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Reversibility Thermodynamics Professor Lee Carkner Lecture 14
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PAL # 13 Entropy Work needed to power isentropic compressor Input is saturated vapor at 160 kPa From table A-12, v = 0.12348, h 1 = 241.11, s 1 = Output is superheated vapor, P = 900 kPa, s 2 = s 1 = 0.9419 From table A-13, h 2 = Mass flow rate m’ = V’/v = 0.033 / 0.12348 = Get work from W = m’ h W = (0.27)(277.06-241.11) =
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Ideal Gas Entropy From the first law and the relationships for work and enthalpy we developed: We need temperature relations for du, dh, d v and dP ds = c v dT/T + R d v/v ds = c p dT/T + R dP/P
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Solving for s We can integrate these equations to get the change in entropy for any ideal gas process s 2 -s 1 = ∫ c p dT/T + R ln (P 2 /P 1 ) Either assume c is constant with T or tabulate results
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Constant Specific Heats If we assume c is constant: s 2 -s 1 = c v,ave ln (T 2 /T 1 ) + R ln ( v 2 / v 1 ) s 2 -s 1 = c p,ave ln (T 2 /T 1 ) + R ln (P 2 /P 1 ) since we are using an average
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Variable Specific Heats s o = ∫ c p (T) dT/T (from absolute zero to T) s o 2 – s o 1 = ∫ c p (T) dT/T (from 1 to 2) s 2 – s 1 = s o 2 – s o 1 – R u ln(P 2 /P 1 ) Where s o 2 and s o 1 are given in the ideal gas tables (A17-A26)
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Isentropic Ideal Gas Approximately true for low friction, low heat processes c v,ave ln (T 2 /T 1 ) + R ln ( v 2 / v 1 ) = 0 ln (T 2 /T 1 ) = -R/c v ln ( v 2 / v 1 ) ln (T 2 /T 1 ) = ln ( v 1 / v 2 ) R/cv (T 2 /T 1 ) = ( v 1 / v 2 ) k-1
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Isentropic Relations We can write the relationships in different ways all involving the ratio of specific heats k (T 2 /T 1 ) = (P 2 /P 1 ) (k-1)/k (P 2 /P 1 ) = ( v 1 / v 2 ) k Or more compactly P v k =constant Note that: R/c v = k-1
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Isentropic, Variable c Given that a process is isentropic, we know something about its final state P r = exp(s o /R) T/P r = v r (P 2 /P 1 ) = (P r2 /P r1 ) ( v 2 / v 1 ) = ( v r2 / v r1 )
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Isentropic Work We can find the work done by reversible steady flow systems in terms of the fluid properties But we know - w = v dP + dke + dpe w = -∫ v dP – ke – pe Note that ke and pe are often zero
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Bernoulli We can also write out ke and pe as functions of z and V (velocity) -w = v (P 2 -P 1 ) + (V 2 2 -V 2 1 )/2 + g(z 2 -z 1 ) Called Bernoulli’s equation Low density gas produces more work
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Isentropic Efficiencies The more the process deviates from isentropic, the more effort required to produce the work The ratio is called the isentropic or adiabatic efficiency
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Turbine For a turbine we look at the difference between the actual (a) outlet properties and those of a isotropic process that ends at the same pressure (s) T = w a / w s ≈ (h 1 – h 2a ) / (h 1 – h 2s )
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Compressor C = w s / w a ≈ (h 2s – h 1 ) / (h 2a – h 1 ) For a pump the liquids are incompressible so: P = w s / w a ≈ v (P 2 -P 1 ) / (h 2a – h 1 )
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Nozzles For a nozzle we compare the actual ke at the exit with the ke of an isentropic process ending at the same pressure N = V 2 2a / V 2 2s ≈ (h 1 – h 2a ) / (h 1 – h 2s ) Can be up to 95%
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Entropy Balance The change of entropy for a system during a process is the sum of three things S in = S out = S gen = We can write as: S sys is simply the difference between the initial and final states of the system Can look up each, or is zero for isentropic processes
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Entropy Transfer Entropy is transferred only by heat or mass flow For heat transfer: S = For mass flow: S = n.b. there is no entropy transfer due to work
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Generating Entropy friction, turbulence, mixing, etc. S sys = S gen + (Q/T) S gen = S sys + Q surr /T surr
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S gen for Control Volumes The rate of entropy for an open system: dS CV /dt = (Q’/T) + m’ i s i – m’ e s e + S’ gen Special cases: Steady flow (dS CV /dt = 0) S’ gen = - (Q’/T) - m’ i s i + m’ e s e Steady flow single stream: S’ gen = Steady flow, single stream, adiabatic: S’ gen =
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Next Time Read: 8.1-8.5 Homework: Ch 7, P: 107, 120, Ch 8, P: 22, 30
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