1 Lec 8: Real gases, specific heats, internal energy, enthalpy.

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Lec 8: Real gases, specific heats, internal energy, enthalpy

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• For next time:– Read: § 4-1 to 4-4

• Outline:– Real gases (Compressibility factor)– Specific heats– Special relationships for ideal gases

• Important points:– How to manipulate the ideal gas law– Energy relationships to specific heats– How to evaluate properties of ideal

gases

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TEAMPLAYTEAMPLAY

Under what conditions is it appropriate to apply the ideal gas equation of state?

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Besides compressibility factor, we can also use more complex

equations of state

• Van der Waals

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•Beattie-Bridgeman:

where

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Specific Heats

Another set of properties that is a common combination of properties are the specific heats. They show up often.

For simple compressible systems, these are:

Rlb

Btuor

Kkg

kJ are units theand

T

uc

mvv

vp

p c as units same thehas and T

hc

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Specific Heats

• cv is called the “constant volume” specific heat

• cp is called the “constant pressure” specific heat

• These names tell you how they are determined or measured.

• These names do not limit the applicability of them to either constant volume or constant pressure processes.

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Specific Heats

• In general, the specific heats are functions of two variables for simple, compressible systems.

• However, we will show that for ideal gases, solids and liquids, they are functions of temperature alone

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Specific Heats and Ideal Gases:

• Joule conducted some experiments where he found that the internal energy, u, was only a function of temperature, u = u(T).

• It was independent of P or v.

• This implies that cv is also only a function of temperature for an ideal gas:

dTcdu v

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We can start with du and integrate to get the change in u:

and dT,c du v

(T)dTcuu2

1

T

T

v12

Note that cv does change with temperature and cannot be automatically pulled from the integral.

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Let’s look at enthalpy for an ideal gas:

• h = u + pv where pv can be replaced by RT because pv = RT.

• Therefore, h = u + RT => since u is only a function of T, R is a constant, then h is also only a function of T

• so h = h(T)

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Similarly, for a change in enthalpy for ideal gases:

and dT,c dh p

(T)dTchh2

1

T

T

p12

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For an ideal gas,

• h = u + RT

RdT

du

dT

dh

Rcc vp

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Ratio of specific heats is given the symbol, k

k(T)

(T)c

(T)c

c

ck

v

p

v

p

kc

R1

c

c

vv

p

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Other relations with the ratio of specific heats which can be easily

developed:

1-k

Rcv

1-k

kRc and p

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For monatomic gases,

constant. isit and R,2

5cp

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For all other gases,

• cp is a function of temperature and it may be calculated from equations such as those in Table A-2 and A-2E in the appendices

• cv may be calculated from cp=cv+R.

• Next figure shows the temperature behavior….many specific heats go up with temperature.

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Variation of Specific Heats with Temperature

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Tabular specific heat data for cv, cp, and k are found in Tables A-2 and A-2E

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Assumption of constant specific heats: when can you use it?

?TTcdTcThTh 12p

T

T

p1122

2

1

where

2

TTccor

2

TcTcc 21

pp2p1p

p

Either formulation for cp will be adequate because cp is fairly linear with T over a narrow temperature range. Take your choice.

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Rule of thumb

• Specific heats for ideal gases may be considered to be constant when T2-T1 200 K or 400 °R.

• (Note in many cases the temperature range can be significantly larger.)

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Changes in enthalpy and internal energy can be calculated from

tabular data:

• Frequently, we wish to know h2-h1 or u2-u1 and we do not want to go to the trouble to integrate

• where cp or cv is a third-degree polynomial in T, as shown in Tables A-2 and A-2E.

dT).c(or dTc vp

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The integration is done for us in the ideal gas tables:

Reference temperature is = 0 K and h = 0 @ Tref = 0 K for ideal gas tables.

kg

kJ

Tables A-17 and A-17 are for air.

Units are mass-based for both h and u.

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Example Problem

Calculate the change in enthalpy of air for a temperature rise from 300 to 800 K.

a) assuming constant specific heats

b) using the ideal gas tables

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Solution

For part a), we calculate the enthalpy difference using:

Tch p

Where,

kgK

kJ040.1)K550(c)(Tcc pavgpp

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Solution - Page 2

K300800kgK

kJ040.1h

For constant specific heats:

kg

kJ520h

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Solution - Page 3

For variable specific heats, we’ll use the ideal gas air tables

kg

kJ95.821h800

kg

kJ19.300h300

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Solution - Page 4

So for variable specific heats:

kg

kJ76.521h

Recall for constant specific heats, h = 520 kJ/kg, which is less than 0.5% difference.

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Consider incompressible substances

• What’s an incompressible substance?– Liquid– Solid

• For incompressible substances

v = constant dv = 0

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Incompressible substances

du

dTT

u du

v

Express u = u(T,v)

But dv = 0 for an incompressible substance, so

We can take the derivative of u:

dTT

u

v

dvv

u

T

0

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Incompressible substances

dTc du v

The right hand side is only dependent on temperature. Thus, u = u(T) only for an incompressible substance.

Recall that vv

cT

u

Thus:

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Enthalpy of incompressible substances

h = u + pv

vpp

cT

u

T

h

•For an incompressible substance, v=const as before.

•If we hold P constant, then we can take (h/T)p and show:

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Specific heats of incompressible substances:

dT

du

T

u

T

h

pp

cp cv

Bottom line: cp = cv = c for an incompressible substance.

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Relationships for incompressible substances.

du = c(T) dT

2

1

2

1

T

T

12

u

u

c(T)dTuudu

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Relationships for incompressible substances.

We can also show that:

)pv(pc(T)dThh 12

T

T

12

2

1

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Relationships for incompressible substances

• Now, if the temperature range is small enough, say up to about 200 K (400 °F), then c may be regarded as a constant, and

• u2 - u1=c(T2 - T1)

• and h2 - h1=c(T2 - T1)+v(p2 - p1)