Fundamentals of mechanical engineering [email protected]
MOHAMMED ABASS ALI What is thermodynamics Thermodynamics is the branch of physics that studies the effects of temperature and heat on physical systems at the macroscopic scale. In addition, it also studies the relationship that exists between heat, work and energy. Thermal energy is found in many forms in today’s society including power generation of electricity using gas, coal or nuclear, heating water by gas or electric. THE BASICS OF THERMODYNAMICS
Basic concepts Properties are Features of a system which include mass, volume, energy, pressure and temperature. Thermodynamics also considers other quantities that are not physical properties, such as mass flow rates and energy transfers by work and heat. Energy forms Fluids and solids can possess several forms of energy. All fluids possess energy due to their temperature and this is referred to as ‘internal energy’. They will also possess ‘ potential energy’ (PE) due to distance (z) above a datum level and if the fluid is moving at a velocity (v), it will also possess ‘kinetic energy’. If the fluid is pressurised, it will possess ‘flow energy’ (FE). Pressure and temperature are the two governing factors and internal energy can be added to FE to produce a single property called ‘enthalpy’. Internal energy The molecules of a fluid possess both kinetic energy (KE) and PE relative to an internal datum. Generally, this is regarded simply as the energy due to its temperature and the change in internal energy in a fluid that undergoes a temperature change is given by ΔU = mcΔT The total internal energy is denoted by the symbol ‘U’, which has values of J, kJ or MJ; also the specific internal energy ‘u’ has the values of kJ/kg. Note: The change in temperature is in either degrees Celsius or Kelvin. potential energy When a mass ‘m’ kg is raised to a height of ‘z’ metres above a datum level, a lifting force is required. The work done in raising the mass is force × distance moved, so PE = mgz = Work where g is the gravity constant Kinetic energy Energy has been expended in doing this work and has therefore been stored in the mass and will be carried along with it. This energy is called ‘kinetic energy’. KE = m.v2/2 Enthalpy, Enthalpy (H) requires both pressure and temperature; it therefore must possess both flow (FE) and internal energy (U). These two energies are added together FE=PV H = FE + U The units are J for the total enthalpy or kJ/kg for the specific enthalpy. Boyle’s law Boyle’s law states that provided the temperature ‘T’ of a perfect gas remains constant, the volume ‘V’ of a given mass of gas is inversely proportional to its pressure ‘P’ of the gas, that is, P∝1/V or P × V = constant if the temperature remains constant. If the gas experiences a change in state during an isothermal process, then
P1V1 = P2V2 = constant EXAMPLE 1 A certain perfect gas is heated at a constant temperature from an initial state of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2. Determine the final volume of the gas. 2 solution State 1: P1 = 325 kN/m and V1 = 0.22 m3. 2 State 2: P2 = 170 kN/m and V2 = ? From the equation, P1 ⋅ V1 = P2 ⋅ V2 P1 ⋅ V1 = P2 ⋅ V2
P2= P1. V1/V2 V2 = 0.4206 m3 Charles’s law Charles’s law states that provided the pressure ‘P’ of a given mass of gas remains constant, the volume ‘V’ of the gas will be directly proportional to the absolute temperature ‘T’ of the gas, that is, V ∞ T, or V = constant × ‘T’. Therefore, V/T constant = for a constant pressure ‘P’. If the gas experiences a change in state during a constant pressure process, then V1/T1 = V2/T2 = constant EXAMPLE 2 A quantity of gas is subjected to a constant pressure process causing the volume of gas to reduce from 0.54 m3 at a temperature of 345°C to 0.32 m3. Calculate the final temperature of the gas at the end of this process. solution From the question V1 = 0.54 m3 T1 = 345 + 273 K V2 = 0.32 m3
V1/T1 = V2/T2 = constant T2= 366.22K Universal gas law The universal gas equation combines pressure, volume and temperature and the relation between Boyle’s and Charles’s laws is expressed in Equation P . V/ T = R (constant) where R is known as the universal gas constant. That is
P1V1/T1 = P2V2/T2 or for ‘m’ kg, occupying V m3 PV = mRT From the definition of the kilogramme-mole, for ‘m’ kg of the gas m = nM where n is the number of moles where M is the molecular weight of the gas. Oxygen has a molecular weight of 32 EXAMPLE A volume of gas at a pressure of 325 kN/m2 is 0.22 m3 and temperature of 618 K is compressed to a volume 0.16 m3 and a pressure of 380 kn/m2. Determine the final temperature of the gas.
Solution State 1:
P1 = 325 kN/m3, V1 = 0.22 m3 and T1 = 618 K. State 2: P2 = 380 kN/m3, V2 = 0.16 m3 and T2 = ? From Equation P1V1/T1 = P2V2/T2
T2 = 525.52 K. SPECIFIC HEAT CAPACITY
The specific heat capacity of any substance is defined as the amount of energy required to raise a unit mass through one degree temperature rise. In thermodynamics, there are two specified conditions used: 1. Constant volume (Cv) 2. Constant pressure (Cp) The two specific heat capacities do not have the same value, and it is very important to distinguish between them. specific heat capacity at constant volume (cv) Consider one kg of a gas supplied with an amount of heat energy sufficient to raise its temperature by 1 K while the volume of the gas constant, the amount of heat energy supplied, is known as the specific heat capacity at constant volume and is denoted by Cv. The basic unit of Cv is J/kg K. For a reversible non- flow process at constant volume: dQ = mCvdT For a perfect gas, the value of Cv will be constant for any one gas at all pressures and temperatures. From equation can be expanded as follows. Heat flow in a constant volume process between two states:
Q12 = mCv(T2 − T1) From the non-flow energy equation: Q − W = (U2 − U1) mCv(T2 − T1) − 0 = (U2 − U1) (U2 − U1) = mCv(T2 − T1) that is dU = Q Note: In a reversible constant volume process, there will be no work energy transfer as the piston will be unable to move; therefore, W = 0 EXAMPLE A quantity of 4.5 kg of gas is heated at a constant volume of 1.5 m3 and temperature 20°C until the temperature rose to 150°C. If the gas is assumed to be perfect, determine 1. The heat flow during the process 2. The pressure at the beginning of the cycle 3. The final pressure Given: Cv = 0.72 kJ/kg K and R = 0.287 kJ/kg K. solution m = 4.5 kg V1 = 1.5 m3 V2 = 1.5 m3 T1 = 20 + 273 = 293 K T2 = 150 + 273 = 423 K Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K Q12 = mCv(T2 − T1) = 4.5 kg × 0.72 kJ/kg K × (423 – 293) K = 421.2 kJ 2. From Equation, PV = mRT For state 1:
P1V1 = mRT P1=mRT 1/V1
4.5kg0.287kJ/kg K 293K/ 1.5m1 3= P1 = 252.27 kN/m2 3. For state 2: P2V2 = mRT2 P2= 4.5kg0.287kJ/kg K 423K/ 1.5m3 P2 = 364.20 kN/m2 specific heat capacity at constant pressure (cp) When 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature by 1 K while the pressure of the gas remains constant, the amount of heat energy that is supplied is known as the specific heat capacity at constant pressure and is denoted by Cp. The unit of Cp is J/ kg K. For a reversible non-flow process at constant pressure: dQ = mCpdT For a perfect gas, the value of Cp is constant for any one gas at all pressures and temperatures. Equation dQ = mCpdT can be expanded as follows: In a reversible constant pressure process, the heat flow Q = mCp(T2 − T1) relationship Between the specific heats Consider a perfect gas being heated at constant pressure from T1 to T2. Referring to the non-flow equation, Q = U2 – U1 + W and the equation for a perfect gas, U2 – U1 = mCv(T2 – T1), combining will give Q = mCv(T2 – T1) + W In a constant pressure process, the work done by the fluid is given by W = P . ΔV that is W = P(V2 – V1) Using the equation PV = mRT: W = mR(T2 – T1) Substituting: Q = mCv(T2 – T1) + mR(T2 – T1) = m(Cv + R)(T2 – T1) mCp(T2 – T1) = m(Cv + R)(T2 – T1) Therefore, Cp = Cv + R This equation may also be written as R = Cp − Cv specific heat ratio ‘γ’ The ratio of specific heat at constant pressure to the specific heat at constant volume is given by the symbol ‘γ’ (gamma). γ= Cp/ Cv From Equation Cp = Cv + R , it is clear that Cp has to be greater than Cv for a perfect gas. It follows, therefore, that the ratio Cp/Cv = γ is always greater than unity. In general, ‘γ’ is 1.4 for diatomic gases such as carbon monoxide (CO), hydrogen (H2), nitrogen (N2) and oxygen (O2). EXAMPLE A particular perfect gas has a specific heat as follows: Cp = 0.846 kJ/kg K and Cv = 0.657 kJ/kg K Determine the gas constant and the molecular weight of the gas. solution From Equation R = Cp – Cv R = Cp – Cv that is R = 0.846 – 0.657 = 0.189 kJ/kg K or R = 189 Nm/kg K. M=Ro/R that is M=8314.4 /189 = 44.0 kg/kmol