Unit 9 – Ideal gas entropy
November 2, 2010
Outline Unit Nine – Entropy Calculations for Ideal Gases
• Quiz Quiz eight eight solut solution ion • Review • Goals Goals for for unit unit nine nine
Mechanical Engineering 370
• Calculati Calculating ng entropy entropy with with ideal gases gases
Thermodynamics
• Constant Constant and and variable variable heat heat capacity capacity
Larry Caretto
• Isentropi Isentropic c calculati calculations ons
November 2, 2010
• Use of equati equations ons and and ideal gas gas tables tables – Air tables versus tables with less data 2
Review
First Unit Nine Goal
• Entro Entropy py is a proper property ty • The maxim maximum um work work is done done in a reversible process
• As a result result of studyin studying g this unit unit you should be able to compute entropy changes in ideal gases for any change of state
– With sign convention for work on an input input device this is the minimum work input
– using constant heat capacities
• In an adiabatic adiabatic reversib reversible le process process the maximum work is done when s = 0
– using equations equations that give heat capacities as a function of temperature
– Find inlet/initial entropy at initial state – Final/outlet state determined by entropy entropy and another final/outlet property
– using ideal gas tables tables
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Second Unit Nine Goal
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Derive Ideal Gas Entropy
• As a result result of studyi studying ng this unit you you should be able to compute the end states of isentropic i sentropic processes in ideal gases
• Entropy Entropy define defined d as ds = (du (du + Pdv)/T Pdv)/T • Multip Multiply ly by by T: Tds Tds = du + Pdv Pdv • Tds Tds – Pdv = du = d(h d(h – Pv) = dh – Pdv Tds = dh dh – vdP vdP – vdP so Tds
– using constant heat heat capacities
• Ideal Ideal gas: gas: Pv = RT, RT, du du = cvdT, dh = cpdT
– using equations that that give heat capacities capacities as a function of temperature
• For For idea ideall gas gases es ds = cvdT/T + Rdv/v • For For idea ideall gas gases es ds = cpdT/T dT/T – RdP/P RdP/P
– using ideal gas tables tables
• Integrate Integrate last last two equat equations ions to get get s 5
ME 370 - Thermodynamics
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Unit 9 – Ideal gas entropy
November 2, 2010
Ideal Gas Entropy Change
Handling cp(T) = cv(T) + R
• dS = cvdT/T + Rdv/v = cpdT/T – RdP/P
• Easiest case is constant heat capacity
T 2 v T P R ln 2 c p ln 2 R ln 2 T 1 v1 T 1 P1
• Integrate between two states (1 and 2) T 2
s2 s1 cv
s2 s1 cv ln
v R ln 2 T v1
dT
T 1
• If we have an equation for cp(T) or cv(T) we can integrate cpdT/T or cvdT/T – We only need one equation since c p(T) and cv(T) are related by c p(T) = cv(T) + R
P dT s2 s1 c p R ln 2 T P1 T T 2
• Ideal gas tables give integral of cpdT/T called so as described next
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Ideal Gas Entropy Tables T
• Define
s o (T ) c p (T ' ) T 0
• So that
o
T 2
o
s (T 2 ) s (T 1 ) T 2
c p (T ' ) T 0
dT ' T '
Example
dT '
• Air is heated from 300 K to 500 K at constant pressure. What is s?
T '
c p (T ' )
T 1 T 1
c p (T ' ) T 0
• Compute result using air tables and repeat with constant heat capacity
dT ' T '
T 2 v T P R ln 2 c p ln 2 R ln 2 T 1 v1 T 1 P1
s2 s1 cv ln
dT ' T '
P2 P 1
P v T s s o (T 2 ) s o (T 1 ) R ln 2 s o (T 2 ) s o (T 1 ) R ln 2 1 P1 v1T 2
s2 s1 s (T 2 ) s (T 1 ) R ln o
o
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Example Answers
Second Example
• Air is heated from 300 K to 500 K at constant pressure. What is s? • From table A-17, page 936, so(300 K) = 1.70203 kJ/kg·K and so(500 K) = 2.21952 kJ/kg·K; s = 0.51749 kJ/kg·K • Assuming constant cp = 1.005 kJ/kg·K gives s = cpln(T2/T1) = 1.005 ln(500/300) = 0.51338 kJ/kg·K
• Air is heated from 300 K, 100 KPa to 500 K, 200 KPa. What is s using the air tables? 0.287 kJ 200 kPa s s o (500 K ) s o (300 K ) ln 100 kPa kg K
s
2.21952 kJ 1.70203 kg K
kg K
0.287 kJ kg K
ln 2
0.31856 kJ
s 11
ME 370 - Thermodynamics
kg K
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Unit 9 – Ideal gas entropy
November 2, 2010
Derive Isentropic Ideal Gas
Third Example • Air is heated from 300 K, 100 KPa to 500 K, 200 KPa. What is s assuming constant heat capacity?
• Start with ds = cpdT/T – RdP/P • For ds = 0, cpdT/T = RdP/P
500 K 0.287 kJ 200 kPa T P 1.005 kJ s c p ln 2 R ln 2 ln ln T 1 P1 kg K 300 K kg K 100 kPa
s
0.51338 kJ .19893 kJ kg K
• Air table result
kg K
0.31444 kJ kg K
kg K
• R/cp = (k – 1)/k, where k = cp/cv 13
Ideal Gas Isentropic Processes –
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Variable Heat Capacity • Set s2 – s1 = 0 in equations below for ideal gas isentropic process
• Final result: T2/T1 = (P2/P1)(k-1)/k T/P(k-1)/k
• ln(T2/T1) = ln(P2/P1)R/Cp or T2/T1 = (P2/P1)R/Cp • R/cp = (cp – cv)/cp = (cp/cv – 1)/ (cp/cv)
0.31856 kJ
s
• Integrate for ds = 0 and constant cp to get cpln(T2/T1) = R ln(P2/P1) giving
= constant
T 2
• Can derive similar equations for other variables
s2 s1 cv
– T2/T1 = (v1/v2)(k-1) => Tvk-1 = constant
T 1
– P2/P1 = (v1/v2)k => Pvk = constant
T 2
s2 s1 c p
• Apply only to ideal gases with constant heat capacities in isentropic processes
T 1
v R ln 2 0 T v1
dT
P R ln 2 0 T P1
dT
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Variable Heat Capacity II • What if we have an equation for cP(T) or cv(T) = cP(T) – R? T 2
T 1
cv
v R ln 2 T v1
dT
T 2
c
p
T 1
P R ln 2 T P1
dT
• Can solve for volume or pressure ratios if T1 and T2 are given • A trial-and-error solution is required if T1 or T2 are unknown 17
ME 370 - Thermodynamics
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so for Isentropic Processes s2 s1 s o (T 2 ) s o (T 1 ) R lnP2 P1 • s2 – s1 = 0 for isentropic process o o s (T 2 ) s (T 1 ) R lnP2 P1 • Given T1 and P2/P1 – find so(T1) in tables – Compute so(T2) = so(T1) + R ln(P 2/P1) – Interpolate in tables to find T 2 that gives s o
• Need to solve by iteration if given v2/v1 instead of P2/P1 – Use so(T2) = so(T1) + R ln(v 1T2/v2T1) 18
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Unit 9 – Ideal gas entropy
•
November 2, 2010
Pr (T) for Isentropic Process For s2 = s1 s o (T 2 ) s o (T 1 ) R ln P2 P1
• Solve for P2/P1 and define Pr (T) = Aexp[so(T)/R]
s (T 2 ) s (T 1 )
– A is an arbitrary P2 e constant that P1 cancels
o
o
R s o (T 2 ) s o (T 1 ) R
ln
P2 P1
s o (T 2 )
e
R s o (T 1 )
e
R
Pr T 2 Pr T 1
• We can tabulate the function Pr (T) such that Pr (T2) = (P2/P1)Pr (T1) relates the end states for s2 = s1
vr (T) for Isentropic Process • We can tabulate the function Pr (T) such that Pr (T2) = (P2/P1)Pr (T1) relates the end states for s2 = s1 • If we know v2/v1 = (T2/T1)(P1/P2) can use T 1
Pr T 2 Pr T 1
P2 P1
T 2 v1 T 1 v2
v1 v2
T 1 Pr T 2 T 2 Pr T 1
Pr T 1 vr T 1 T 2 vr T 2 Pr T 2
• Tabulate function vr (T) such that vr (T2) = (v2/v1)vr (T1) gives end states for s2 = s1 19
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Variable Heat Capacity Tables
Example Problem
• For ideal gas, isentropic processes, with variable heat capacities we can define Pr (T), and vr (T) such that
• 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 input?
– v 2/v1 = vr (T2)/ vr (T1) – P2/P1 = Pr (T2)/ Pr (T1)
Relates end states such that s(T2,P2) = s(T1,P1)
• Minimum work input for adiabatic process is when process is isentropic
• Values of Pr (T), and vr (T) given in ideal gas tables
• Steady first law for one inlet, one outlet, negligible KE and PE:
• We still use Pv = RT at individual data points 21
W m (hout hin ) Q u
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Example Continued
Example Concluded
• What is outlet state for maximum work? Use ideal gas tables for air on page 934
• Use enthalpy data from ideal gas tables
– Pr (300 K) = 1.3860
– hout = h(574.1 K) = 579.86 kJ/kg (interpolated)
– P2/P1 = Pr (T2)/ Pr (T1) so that
m (h h ) W u in out
– Pr (T2) = Pr (T1) P2/P1 = 1.3860(1000/100)
10 kg 300.19 kJ
579.87 kJ 1 MW s 2.797 MW s kg kg 1000 kJ • Negative value shows work input
– What is T with Pr = 13.860 – Pr (570 K) = 13.50; P r (580 K) = 14.38 – Interpolate to get P r (T2) = 13.86 at T 2 = 574.1 K
• Work input of 2.797 MW is minimum 23
ME 370 - Thermodynamics
– hin = h(300 K) = 300.19 kJ/kg
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Unit 9 – Ideal gas entropy
November 2, 2010
Repeat with Constant c p
Class Exercise
• Here we use data for air at 300 K: k = 1.4, cp = 1.005 kJ/kg·K (page 911)
• Air in an adiabatic piston-cylinder device with V1 = 0.1 m3, P1 = 50 kPa, and T1 = 300 K is compressed to V2 = 0.01 m3. Find the minimum work input
• T2 = T1(P2/P1)(k – 1)/k = (300 K)[(1000 kPa)/(100 kPa)](1.4 – 1)/1.4 = 579.2 K u m (hin hout ) m c p (T in T out ) W 10 kg 1.005 kJ 1 MW s 300 K 579.2 K 2.806 MW s kg K 1000 kJ
• Do the calculation for variable heat capacity and repeat with constant c p – Variable: v2/v1 = vr (T2)/ vr (T1) and w = – u – Constant: T2/T1 = (v1/v2)(k-1) and w = – cvT
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Constant Heat Capacity Solution
Constant Heat Capacity Solution II
• State data: V1 = 0.1 m3, P1 = 50 kPa, T1 = 300 K, V2 = 0.01 m3 • First law: Q = U + W = m(u2 – u1) + W so W = m(u1 – u2) for Q = 0 • Second law: Maximum work in adiabatic process is when s = 0 • Properties: Air as ideal gas with constant heat capacity (cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K,k = 1.4 at 300K, p 911)
• Ideal gas equations : PV = mRT, du = cvdT, dh= cpdT; for s = 0 with constant heat capacity, T2 = T1(v1/v2)k-1 • Solution:
v T 2 T 1 2 v1
k 1
1.4 1
0.1 m3 (300 K ) 3 0.01 m
753.6 K
T 1
W mu1 u2 m cv dT mcv T 1 T 2 T 2
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Constant Heat Capacity Solution III • Solution concluded: m
P1V 1 RT 1
50 kPa 0.1 m
3
0.05807 kg 0.287 kPa m3 300 K kg K T 1
W mu1 u2 m cv dT mcv T 1 T 2 T 2
0.05807 kg
0.718 kJ kg K
300 K 753.6 K 18.9 kJ
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Variable Heat Capacity Solution • State data: V1 = 0.1 m3, P1 = 50 kPa, T1 = 300 K, V2 = 0.01 m3 • First law: Q = U + W = m(u2 – u1) + W so W = m(u1 – u2) for Q = 0 • Second law: Maximum work in adiabatic process is when s = 0 • Properties: Use ideal gas tables for air on page 936
Minimum work input is 18.9 kJ 29
ME 370 - Thermodynamics
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Unit 9 – Ideal gas entropy
November 2, 2010
Variable Heat Capacity Solution II • Ideal gas equations : – PV = mRT, – u = u(T2) – u(T1)
W mu1 u 2 mu T 1 u T 2
• Solution concluded: m
P1V 1 RT 1
50 kPa 0.1 m3
0.287 kPa m 3 kg K
– h = h(T2) – h(T1) – For s = 0, v r (T2) = vr (T1)(v2/v1) = vr (T1)(V2/V1)
• Solution: At T1 = 300 K vr = 621.2 – v r (T2) = vr (T1)(V1/V2) = 621.2(0.01 m 3) / (0.1 m3) = 62.12 – In tables vr (730 K) = 62.13 so T 2 = 730 K
0.05807 kg
300 K
• Air tables: u(T1 = 300 K) = 214.07 kJ/kg·K and u(T2 = 730 K) = 536.07 kJ/kg·K and
214.07 kJ 536.07 kJ kg kg 18.7 kJ
W mu T 1 uT 2 0.05807 kg
Minimum work input is 18.7 kJ 31
ME 370 - Thermodynamics
Variable Heat Capacity Solution III
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