Lecture 14: Ideal Gas Properties Polytropic processescc.sjtu.edu.cn/Upload/20160313223422438.pdf1.1...

Preview:

Citation preview

1.1

ME 200 –Thermodynamics I Spring 2016

Lecture 14: Ideal Gas PropertiesPolytropic processes

Yong LiShanghai Jiao Tong University

Institute of Refrigeration and Cryogenics800 Dong Chuan Road Shanghai, 200240, P. R. China

Email : liyo@sjtu.edu.cnPhone: 86-21-34206056; Fax: 86-21-34206056

1.2

u, h, c of Ideal Gases specific internal energy depends only on T

specific enthalpy depends only on T

Important relation

1.3

Notes on Ideal Gas

Using statistical thermodynamic viewpoints regarding the translational, vibrational, and rotational modes:» For monoatomic gases: k = 1.67» For diatomic gases: k = 1.4» For polyatomic gases: k ~ 1.33

u and h depends only on T

cv and cp can be expressed in terms of k and R

Incompressible Substances

v =const

cp cv

1.4

increases with T for all gases,except for the monatonic gases Ar, Ne,and He.

Ideal Gas Specific Heats as Functions of T, cv(T), cp(T)

A polynomial form equation:

of monatonic gases is closelyconstant at the value predicted bykinetic theory:

1.5

Ideal Gas Properties (contd.)

How to evaluate changes in internal energy and enthalpy for IG?» Use tabulated data (most accurate; remember to make sure that reference

states are consistent)» Use known dependence of cv and cp on temperature and integrate

(reasonably accurate; very tedious)

» Use constant cv and cp neglecting temperature dependence (works well for small temperature ranges; otherwise cv and cp at average temperature)

1.6

Evaluating u and h Using Ideal Gas Tables

Table A-17 ideal gas properties of air

Table A-2, ideal gas properties of selected gases

h =0 at Tref = 0 K

1.7

Assuming Constant cv , cp

T is slight, » use arithmetic average of the cv , cp

» use the cv , cp at the average T over the interval

cv=constant cp=constant

1.8

Example 1: Evaluation of the u of an Ideal Gas

Air at 300 K and 200 kPa is heated at constant pressure totemperature of 600 K. Determine the change in internal energy ofair per unit mass, using (a) data from the air table (Table A–17), (b)the functional form of the specific heat (Table A–2c), and (c) theaverage specific heat value (Table A–2b).

(a) From Table A–17

1.9

Example 1: (contd.)

(b) functional form of the specific heat Table A–2c

data from the air table

1.10

Example 1: (contd.)

(c) the average specific heat value Table A–2b

data from the air table

functional form of the specific heat

1.11

Polytropic Process polytropic process of a closed system

.» n=0, the process is an isobaric (constant-pressure) process, » = , the process is an isometric (constant-volume) process.

For a polytropic process

For an ideal gasExample 2.1

1.12

Example

Find: W & Q in kJ/kg.

287.211 (1.401 1)

0.716 /

vRc

kkJ kg K

2 1( ) 127.02 0.716(428 295) 32.02 /vQ W c T T kJ kgm m

2 1( ) Q W u um m

Recommended