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What Thermo’s all about? Imagine how the world changed: - Four horses pulling a city Omnibus - Wind pushing ships a cross the ocean - Bucket and rope to lift water out of mine. Instead all of this “engines” existed that could do all these jobs at much lower cost. In the 19th century, it come “Industrial Revolution”. Steam Engines, Internal Combustion Engines, Electric Motors, and many other types of engines. Goal: “Ability to do Work” Thermo was born here to describe the heat engines “Power developed from heat”. Thermo was then developed to study energy. How much of it is in coal, wood, running water, in steam at low and at high pressure, and at low and high temperature? This is the concern of the “First Law of Thermodynamics” It was discovered early that you couldn’t always transfer all of one kind to another. (Energy to work is not possible). How efficiently can energy be converted from one form to another? “Sadi Carnot” answered leading to the “Second Law of Thermodynamics” 1st and 2nd laws are the main business of thermodynamics. Thermo: Is the science that deals with the transformation of all kinds of energies from one form to another. It is important to note that these laws have No proof in the mathematical sense. Their validity lies in the absence of contrary results. 1st law: Energy interchange. (How much of this kind of energy is equivalent to that kind of energy?). 2nd law: Energy amount and direction possible. (What changes are possible, what are not, and which direction?). READ Ch.1, Introduction to Chemical Eng. Thermodynamics / Smith. Chapter 2 First Law of Thermodynamics Energy can’t be created or destroyed. You can only change it from one form to another, or you can only add it to the system from the surroundings. Other forms: Introduction to Chemical Eng. Thermodynamics / Smith / p22. Energy Balance Closed system (No composition change) U + K + P = Q + W Units: Energy (J) SI (ft.lbf) EE System. For units check: App. A W Q +VE System System W Q -VE Surroundings Thermodynamics State And State Functions State Functions: “Exact Differential” F= state1 = F(state2) – F(state1) Independent of path depends only on the state of the material, like “height”. state2 dF (State Function = Point Function) Any state function can be expressed mathematically as a function of thermodynamic coordinates (P, T). Their value can be identified with points on a graph. Path Function: Depend on the path, like distance. Work and Heat, are not properties, they represent the energy changes between surr. and system. They are associated with areas rather than points on the graph. They considered as path functions. Intensive and Extensive Properties: Extensive Property: Value of property depends on size of the system (Mass, Vol., Internal Energy). (Additive) Intensive Property: Value of property does not depend on size of the system (P, T, Specific Vol.). (Not additive) Properties are usually tabulated on a molar basis. Equilibrium “Static Conditions” or No change. Thermodynamically, indicates the absence of any tendency toward change on a macroscopic scale. If all forces in “exact” balance, then the system at equilibrium. If a driving force(s) exist but due to a resistance, no change in state occurs, the system is still considered in a non-equilibrium state. Different driving forces to bring different kinds of changes: 1. Mechanical forces: (Equal pressure, mech. equilibrium) 2. Thermal forces: ( Equal temp., Thermal equilibrium) 3. Chemical potentials: (Chemical rxn, Chemical equilibrium) (Phase change, Phase equilibrium) At equilibrium, all such forces are in balance. The Phase Rule Degrees of freedom (F): The number of independent variables that must be fixed to establish the intensive state of any system. F = 2- + N Where: = No. of phases. N= No. of Components. If F = 0, the system is Invariant max= 2 + N If N =1, then max = 3 (Triple point) Ex. 2.5 The Reversible Process A process is reversible when its direction can be reversed at any point by an infinitesimal change in external conditions. Ex.: Removing powder particles from piston-cylinder arrangement. (No friction). dW = -P dV PdV v2 Wrev = The reversible Process = Ideal Process Ex. 2.6 v1 Constant–V and Constant-P Processes For a closed system d(nU) = dQ + dW For a mechanically reversible, non flow process dW= -P d(nV) Combine d(nU) = dQ - P d(nV) (2.6) (1.2) (2.8) Constant Volume process Work is zero dQ = d(nU) Integrate Q= n U (const V) (const V) (2.9) (2.10) For mechanically reversible, constant volume, non flow process Heat transfer = Internal energy change of the system Constant Pressure process From definition of enthalpy (H) H = U + PV (2.11) Enthalpy is an extensive property (like V and U) depends on mass. Its unit is (J). nH = nU + P(nV) For infinitesimal constant pressure change d(nH) = d(nU) + P d(nV) Substitute d(nU) from eq’n 2.8, you get; dQ = d(nH) (const P) (2.12) Integrate Q = n H (const P) (2.13) For mechanically reversible, constant pressure, non flow process Heat transfer = Enthalpy change of the system Heat Capacity The amount of heat to be added or removed from the body to change its temperature by 1oC (1k). C = dQ/dT Heat Capacity at Constant Volume CV = (dU/dT)V dU = CV dT Integrate U = (const V) C dT T2 T1 V For rev., const.- volume process Q = n U = n C dT T2 T1 V (const V) Heat Capacity at Constant Pressure CP = (dH/dT)P dH = CP dT Integrate H = For rev., const.- pressure process (const P) C dT T2 T1 Q = n H = n P C dT T2 T1 P (const P) Mass And Energy Balances for Open Systems Mass Balance for open system dmcv/dt + (m.)fs = 0 (m.)fs = m.3 – m.1 - m.2 m. = u A dmcv/dt + ( u A )fs = 0 Continuity Equation Accumulation term Where Special case: Steady State (Conditions within the control volume do not change with time). ( u A )fs = 0 For a single entrance and single exit streams, the mass flow rate is the same: m. = u2 A2 u1 A1 Because specific volume is reciprocal of density m. = u1 A1 /V1 u2 A2 /V2u A /V Continuity Equation Energy Balance for Steady-State Flow system Open system (steady state d(mU)cv/dt) = 0) H + u2/2gc + g/gc z = Q + W Enthalpy H = U + PV Work = Ws = Shaft work + PV work W = Ws + P1V1 - P2V2 U + K + P = Q + Ws + P1V1 - P2V2 Arrange (U2+ P2V2) - (U1+ P1V1) + K + P = Q + Ws (H) + K + P = Q + Ws K= ½ u2 P = g Z EES Special case: H + u2/2 + g z = Q + Ws H + u2/2gc + g/gc z = Q + Ws If K = 0 P = 0 H = Q + Ws