2: Thermodynamics Review

Fundamental Concepts

1. Objects are at equilibrium with each other only when they have the same "degree of hotness" : concept of Temperature:

2. The impossibility of perpetual motion machines : concept of Internal Energy.

3. The impossibility of reversing any natural process in its entirety : concept of Entropy:

Closed System: A collection of matter of fixed identity. The boundaries may change, but there can be no mass transfer across them. Example: a nuclear submarine moving under the sea.

Control Volume: A region of fixed shape, location, and volume. Matter can enter and leave the boundaries. Example: the inside of a classroom, or a jet engine.

Note: The laws of thermodynamics are usually found written for closed systems. However, jet engines are more easily modeled as control volumes, since we don't usually care which particular molecules of air are inside the engine at any given time.

First Law of Thermodynamics

Q = DE + W for a closed system,

where Q is the heat transferred into the system, W is the work done by, or taken out, of the system, and DE is the increase in the internal energy E of the system.

Note that this precludes the possibility of "perpetual motion" machines, whose work output is always greater than the work obtained by converting the heat put into them. At some point, there will be no more internal energy available to convert to work.

Derivation of the Energy Equation for a Control Volume

We can also relate the rates of heat transfer, energy change, and work as:

Consider each of these rates for a control volume in a fluid:

^{where is the mass flow rate out of the control surface through its surface, and e is the energy per unit mass of the fluid.}

where is specific volume (volume per unit mass), and is the power taken out through shear forces and through rotating shafts.

^{The energy per unit mass of the fluid can be written as}

^{where e}_int ^{is internal energy, u is speed, and P.E. is potential energy per unit mass. Thus,}

where the integral is taken over the surface of the control volume. Now

, the enthalpy per unit mass.

can be written as , and can be written as

where is the component of the velocity directed outward from the surface of the control volume.

Thus,

Special Cases

Adiabatic: q = 0

Steady:

No body forces: P.E. = 0

No work done except that of expansion: = 0

If all of the above are true,

^so

^{is a constant.}

Stagnation enthalpy is constant along a streamline in steady adiabatic flow, where body forces are negligible and the only work is that of expansion.

Energy Equation for Steady One-Dimensional Flow

Consider an engine represented by a control volume as shown below. Fluid enters from the left at the rate of per unit time, and leaves from the right boundary. Work is extracted from the fluid per unit mass flow rate, and heat is added to the fluid per unit mass flow rate.

Neglect changes in potential energy of the fluid. Assume that steady-state conditions exist, so that time derivatives are zero in the frame of reference moving with the control volume. Ignore variations across the streamlines, so that integrals such as can be replaced by . Thus, the energy equation reduces to

or

Note:

Stagnation enthalpy increases when heat is added, and decreases when work is taken out.

Second Law of Thermodynamics

For any process in a closed system,

where dQ is the heat transferred at temperature T to the system, and ds is the resulting increase in entropy of the system.

For a control volume, the entropy inequality may be written as

For a change in an isolated system (no energy or mass transfer),

Reversible Process

Rearranging terms,

from the 1^st Law of Thermodynamics.

If the only work being that is that due to volume change against pressure,

Applicable to a closed system in internal equilibrium with the only work being that due to volume change against a pressure.

Equations of State

For a system composed of a pure substance at equilibrium, only two independent static properties need be specified to describe the thermodynamic state of the system. The relation between any two properties of such a system is called an equation of state.

Thermal Equation of State

For a thermally perfect gas,

where p is absolute pressure, r is density, T is absolute temperature, and R is the gas constant for the particular gas.

where is the "universal gas constant", a constant for all gases, and is the molecular weight of the gas. has the value of 8314 Joules per Kelvin per kilogram-mole.

Note: At extremely high temperatures, such as those encountered in flames, and in hypersonic flight (> 2500K) nitrogen and oxygen start dissociating into single atoms, so that the effective molecular weight of air decreases.

Example using the perfect gas relation:

Calculate the density of air at a pressure of 1 atmosphere and a temperature of 25 deg. C.

Solution:

Pressure: 1 atmosphere = 101300 Newtons per square meter

Temperature: 25 deg. Celsius = 273.15 + 25

= 298.15 deg. Kelvin

The thermal equation of state relates the pressure P, absolute temperature T, and density r of a gas:

P = r RT, where

R = R_u / MW,

R_u being the Universal Gas Constant (8314.3 in SI units ), and MW the molecular weight of the gas. Air is composed of 79% Nitrogen, and 21% Oxygen. The molecular weight of Nitrogen (N₂) is 28, and that of Oxygen (O₂) is 32. Thus the mean molecular weight is

MW = 0.79 * 28 + 0.21*32 = 28.84

Thus, the gas constant for air, R = 8314.3 / 28.84 = 288.29 mK^-1s^-2

r = 101300 / (298.15 * 288.29) = 1.1785 kg/m³.

( Note: A more exact representation of air at sea level is: 79% nitrogen, 20% oxygen, and 1% argon (MW=44), giving a molecular weight of 28.96. This makes R = 287.04 mK^-1s^-2)

It is useful to remember that in SI units, atmospheric pressure at sea-level is approximately 100,000, temperature is 300, and density is 1.2.

Caloric Equations of State

For a calorically perfect gas, internal energy per unit mass depends only on temperature.

e = e(T)

Enthalpy per unit mass

h = e + p/r,

so that h = h(T) as well.

Specific Heats

Specific heat at constant volume

or

Specific heat at constant pressure

Now,

, and h = e + p/r,

so that

At constant pressure, dp = 0, so that dq = dh

or

Ratio of Specific Heats

, so that

Define

Thus,

Note: At low to moderate temperatures, (200 to 700K), the ratio of specific heats of diatomic gases such as nitrogen and oxygen is equal to 1.4. In the high temperature regions of engines, this value changes because the specific heats change and the gas composition is different from that of cold air. For most calculations of flows in jet engines, it is usual to assume g = 1.4 in the cooler regions, and g = 1.33 in the hot gases in the turbine and hot nozzle.

Variation of specific heats over a large range of temperatures

Where large changes in temperature occur, such as those encountered across a shock wave in hypersonic flow, or through the combustion chamber and turbine of a jet engine, cp, cv and g cannot be assumed to be equal to their low-temperature values.