The energy transfer between different systems can be expressed as:
E1 = E2 (1)The internal energy encompasses:
where
E1 = initial energy
E2 = final energy
- The kinetic energy associated with the motions of the atoms
- The potential energy stored in the chemical bonds of the molecules
- The gravitational energy of the system
The first law is the starting point for the science of thermodynamics and for engineering analysis.
Based on the types of exchange that can take place we will define three types of systems:
- isolated systems: no exchange of matter or energy
- closed systems: no exchange of matter but some exchange of energy
- open systems: exchange of both matter and energy
Internal Energy - Internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale. For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic . But on the microscopic scale it is a seething mass of high speed molecules. If the water were tossed across the room, this microscopic energy would not necessarily be changed when we superimpose an ordered large scale motion on the water as a whole. |
Heat - Heat may be defined as energy in transit from a high temperature object to a lower temperature object. An object does not possess "heat"; the appropriate term for the microscopic energy in an object is internal energy. The internal energy may be increased by transferring energy to the object from a higher temperature (hotter) object - this is called heating. |
Work - When work is done by a thermodynamic system, it is usually a gas that is doing the work. The work done by a gas at constant pressure is W = p dV, where W id work, p is pressure and dV is change in volume. For non-constant pressure, the work can be visualized as the area under the pressure-volume curve which represents the process taking place. |
Heat Engines -Refrigerators, Heat pumps, Carnot cycle, Otto cycle |
dE = Q - W (2)1st law does not provide the information of direction of processes and does not determine the final equilibrium state. Intuitively, we know that energy flows from high temperature to low temperature. Thus, the 2nd law is needed to determine the direction of processes.
where
dE = change in internal energy
Q = heat added to the system
W = work done by the system
Enthalpy is the "thermodynamic potential" useful in the chemical thermodynamics of reactions and non-cyclic processes. Enthalpy is defined by
H = U + PV (3)Enthalpy is then a precisely measurable state variable, since it is defined in terms of three other precisely definable state variables.
where
H = enthalpy
U = internal energy
P = pressure
V = volume
There are two classical statements of the second law of thermodynamics:
Kelvin & Planc "No (heat) engine whose working fluid undergoes a cycle can absorb heat from a single reservoir, deliver an equivalent amount of work, and deliver no other effect" |
Clausius "No machine whose working fluid undergoes a cycle can absorb heat from one system, reject heat to another system and produce no other effect" |
The second law is concerned with entropy (S). Entropy is produced by all processes and associated with the entropy production is the loss of ability to do work. The second law says that the entropy of the universe increases. An increase in overall disorder is therefore spontaneous. If the volume and energy of a system are constant, then every change to the system increases the entropy. If volume or energy change, then the entropy of the system actually decrease. However, the entropy of the universe does not decrease.
For energy to be available there must be a region with high energy level and a region with low energy level. Useful work must be derived from the energy that would flows from the high level to the low level.
- 100% of the energy can not be transformed to work
- Entropy can be produced but never destroyed
Efficiency of a heat machine
The efficiency of a heat machine working between two energy levels is defined in terms of absolute temperature:η = ( Th - Tc ) / Th = 1 - Tc / Th(1)As a consequence, to attain maximum efficiency the Tc would have to be as cold as possible. For 100% efficiency the Tc would have to equal 0 K. This is practically impossible, so the efficiency is always less than 1 (less than 100%).
where
η = efficiency
Th = temperature high level (K)
Tc = temperature low level (K)
Change in entropy > 0 irreversible process | Change in entropy = 0 reversible process | Change in entropy < 0 impossible process |
- For the universe as a whole the entropy is increasing!
Entropy definition
Entropy is defined as :S = H / T (2)A change in the entropy of a system is caused by a change in its heat content, where the change of entropy is equal to the heat change divided by the average absolute temperature (Ta):
where
S = entrophy (kJ/kg K)
H = enthalpy (kJ/kg)
T = absolute temperature (K)
dS = dH / Ta (3)The sum of (H / T) values for each step in the Carnot cycle equals 0. This only happens because for every positive H there is a countering negative H, overall.
Carnot Heat Cycle In a heat engine, a gas is reversibly heated and then cooled. A model of the cycle is as follows: State 1 --(isothermal expansion) --> State 2 --(adiabatic expansion) --> State 3 --(isothermal compression) --> State 4 --(adiabatic compression) --> State 1 State 1 to State 2: Isothermal Expansion Isothermal expansion occurs at a high temperature Th, dT = 0 and dE1 = 0. Since dE = H + w, w1 = - H1. For ideal gases, dE is dependent on temperature only. State 2 to State 3: Adiabatic Expansion The gas is cooled from the high temperature, Th, to the low temperature, Tc. dE2 = w2 and H2 = 0 (adiabatic). State 3 to State 4: Isothermal Compression This is the reverse of the process between states 1 and 2. The gas is compressed at Tc. dT = 0 and dE3 = 0. w3 = - H3 State 4 to State 1: Adiabatic Compression This is the reverse of the process between states 2 and 3. dE4 = w4 and H4 = 0 (adiabatic). The processes in the Carnot cycle can be graphed as the pressure vs. the volume. The area enclosed in the curve is then the work for the Carnot cycle because w = - integral (P dV). Since this is a cycle, dE overall equals 0. Therefore, -w = H = H1 + H2 + H3 + H4 If you decrease Tc, then the quantity -w gets larger in magnitude. if -w > 0 then H > 0 and the system, the heat engine, does work on the surroundings. |
- Entropy is not conserved like energy!
Example - Entropy Heating Water
A process raises 1 kg of water from 0 to 100oC (273 to 373 K) under atmospheric conditions.Specific enthalpy at 0oC (hf) = 0 kJ/kg (from steam tables) (Specific - per unit mass)
Specific enthalpy of water at 100oC (hf) = 419 kJ/kg (from steam tables)
Change in specific entropy:
dS = dH / Ta
= ((419 kJ/kg) - (0 kJ/kg)) / ((273 K + 373 K)/2)
= 1.297 kJ/kgK
Example - Entropy Evaporation Water to Steam
A process changes 1 kg of water at 100oC (373 K) to saturated steam at 100oC (373 K) under atmospheric conditions.Specific enthalpy of steam at 100oC (373 K) before evaporating = 0 kJ/kg (from steam tables)
Specific enthalpy of steam at 100oC (373 K) after evaporating = 2 258 kJ/kg (from steam tables)
Change in specific entropy:
dS = dH / TaThe total change in specific entropy from water at 0oC to saturated steam at 100oC is the sum of the change in specific entropy for the water, plus the change of specific entropy for the steam.
= (2 258 - 0) / ((373 + 373)/2)
= 6.054 kJ/kgK
Example - Entropy Superheated Steam
A process superheats 1 kg of saturated steam at atmospheric pressure to 150oC (423 K).Specific total enthalpy of steam at 100oC (373 K) = 2 675 kJ/kg (from steam tables)
Specific total enthalpy of superheated steam at 150oC (373 K) = 2 777 kJ/kg (from steam tables)
Change in specific entropy:
dS = dH / Ta
= ((2777 kJ/kg) - (2675 kJ/kg)) / ((423 K + 373 K)/2)
= 0.256 kJ/kgK
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