AE4451 Seitzman Fall 2000 Handout 3 Entropy Change with Heat Addition In the Brayton cycle we replaced the actual combustion process with ideal reversible heat transfer i e a reversible heat exchanger It was stated then that reversible heat addition with no work implied a constant pressure process This handout examines in more detail the conditions required Temperature Constraints for Reversible Heat Transfer We begin by examining a device that adds heat to a flow reversibly but does no work As illustrated in Fig 1 imagine heat transfer from one body to another occurring along the adjoining be the heat transfer rate along length of two control volumes one surrounding each body Let Q an infinitesimal length of the walls To produce reversible heat transfer the first condition is that the heat transfer must occur across an infinitesimally small temperature difference or equivalently the local wall temperature of the two bodies must equal be equal at any given T 1 distance along the walls Otherwise the entropy transferred from one body e g Q wall 1 T would not be the same as the entropy transferred to the second body e g Q wall 2 Body1 Q wall 1 wall 2 Body2 Figure 1 Heat transfer from body 1 to body 2 In other words consider two bodies with a finite temperature difference and with Twall 1 Twall 2 2 In this case the entropy transfer out of body 1 would be less than the entropy T T transferred into body 2 Q Q This means that entropy must have been wall 1 wall 2 produced in the process thereby making the process irreversible Thus we see that a reversible heat exchanger must have an infinitesimally small temperature difference between the bodies fluids exchanging energy 1Recall 2The that heat transfer into or out of a control volume also transports entropy into or out of the control volume condition Twall 1 Twall 2 is impossible since it would require heat transfer from a colder body to a warmer body without doing work Pressure Constraints for Reversible Heat Transfer to a Fluid Low Velocity Now consider a fluid going through the heat exchanger exchanging energy with the walls of the device Fig 2 but again assuming no work except flow work is done As we just saw the local wall temperature must equal the temperature of the fluid at a given distance along the device Q Temp T3 Twall T3 Twall T2 T2 p2 s2 dT T3 p3 s3 Q T2 Distance Along Flow Figure 2 Heat transfer into a fluid in a reversible heat exchanger For such a control volume which again is an idealized model of our combustor the conservation equations assuming a calorically and thermally perfect gas with negligible velocity are Mass if this was a combustor we would be ignoring mass addition due to the fuel m 2 m 3 m Energy constant cp no work but flow work m h2 Q m h3 3 m h2 Q m h3 2 3 therefore 3 Q m c p dT 2 2 Q m c p dT or Entropy reversible no entropy production Q m s 3 T 2 3 m s 2 Q 2 T m s3 s 2 3 Replacing the heat transfer term on the left hand side of this equation with the result from the energy equation and using our state relation for s on the right hand side we find 2 3 2 m c p dT T T p m c p ln 3 R ln 3 T2 p2 and then solving the integral for cp const T3 T p m c p ln 3 R ln 3 T2 T2 p2 p3 0 or ln p2 m c p ln The only way this can be true is for p3 p2 1 i e reversible heat addition for a flow with negligible velocity must take place at constant pressure Pressure Constraints for Reversible Heat Transfer to a Fluid Arbitrary Velocity Now let us consider the case of reversible heat transfer no work to a fluid with nonnegligible velocity Again writing the conservation equations Mass same as above Energy again constant cp no work but now use stagnation values to include kinetic energy m ho 2 Q m ho 3 3 m ho 2 Q m ho 3 2 3 3 2 2 Q m c p dTo Q m c p dTo Entropy again no entropy production Q m so 3 T 2 3 m so 2 Q m so 3 so 2 T 2 3 Again using the energy equation to replace the heat transfer and using the s equation of state 3 m c p dTo To 3 po 3 2 T m c p ln To2 R ln po 2 m c pTo dTo To 3 po 3 2 T To m c p ln To2 R ln po 2 3 3 T p 1 2 dTo m c p 1 M m c p ln o 3 R ln o 3 To 2 2 po 2 To 2 3 where To T was replaced in the last step with the function of Mach number in parentheses Again solving the integral and assuming some average value of M since M could change as heat is added T T p 1 2 m c p 1 M av ln o 3 m c p ln o 3 R ln o 3 2 To 2 po 2 To 2 We see that if Mav is not zero po3 po2 The general result for po3 po2 is found by solving the above equation T R po 3 1 2 M av ln o 3 ln c p po 2 2 To 2 p T ln o 3 ln o 3 po 2 To 2 1 cp M av 2 2 R po 3 To 2 2 po 2 To 3 T ln o 3 To 2 1 M av 2 2 1 M av 2 where it is clear that po3 po2 1 only if a To3 To2 1 i e no heat addition or b Mav 0 i e heat addition at negligible velocity Otherwise po3 po2 1 i e reversible heat addition in high speed flows leads to a loss in stagnation pressure 4