EGR 334 ThermodynamicsChapter 3: Section 9-10

Lecture 08: Specific Heat Capacity

Quiz Today?

Today’s main concepts:• Introduce Specific Heats, cv and cP.• Understand when it is acceptable to apply specific heat.• Calculate changes of energy using specific heats • For liquids and solids, the saturated properties values may be

used to approximate properties values for supercooled property values, if the model can be treated as incompressible.

• For incompressible liquids and solids, cv ≈ cp • the specific heat ratio is defined as k = cp/cv

Reading Assignment:

Homework Assignment: • Read Chap 3: Sections 11

From Chap 3: 49, 55,68, 78

3

Think back to those simple, happy, carefree days of your youth, back when you were in Physics lab….

Q cm T

In one experiment you dropped hot aluminum pellets into a container of water and measured the change of temperature the water underwent.

You performed experiments to try and figure out how the world around you worked.

Using that information you found a property of aluminum called

specific heat capacity.

… it requires 1 calorie of heat to raise the temperature of 1 gram of water by 1 deg C.

4

In Thermodynamics, two common forms of specific heat capacity are used.

Sec 3.9 : Specific Heat

-- Used to calculate changes of energy between states-- Defined for pure, simple compressible substances-- May be used only under certain “special conditions”-- Usually applied to “ideal gas” model

When it can be applied: (if cv and cp are often treated as constants)

vv

ucT

p

p

hcT

2 1 2 1( )vu u c T T

and

Constant Volume process

Constant Pressure process

2 1 2 1( )ph h c T T and

5

Specific Heatsthe internal energy change or enthalpy change when heat is added at constant pressure or constant volume.

Sec 3.9 : Specific Heat

VTfU ,At constant volume

This is an exact differential

dVVUdTcdV

VUdT

TUdU

TV

TV

whereV

V

cTU

At constant volume, dV=0; thus, 0

dVVU

T

Finally: 2

1

T

T VV dTcUdTcdU

T

v

isobars

T1

T2

6Sec 3.9 : Specific Heat

PTfH ,At Constant Pressure

This is an exact differential

dPPHdTc

dPPHdT

THdH

TP

TP

where PP

cTH

At constant pressure, dP=0; thus, 0

dPPH

T

Finally, 2

1

T

T PP dTcHdTcdH

isobars

T

vT1

T2

7Sec 3.9 : Specific Heat

Specific Heat Ratio:

For air (at 68oF (20oC) and 14.7 psia (1 atm)):

V

P

cck

0.24 /1.01 / 1.40.718 / 0.17 /

om

air om

Btu lb FkJ kg KkkJ kg K Btu lb F

Fig03_09Fig. 3.9 Shows that cv and cp for water/steam vary with temperature and pressure.

Fig03_09In practice, specific heats will be used as constants which are looked up on tables based on standard values of temperature and pressure.

Gases - Specific Heats and Individual Gas Constants Example of Table:

10

Evaluating Properties of Liquids and Solids:

For liquids and solids, it is acceptable practice to approximate

( , ) ( )fv T p v T ( , ) ( )fu T p u T

( , ) ( )fh T p h Tand

While Appendix A does have a table for super-cooled water, for many other liquids, a super-cooled table is not available.

What to do?

if a super cooled table is not available.

11Sec 3.10.1 : Approximations for Liquids using Saturated Data

For a liquid, there is little change in v, u, h, s at different pressure and fixed T. Therefore,

ThPTh

TuVTu

TvPTv

ff

ff

ff

,

,

,Evaluate liquids at the saturated state

Since these properties are essentially only a function of T and not P, we call them Incompressible.

12

Example:

What is the enthalpy for Refrigerant 22 atT = 10 deg F. and p = 40 psi.(Refer to Table A-7E)

p = 40 psiTem

pera

ture

spec. vol.

*

pres

sure

spec. vol.

*

T=10 deg C

( , ) ( )fh T p h T

o

at T=10 deg….hf=13.33 Btu/lbm

13Sec 3.10.2 : Incompressible Substance Model

,

,

,

constantf s

f s

f s

v

u u T

h h T

Thus

Incompressible Substance: Includes any substance whose properties do not change with pressure.

For liquids and solids:

dTdUc

TUc V

VV

But, PVTuPTh ,

So now what? Take partial with respect to T

0

PPPP TU

TPV

TU

TH

Thus,ccc

dTdU

TH

vPP

14Sec 3.10.2 : Incompressible Substance Model

If c = constant then that means,

Use the heat capacity to calculate the change in internal energy.

12121212

12

2

1

2

1

PPvdTTcPPvuuhhh

dTTcuuu

T

T

T

T

12 PPv is small and can usually be dropped

Therefore, 1212

2

1

TTcdTTchhhT

T

15

Table of specific heat for incompressible materials.

See the course website for the complete tables of specific heats for both compressible and incompressible materials.

16

End of Slides for Lecture 08