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

2 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

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

4 Besides compressibility factor, we can also use more complex equations of state Van der Waals

5 Beattie-Bridgeman: where

6 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:

7 Specific Heats c v is called the “constant volume” specific heat c p 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.

8 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

9 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 c v is also only a function of temperature for an ideal gas:

10 We can start with du and integrate to get the change in u: Note that c v does change with temperature and cannot be automatically pulled from the integral.

11 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)

12 Similarly, for a change in enthalpy for ideal gases:

13 For an ideal gas, h = u + RT

14 Ratio of specific heats is given the symbol, k

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

16 For monatomic gases,

17 For all other gases, c p 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 c v may be calculated from c p =c v +R. Next figure shows the temperature behavior….many specific heats go up with temperature.

18 Variation of Specific Heats with Temperature

19 Tabular specific heat data for c v, c p, and k are found in Tables A-2 and A-2E

20 Assumption of constant specific heats: when can you use it? where Either formulation for c p will be adequate because c p is fairly linear with T over a narrow temperature range. Take your choice.

21 Rule of thumb Specific heats for ideal gases may be considered to be constant when T 2 -T 1  200 K or 400 °R. (Note in many cases the temperature range can be significantly larger.)

22 Changes in enthalpy and internal energy can be calculated from tabular data: Frequently, we wish to know h 2 -h 1 or u 2 -u 1 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.

23 The integration is done for us in the ideal gas tables: Reference temperature is = 0 K and h = T ref = 0 K for ideal gas tables. Tables A-17 and A-17 are for air. Units are mass-based for both h and u.

24 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

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

26 Solution - Page 2 For constant specific heats:

27 Solution - Page 3 For variable specific heats, we’ll use the ideal gas air tables

28 Solution - Page 4 So for variable specific heats: Recall for constant specific heats, h = 520 kJ/kg, which is less than 0.5% difference.

29 Consider incompressible substances What’s an incompressible substance? –Liquid –Solid For incompressible substances v = constant dv = 0

30 Incompressible substances Express u = u(T,v) But dv = 0 for an incompressible substance, so We can take the derivative of u: 0

31 Incompressible substances The right hand side is only dependent on temperature. Thus, u = u(T) only for an incompressible substance. Recall that Thus:

32 Enthalpy of incompressible substances h = u + pv For an incompressible substance, v=const as before. If we hold P constant, then we can take (h/T) p and show:

33 Specific heats of incompressible substances: cpcp cvcv Bottom line: c p = c v = c for an incompressible substance.

34 Relationships for incompressible substances. du = c(T) dT

35 Relationships for incompressible substances. We can also show that:

36 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 u 2 - u 1 =c(T 2 - T 1 ) and h 2 - h 1 =c(T 2 - T 1 )+v(p 2 - p 1 )