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Engg Thermodynamics - Lecture - 14 - Unit - 5 - Gas Mixtures & Psychrometry

 PSYCHROMETRY Psychrometric properties, Psychrometric charts. Property calculations of air vapour mixtures by using chart and expressions. Psychrometric process – ├ Adiabatic saturation, ├ Sensible heating and cooling, ├ Humidification, ├ Dehumidification, ├ Evaporative cooling and ├ Adiabatic mixing. Simple Applications Psychrometry Psychrometry is the science dealing with the physical laws of air – water vapour mixtures. When designing an air conditioning system, the temperature and moisture content of the air to be conditioned, and the same properties of the air needed to produce the desired air conditioning effect. In other words, we can say that Psychrometry is the study of MOIST AIR or mixture of dry air and water vapour. PROPERTIES OF PSYCHROMETRY Dry-bulb temperature Wet bulb temperature Wet bulb depression Dew point temperature Dew point depression Relative humidity (Ñ„) Humidity ratio (or) Specific Humidity (or) Moisture content (ῳ) Degree of saturation (or) Percentage of ...
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Engg Thermodynamics - Lecture - 13 - Unit - 4 - Ideal Gas and Real Gas & TD Relations

 Ideal Gas Imaginary Substance Obeys the law of PV=RT At low Pressure and High Temperature – density of gas Decreases Real Gas At High Pressure – Gas start to Deviate from Ideal Gas Measuring of Deviation – Compressibility Factor PV=ZRT Z=PV / RT Z = Vactual / Videal For Ideal Gas Z = 1 For Real Gas Z > 1 Important Laws (Ideal Gas) Boyle's Law (constant temperature) P = constant / V Charles Law (constant pressure) V = constant x T Gay-Lussac’s Law (constant volume) P = constant x T THE ENTHALPY OF ANY SUBSTANCE h=u+pv for an ideal gas h=u+RT h=f(T) dh=du+RdT since R is constant ∆h=∆u+R∆T =Cv∆T+R∆T = (Cv+R)∆T Since h is a function of T only, Cp=(∂h/∂T)p Entropy change of an ideal gas: (Eqn – 1) From the general property relations Q = W+ U Tds=du+pdv And for an ideal gas, du=CvdT, dh=CpdT, and pv=RT, the entropy change between any two states 1 and 2 may be computed as given below ds=du/T+p/Tdv =CvdT/T+Rdv/v S2-s1=Cv ln T2/T1+R ln v2/v1 . Entropy change of an ideal gas: (Eqn -2) Fr...
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Engg Thermodynamics - Lecture - 12 - Unit - 3 - Steam Power Cycle

 Rankine Large electric power plants typically utilize a vapor power cycle. Regardless of the heat source, be it nuclear or combustion of coal, oil, natural gas, wood chips, etc., the remaining details of these plants are similar. Typically a pure working fluid, usually water, is circulated through a cycle, and that fluid trades heat and work with its surroundings. We sketch a typical power plant cycle for electricity generation in Fig.... The ideal Rankine cycle was first described in 1859 by William John Macquorn Rankine, long after the steam engine wasin wide usage.  The cycle has the following steps: • 1 → 2: isentropic compression in a pump, • 2 → 3: isobaric heating in a boiler, • 3 → 4: isentropic expansion in a turbine, and • 4 → 1: isobaric cooling in a condenser. Two variants of the T − s diagram are given in Fig.. The first is more efficient as it has the appearance of a Carnot cycle. However, it is impractical, as it induces liquid water in the turbine, which can d...
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Thermodynamics - Lecture - 11 - Unit - 3 - Steam Properties and Problems

Problem: 1   A vessel of volume 0.04 m 3 contains a mixture of saturated water and steam at a temperature of 250°C The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume, enthalpy, entropy and internal energy. Given data: V=0.04m 3 T= 250°C m 1 = 9 kg To find: p, m, v, h, S, and U   Solution: From Steam Tables corresponding to 250°C, v f =v 1 = 0.001251 m/kg V g = V s = 0.050037 m/kg p= 39.776 bar   Total volume occupied by the liquid, V 1 = m 1 v 1 = 9 x 0.001251 = 0.0113 m 3 Total volume of the vessel, V = Volume of liquid + Volume of steam V 1 +V s V s =0.0287 m 3   Mass of steam, m s = V s / V s   =0.0287 / 0.050037 =0.574 kg Mass of mixture of liquid and steam, m=m 1 +m s = 9+ 0.574 = 9.574 kg   Total specific volume of the mixture, v = V / m   = 0.04 /9.574 = 0.00418 m 3 /kg   We know that, v=v f +x V fg V fg = v g - v f 0.00418 = 0.001251 + x (0.05003...
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Thermodynamics - Lecture - 10 - Unit - 3 - Steam Properties and Steam Power

  Properties of Pure Substances If any two independent intensive properties of a simple compressible system are defined, other properties automatically assume definite values.  These properties can be expressed in terms of charts, tables or equations. This chapter covers the charts and tables of properties of steam. Pure Substances Substances   of fixed chemical composition are known as pure substances    Example : Water, Helium, Nitrogen, Oxygen etc. Substances exist in any one of the three phases namely solid, liquid and gas. For example, H 2 O may exist in the form ice (solid), Water (Liquid) or Steam (Gaseous). In all these phases it will have the same chemical composition. A   Mixture of two or more phases of a pure substance should also be regarded as pure substance. If Water and Steam Co-exits in a container, the chemical composition of both the Vapour and liquid phases will be identical. Hence this heterogeneous system is also a pure...
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