Transcript Unit Eight Quiz Solutions and Unit Nine Goals Mechanical Engineering 370 Thermodynamics Larry Caretto April 1, 2003

Unit Eight Quiz Solutions and
Unit Nine Goals
Mechanical Engineering 370
Thermodynamics
Larry Caretto
April 1, 2003
Outline
•
•
•
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Solution to quiz eight
Revised schedule
Review second law
Goals for unit eight
– Calculating entropy with ideal gases
– Constant and variable heat capacity
– Isentropic calculations
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Course Schedule Changes
• Midterm on April 24 (Review April 22)
• Move schedule for unit ten on April 22–
24 up to April 8–10
• Unit ten quiz on April 22; no quiz on
April 29 following midterm
• Writing assignment due April 11
• Design project due May 16
3
The Second Law
• There exists an extensive thermodynamic property called the entropy, S,
defined as follows:
dS = (dU + PdV)/T
• For any process dS ≥ dQ/T
• For an isolated system dS ≥ 0
• T must be absolute temperature
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Entropy is a Property
• If we know the state of the system, we can
find the entropy
• We can use the entropy as one of the
properties to define the state
• Use the following if we are given a value of
s and a value of T or P
– if s < sf(T or P) => compressed liquid
– if s > sg(T or P) => gas (superheat) region
– otherwise in mixed region
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Reversible Processes
• This is an idealization; we cannot do
better than a reversible process
• Internal reversibility has dS = dQ/T
• Internal and external reversibility has
dSisolated system = 0
• It is possible to have dS = dQ/T for a
system with dSisolated system > 0
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Maximum Work (Output)
• dS  dQ/T; if reversible dS = dQ/T
• Compare two processes with between
same states (dU = dUrev)
• dS = dQrev/T = [dUrev + dWrev]/T  dQ/T
• [dUrev + dWrev]/T  [dU + dW]/T
• dWrev  dW
• Maximum work in a reversible process
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Maximum Adiabatic Work (DS = 0)
• From given inlet conditions, find the
initial state properties including sinitial
• The maximum work in an adiabatic
process occurs when sfinal = sinitial
• From sfinal = sinitial and one other
property of final state get all final state
properties
• Find work from first law as done in
previous quizzes and exercises
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First Unit Nine Goal
• As a result of studying this unit you
should be able to compute entropy
changes in ideal gases
– using constant heat capacities
– using equations that give heat capacities
as a function of temperature
– using ideal gas tables
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Second Unit Nine Goal
• As a result of studying this unit you
should be able to compute the end
states of isentropic processes in ideal
gases
– using constant heat capacities
– using equations that give heat capacities
as a function of temperature
– using ideal gas tables
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Ideal Gas Entropy
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Entropy defined as ds = (du + Pdv)/T
We can write Tds = du + Pdv
Since du = d(h – Pv) = dh – Pdv – vdP
We can also write Tds = dh - vdP
Ideal gas: Pv = RT, du = cvdT, dh = cpdT
For ideal gases ds = cvdT/T + Rdv/v
For ideal gases ds = cpdT/T – RdP/P
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Ideal Gas Entropy Change
• Integrate ideal gas ds equations to get
 v2 
dT
s2  s1   cv
 R ln 
T
 v1 
T1
T2
 P2 
dT
s2  s1   c p
 R ln 
T
 P1 
T1
T2
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Ideal Gas Entropy Tables
T
• Define
dT '
s (T )   c p (T ' )
T'
T0
o
T2
dT
• So that s (T2 )  s (T1 )  c p
T T
1
 P2 
o
o
s2  s1  s (T2 )  s (T1 )  R ln 
 P1 
o
o
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Example
• Air is heated from 300 K to 500 K at
constant pressure. What is Ds?
• From table A-17, page 849, so(300 K) =
1.71865 kJ/kg∙K and so(500 K) =
2.21952 kJ/kg∙K; Ds = 0.50087 kJ/kg∙K
• Assuming constant cp = 1.005 kJ/kg∙K
gives Ds = cpln(T2/T1) =
1.005 ln(500/300) = 0.51338 kJ/kg∙K
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Ideal Gas Isentropic Processes
• Start with ds = cpdT/T – RdP/P
• For ds = 0, cpdT/T = RdP/P
• Integrate for ds = 0 and constant cp to
get cpln(T2/T1) = R ln(P2/P1) so that
ln(T2/T1) = R ln(P2/P1)R/Cp or T2/T1 =
(P2/P1)R/Cp
• R/cp = (cp – cv)/cp = (cp/cv – 1)/ (cp/cv)
• R/cp = (k – 1)/k, where k = cp/cv
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Ideal Gas Isentropic Processes
• Final result: T2/T1 = (P2/P1)(k-1)/k
• Can derive similar equations for other
variables
– T2/T1 = (v2/v1)(k-1)
– P2/P1 = (v2/v1)k
• Apply only to ideal gases with constant
heat capacities in isentropic processes
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Variable Heat Capacity
• Set s2 – s1 = 0 in equations below for
ideal gas isentropic process
 v2 
dT
s2  s1   cv
 R ln   0
T
 v1 
T1
T2
 P2 
dT
s2  s1   c p
 R ln   0
T
 P1 
T1
T2
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Variable Heat Capacity
• Can solve for volume or pressure ratios
if T1 and T2 are given
 v2 
dT


c


R
ln
v
T T
v 
 1
1
T2
 P2 
dT
T c p T  R ln P1 
1
T2
• A trial-and-error solution is required if T1
or T2 are unknown
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Variable Heat Capacity Tables
• For ideal gas, isentropic processes, with
variable heat capacities we can define
Pr(T), and vr(T) such that
– v2/v1 = vr(T2)/ vr(T1)
– P2/P1 = Pr(T2)/ Pr(T1)
• Values of Pr(T), and vr(T) given in ideal
gas tables
• We still use Pv = RT at points
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Example Problem
• Adiabatic, steady-flow air compressor
used to compress 10 kg/s of air from
300 K, 100 kPa to 1 MPa. What is the
minimum work?
• Minimum work in adiabatic process is
when process is isentropic
  W  m
• First law: Q

(hout  hin )
u
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Example Continued
• What is outlet state for maximum work?
Use ideal gas tables for air on page 849
– Pr(300 K) = 1.3860
– P2/P1 = Pr(T2)/ Pr(T1) so that
– Pr(T2) = Pr(T1) P2/P1 = 1.3860(1000/100)
– What is T with Pr = 13.860
– Pr(570 K) = 13.50; Pr(580 K) = 14.38
– Interpolate to get Pr = 13.86 at T = 574.1 K
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Example Concluded
• Use enthalpy data from ideal gas tables
– hin = h(300 K) = 300.19 kJ/kg
– hout = h(574.1 K) = 579.86 kJ/kg
10 kg  300.19 kJ 579.87 kJ  1 MW  s

Wu  m (hin  hout ) 

 2.797 MW


s 
kg
kg
 1000kJ
• Negative value shows work input
• Minimum input in absolute value
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Repeat Example with Constant cp
• Here we use data for air that k = 1.4
and cp = 1.005 kJ/kg∙K
• T2 = T1(P2/P1)(k – 1)/k = (300 K)[(1000
kPa)/(100 kPa)](1.4 – 1)/1.4 = 579.2 K
Wu  m (hin  hout )  m c p (Tin  Tout ) 
10 kg 1.005 kJ
1 MW  s
300 K  579.2 K 
 2.806 MW
s
kg  K
1000kJ
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