Water vapor can be treated as ideal gas at low

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Water vapor can be treated as ideal gas at low pressures (P < 10 kPa) Error of the assumption is high at high pressures close to saturated vapor line and critical point Air conditioning systems operate at low pressures, thus ideal gas assumption is valid Steam power plants operate at high pressures, thus ideal gas assumption cannot be used 46
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Compressibility Factor (Z) To obtain correct relation, the compressibility effect is considered The compressibility factor is used to correct the ideal gas relation so that it can accurately represent real gas behavior Compressibility factor (Z) is given as: 𝑍 = 𝑃? 𝑅𝑇 or 𝑃? = 𝑍𝑅? ( Z = 1 for ideal gas) Z can also be related to specific volume by: 𝑍 = ? 𝑎???𝑎𝑙 ? 𝑖??𝑎𝑙 = ? ? 𝑖??𝑎𝑙 where ? 𝑖??𝑎? = 𝑅𝑇 𝑃 47
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Specific Heats (C p & C v ) Different substances have different capabilities to transfer or store heat (energy), i.e. they heat up or cool down at different rates For the same mass, the heat needed to change the temperature by 1°C vary for different substances/gas Example: Heat required to raise temperature of air from 20 - 30 °C = 4.5 kJ Heat required to raise temperature of water from 20 - 30 °C = 41.8 kJ Therefore, the specific heats of water is different from that of air Specific heat the energy required to raise the temperature of a unit mass of a substance by one degree The specific heat can be measured at constant pressure (C p ) , or at constant volume (C v ) 48
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C p energy required to raise the temperature of the unit mass of a substance by one degree at constant pressure Equal to the change in enthalpy with temperature at constant pressure ? ? = 𝜕ℎ 𝜕𝑇 ? ?? ??.? or ?? ??.°𝐶 C v energy required to raise the temperature of the unit mass of a substance by one degree at constant volume Equal to the change in internal energy with temperature at constant volume ? ? = 𝜕? 𝜕𝑇 ? ?? ??.? or ?? ??.°𝐶 For incompressible substances such as liquids and solids, the specific volume of solids remain constant, thus: ? ? = ? ? = ? 49 Energy in Energy out = change in energy ? 𝑖? − ? ??? = ∆? ?𝑦???? From, ?𝑝?? = ?ℎ ? ? = 𝜕ℎ 𝜕𝑇 From, ???? = ?? ? ? = 𝜕? 𝜕𝑇
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Ideal Gases The internal energy of an ideal gas is a function of temperature only ? = ? ? From definition of enthalpy: ℎ = ? + 𝑃? From equation of state: 𝑃? = 𝑅? So, enthalpy: ℎ = ? + 𝑅? Since R is constant and u is a function of temperature only, therefore h is also a function of temperature only ℎ = ℎ ? Since u and h only depend on temperature, then C v and C p also depend only on temperature. i.e. at a given temperature, u, h, C v and C p are constants ? ? = ? ? ? and ? ? = ? ? ? 50
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So, from ? ? = 𝜕ℎ 𝜕𝑇 ? and ? ? = 𝜕? 𝜕𝑇 ? ? ? = ?ℎ ?𝑇 ? or ?ℎ = ? ? ?? ? ? = ?? ?𝑇 ? or ?? = ? ? ?? We can integrate the equation to give change in enthalpy and internal energy over temperature ∆ℎ = ℎ 2 − ℎ 1 = 1 2 ?
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