt c(t)dt (4-) The change in internal energy between states and is then obtained by integration: Δu u u c(t)dt (kj/kg) (4-4) For small temperature intervals, a c value at the average temperature can be used and treated as a constant: Δu avg c (T T ) (kj/kg) (4-5) 4
Enthalpy Changes Using the definition of enthalpy h u + Pv and noting that v constant. Differentiate i h: Integrating: 0 dh du+ v dp + Pdv du+ vdp (4-6) Δh Δu+ v dp cavgδt + vδp (4-7) For solids, the term vδp is insignificant and thus: Δh Δu c avg For liquids, id two special cases are commonly encountered: ΔT. Constant-pressure processes, as in heaters(δp0): Δh Δu c ΔT. Constant-temperature processes, as in pumps (ΔT0): Δh vδp avg 4
Comment on the last equation Δh vδp For a process between states and it can be expressed as: h h v(p P ) By taking state to be the compressed liquid state at a given T and P and state to be the saturated liquid state at the same temperature, The enthalpy of the compressed liquid can be expressed as: h h + v (P f@t P ) @P,T f@t sat@t (4-8) As discussed in Ch. the enthalpy of the compressed liquid could be taken h @P,T h f@ T however, the contribution of the last term is often very small, and is neglected. 44
Example: In a manufacturing facility, 5-cm-diameter brass balls (ρ 85 kg/m and c p 0.85 kj/kg C) initially at 0 C are quenched in a water bath at 50 C for a period of min at a rate of 00 balls per minute. If the temperature of the balls after quenching is 74 C, determine the rate at which heat needs to be removed from the water in order to keep its temperature t constant t at 50 C. 45
Solution: Analysis: We take a single ball as the system. The energy balance for this closed system can be expressed as: Then the rate of heat transfer from the balls to the water becomes Therefore, heat must be removed from the water at a rate of 988 kj/min in order to keep its temperature constant at 50 C since energy input must be equal to energy output tfor a system whose energy level lremains constant. t That is, 46