Reversibility

Thermodynamics

Professor Lee Carkner

Lecture 14

PAL # 13 Entropy Work needed to power isentropic

compressor Input is saturated vapor at 160 kPa

From table A-12, v = 0.12348, h1 = 241.11, s1 =

Output is superheated vapor, P = 900 kPa, s2= s1 = 0.9419 From table A-13, h2 =

Mass flow rate m’ = V’/v = 0.033 / 0.12348 =

Get work from W = m’h W = (0.27)(277.06-241.11) =

Ideal Gas Entropy

From the first law and the relationships for work and enthalpy we developed:

We need temperature relations for du, dh, dv and dP

ds = cv dT/T + R dv/v ds = cp dT/T + R dP/P

Solving for s

We can integrate these equations to get the change in entropy for any ideal gas process

s2-s1 = ∫ cp dT/T + R ln (P2/P1)

Either assume c is constant with T or tabulate

results

Constant Specific Heats

If we assume c is constant:

s2-s1 = cv,ave ln (T2/T1) + R ln (v2/v1)

s2-s1 = cp,ave ln (T2/T1) + R ln (P2/P1)

since we are using an average

Variable Specific Heats

so = ∫ cp(T) dT/T (from absolute zero to T)

so2 – so

1 = ∫ cp(T) dT/T (from 1 to 2)

s2 – s1 = so2 – so

1 – Ru ln(P2/P1) Where so

2 and so1 are given in the ideal gas

tables (A17-A26)

Isentropic Ideal Gas

Approximately true for low friction, low heat processes

cv,ave ln (T2/T1) + R ln (v2/v1) = 0

ln (T2/T1) = -R/cv ln (v2/v1)

ln (T2/T1) = ln (v1/v2)R/cv

(T2/T1) = (v1/v2)k-1

Isentropic Relations

We can write the relationships in different ways all involving the ratio of specific heats k

(T2/T1) = (P2/P1)(k-1)/k

(P2/P1) = (v1/v2)k

Or more compactly Pvk =constant

Note that: R/cv = k-1

Isentropic, Variable c

Given that a process is isentropic, we know something about its final state

Pr = exp(so/R) T/Pr = vr

(P2/P1) = (Pr2/Pr1)(v2/v1) = (vr2/vr1)

Isentropic Work

We can find the work done by reversible steady flow systems in terms of the fluid properties

But we know

-w = vdP + dke + dpe

w = -∫ vdP – ke – pe Note that ke and pe are often zero

Bernoulli

We can also write out ke and pe as functions of

z and V (velocity)

-w = v(P2-P1) + (V22-V2

1)/2 + g(z2-z1)

Called Bernoulli’s equation

Low density gas produces more work

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

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 = wa / ws ≈ (h1 – h2a) / (h1 – h2s)

Compressor

C = ws / wa ≈ (h2s – h1) / (h2a – h1)

For a pump the liquids are incompressible so:

P = ws / wa ≈ v(P2-P1) / (h2a – h1)

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 = V22a / V2

2s ≈ (h1 – h2a) / (h1 – h2s)

Can be up to 95%

Entropy Balance

The change of entropy for a system during a process is the sum of three things Sin = Sout = Sgen =

We can write as:

Ssys is simply the difference between the initial and final states of the system Can look up each, or is zero for isentropic processes

Entropy Transfer

Entropy is transferred only by heat or mass flow For heat transfer:

S = S = S =

For mass flow: S =

n.b. there is no entropy transfer due to work

Generating Entropy

friction, turbulence, mixing, etc.

Ssys = Sgen + (Q/T)

Sgen = Ssys + Qsurr/Tsurr

Sgen for Control Volumes

The rate of entropy for an open system:

dSCV/dt = (Q’/T) + m’isi – m’ese + S’gen

Special cases:

Steady flow (dSCV/dt = 0) S’gen = -(Q’/T) - m’isi + m’ese

Steady flow single stream: S’gen =

Steady flow, single stream, adiabatic: S’gen =

Next Time

Read: 8.1-8.5 Homework: Ch 7, P: 107, 120, Ch 8, P: 22,

30