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by: Aric Jast MD

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# THERMODYNAMICS EML 3100

Aric Jast MD
FSU
GPA 3.52

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
9
WORDS
KARMA
25 ?

## Popular in Engineering Mechanical

This 9 page Class Notes was uploaded by Aric Jast MD on Thursday September 17, 2015. The Class Notes belongs to EML 3100 at Florida State University taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/205481/eml-3100-florida-state-university in Engineering Mechanical at Florida State University.

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Date Created: 09/17/15
The First Law and the Stirling Cycle Recall the expression for the First Law for a process of a closed system in terms of intensive variables A 12912 W12 p We have shown previously that the work for a process is given by 2 W12 2 IPdV 1 where p must be expressed as a function of v We have just shown that the change in internal energy is given by 2 2 Ann1du1cvdTcvltTrmw 1 1 We can now evaluate the heat transfer for a process From the First Law we write 2 2 q12 Au12 W12 2 IcvdT Ipdv 1 1 Recall the Stirling cycle and the work calculations for each process P 3 39 Heatmg at constant Isothermal expans1on v volume W34 RT3 lni4 W23 0 9 V3 2 Cooling at constant Isothermal compression 1 VOIume39 W41 0 w12 RT1 In 2 1 V3v2 v1v4 Calculate the heat transfer for each process and the cycle as a whole Assume a constant speci c heat Recall again 2 412 cvT2 T1 IPdV 1 Process 12 Isothermal compression T2 T1 Recall also that plvl pzvz constant 412 cvT2 T1RTI lnvi RT11nvi2 V1 V1 Process 23 Heat addition at constant volume v v1 constant 3 423 CVT3 T2 lPdV CVT3 T2 2 Process 34 Isothermal expansion T4 T3 Recall also that p4V4 p3V3 constant v v q34 cvT4 T3 RT3 11174 RT3 11174 V3 V3 Process 41 Cooling at constant volume v v1 constant 1 q cvT1 T4IpdvcvTl T4 4 Cycle Calculation There are three quantities of primary interest 0 Net work of the cycle 0 Total heat input to the cycle 0 Thermal ef ciency of the cycle Net work of the cycle wnet Zwij 2 W12 W23 W34 W41 wne RT1 mi2 0 RT3 mi4 V1 V3 v v Recall that V3 v and V4 v1 also note that v1gt vz thus lni2 lni1 and V1 V2 WW RT11nViRT31nil RT1 1111RT3 mil V1 V2 V2 V2 wnet Tllni1 2 Since v1 gt v and T3 gt T1 wnet gt 0 the work is positive indicating a net work output Recall our sign convention that positive work represents work output and negative work represents work input Heat input to the cycle According to our sign convention positive values of heat represent heat input and negative values represent heat output We therefore must examine the sign of the heat transfer in each ofthe processes Process 12 qlz RT1 lnvi2 RT1 In1 lt 0 3 heat output V1 V2 Process 23 q23 CV T3 T1gt 0 3 heat input Process 34 q34 RT3 lnvi4 RT3 lnvi1 lt 0 3 heat input V3 V2 Process 41 q41 CVT1 T4 CV T3 T1 gt 0 3 heat output Total heat input qin Q23 9341 1m 2 CVT3 T1RT31nV 1 V2 Thermal ef ciency of the Stirling Cycle 1 net work out ut w 77 thermal ef ciency p i heat input qm RT3 T1ln 1 V2 7 cVT3 T1RT31ni1 V2 Look at the Stirling cycle processes 23 and 41 heat transfer in terms of their magnitudes P N Q41 m3 Note that the magnitude of the heat output q41 is equal to the magnitude of the heat input qz3 Is it possible to use the heat output q41 to supply the heat input ng Yes in principle it can If this is done then the only heat input required will be q34 The use of heat output to supply heat input is called regeneration Since the only external heat supplied is now q34 the thermal ef ciency with regeneration becomes V RT T lnil 3 1 V2 T341 T1 77mg VT1Ti RT31n71 3 3 V2 Summary of the energy calculations A summary of the energy calculation is represented on the p v diagram below P Isothermal expansion 3 A1134 0 Heating at constant volume W23 0 934 W34 RT31nvi4 v m3 A1123 CvT3 T2 gt 3 Cooling at constant volume W41 0 Q41 A1141 CvT1 T4 CvT3 T2 Isothermal compression A1112 0 912 W12 RT1 111 V1 V3v2 Calculation of Work in the Stirling Cycle We have shown in the previous class that the Stirling cycle consists of four processes The four processes are I Isothermal expansion II Cooling at constant volume III Isothermal compression IV Heating at constant volume Let us number each of the states such that the initial state state 1 corresponds to the ambient condition of pressure temperature and volume Process 1 2 Isothermal compression Process 2 3 Heating at constant volume Process 3 4 Isothermal expansion Process 4 1 Cooling at constant volume The Stirling cycle is represented on a pressurevolume p v diagram as Heating at constant Isothermal expansion a volume 2 Cooling at constant volume 1 Isothermal compression l V3 V2 V1 V V We shall now compute the work for each of the processes and for the cycle as a whole Recall the work expression for a pistoncylinder system V2 2 m21pdVjpdV V1 1 V To evaluate this integral we must have an expression for p as a function of V A relationship among the p V and Tvariables is called an equation of state We shall now obtain the expressions for work for each of the four processes of the Stirling engine PROCESS l 2 ISOTHERMAL COMPRESSION The process is at constant temperature Assume an ideal gas pV mRT Since the temperature is constant for this process T1 T2 we have the following relationship between the pressure and volume pV mRT1 constant In particular mRz V Substitute for p in the expression for work V2 2 de 2 V W12 ipdVIpdVmRilEf7mR711fdanmR lni Note that since V2 lt V1 the term In2 lt 0 and W12 lt 0 This is consistent with our sign 1 convention that work done on the system to compress the gas is negative PROCESS 2 3 HEATING AT CONSTANT VOLUME For this constant volume process V3 V2 it follows that dV 0 The expression for work becomes 3 W23jpdV0 2 ISOTHERMAL EXPANSION PROCESS 3 4 For this constant temperature process we have T3 T4 constant thus pV mRT3 constant The expression for work becomes W34 2 jpdV mRledan W34 mRT31n5 3 COOLING AT CONSTANT VOLUME PROCESS 4 1 For this process V4 V1 hence dV 0 and the expression for work becomes NET WORK FOR THE CYCLE The net work for the cycle is the sum of the work for each of the processes n22 W WIZW23W23W41mR711n0mR7ln0 1 3 Since V4 V1 and V3 V2 the above expression can be simplified to W r12 mRT1 an mRT3 ln5 mRT1 ln5 mRT3 lnE V1 V3 V2 V 2 Finally we obtain M W t mRlt2 Jam Note that since T3 gt T1 we have Wnet gt 0 Thus the net work is positive indicating that the overall effect of the cycle was to have done work on the surroundings

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