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ENGR 2213 Thermodynamics

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Title: ENGR 2213 Thermodynamics


1
ENGR 2213 Thermodynamics
  • F. C. Lai
  • School of Aerospace and Mechanical
  • Engineering
  • University of Oklahoma

2
Two Important Relations
(dQ)rev T dS
(dQ)rev dU (dW)rev
First Law of Thermodynamics
T dS dU p dV
T ds du p dv
H U pV
dH dU p dV V dp
T dS dH - V dp
T ds dh - v dp
3
Entropy Change of an Ideal Gas
T ds du p dv
For an ideal gas, du cv dT, pv RT
4
Entropy Change of an Ideal Gas
T ds dh - v dp
For an ideal gas, dh cp dT, pv RT
5
Entropy Change of an Ideal Gas
Standard-State Entropy
Reference state 1 atm and 0 K
6
Isentropic Processes of Ideal Gases
1. Constant Specific Heats
(a)
(b)
7
Isentropic Processes of Ideal Gases
1. Constant Specific Heats
(a)
R cp cv k cp/cv R/cv k 1
8
Isentropic Processes of Ideal Gases
1. Constant Specific Heats
(b)
R cp cv k cp/cv R/cp (k 1)/k
9
Isentropic Processes of Ideal Gases
1. Constant Specific Heats
p1V1k p2V2k
Polytropic Processes pVn
constant
n 0 constant pressure isobaric
processes
n 1 constant temperature isothermal
processes
n k constant entropy isentropic
processes
n 8 constant volume isometric
processes
10
Isentropic Processes of Ideal Gases
2. Variable Specific Heats
Relative Pressure pr expsº(T)/R ? is not
truly a pressure ? is a function of
temperature
11
Isentropic Processes of Ideal Gases
2. Variable Specific Heats
Relative Volume vr RT/pr(T) ? is not truly
a volume ? is a function of temperature
12
Example 1
An insulated tank is filled initially with 5 kg
of air at p1 500 kPa and T1 500 K. A leak
develops and air slowly escapes until p2 100
kPa. Determine the final mass of air in the
tank and its temperature.
no heat transfer, Q 0
Insulated
Slow leakage
can be approximated as a reversible process
The process can be assumed to be an
isentropic process.
13
Example 1 (continued)
Table A-17 T1 500 K pr1 8.411
(8.411)(100/500) 1.6822
Table A-17 pr2 1.6822 T2 317 K
14
Example 2
Air is compressed in an adiabatic piston-cylinder
device from p1 95 kPa and T1 22 ºC in a
reversible manner. If V1/V2 8, Find the
final temperature of air.
no heat transfer, Q 0
Adiabatic
Reversible
The process can be assumed to be an
isentropic process.
15
Example 2 (continued)
Table A-17 T1 295 K vr1 647.9
Table A-17 vr2 80.99 T2 662.7 K
(647.9)(1/8) 80.99
Alternative approach
Table A-2 Tave 450 K k 1.391
(295)(8)0.391 665.2 K
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