Conservation of Energy for a Closed System First Law of Thermodynamics Dr. Md. Zahurul Haq rofessor Department of Mechanical Engineering Bangladesh University of Engineering & Technology BUET Dhaka-000, Bangladesh First Law of Thermodynamics When a system undergoes a cyclic change, the net heat to or from the system is equal to the net work from or to the system. J = δw zahurul@me.buet.ac.bd http://teacher.buet.ac.bd/zahurul/ ME 20: Basic Thermodynamics Mechanical equivalent of heat, J = { 4.868 kj/kcal.0 in SI unit c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 / 9 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 2 / 9 First Law of Thermodynamics for a Change in State T363 δw = J = δw [J =.0 in SI unit] A + 2 C = δw A + 2 δw C B + 2 C = δw B + 2 δw 2 C 2 : A B = δw A δw B A δw A = B δw A = δw B = δw B δw is independent of the path and dependent only on the initial and final states; hence, it has the characteristics of a property and this property is denoted by energy, E. δw = de Q 2 W 2 = E c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 3 / 9 hysical significance of the property E: it represents all the energy of the system in the given state. This energy might be present in a variety of forms, such as: Kinetic Energy KE: energy of a system associated with motion. otential Energy E: energy associated with a mass that is located at a specified position in a force field. Internal Energy U: some forms of energy, e.g., chemical, nuclear, magnetic, electrical, and thermal depend in some way on the molecular structure of the substance that is being considered, and these energies are grouped as the internal energy of a system, U. KE & E are external forms of energy as these are independent of the molecular structure of matter. These are associated with the selected coordinate frame and can be specified by the macroscopic parameters of mass, velocity & elevation. Internal energy, like kinetic and potential energy, has no natural zero value. Therefore, it is necessary to arbitrarily define the specific internal energy of a substance to be zero at some state that is referred to as the reference state. c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 4 / 9
Internal Energy U: A Thermodynamic roperty E = U + KE + E + δw = du + dke+de+ δw = du + dke + de + = decm de cm = Q W du = du = U 2 U = mu 2 u dke = mvdv = dke = 2 mv2 2 V2 de = mgdz = de = mgz 2 Z = mgh T33 T34 Q 2 W 2 = [ U 2 U + 2 mv2 2 V2 +mgz 2 Z ] Various forms of microscopic energies that make up sensible energy. Internal energy of a system is the sum of all forms of the microscopic energies. c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 5 / 9 q 2 w 2 = [ u 2 u + 2 V2 2 V2 +gz 2 Z ] u 2 u c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 6 / 9 Enthalpy H: A Thermodynamic roperty H U + V h u + v Q 2 W 2 = E T29 Q 2 W 2 = U if KE 0, E 0 W 2 = dv = V Q 2 = U 2 U + V Q 2 = U 2 + 2 U + V T7 Q 2 = H 2 H The heat transfer in a constant-pressure quasi-equilibrium process is equal to the change in enthalpy, which includes both the change in internal energy and the work for this particular process. T30 T3 T32 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 7 / 9 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 8 / 9
Joule s Free Expansion Experiment Specific Heats C m C V m V C m T39 Valve is opened and allowed to equilibrate. No change in water temperature. So, no heat transfer takes place. st Law: Q 2 = 0, W 2 = 0, = U = 0. & V changed during this process, but without any change in U. So, U f,v = U = ft for ideal gas. h = u + v = u + RT H = ft for ideal gas. = du +δw = du + dv c V = δu V + dv = c V V = du dt = δu V = δu + dv V + 0 as u = ft for ideal gas = du +δw = dh v+dv = dh vd = δh v d c = δh v d = δh 0 = c as h = ft for ideal gas = dh dt h = u+dv = u+rt dh = du+rdt c dt = c V dt+rdt Specific heat ratio, k c c V = c c V = R c V = R k c = kr k c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 9 / 9 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 0 / 9 Ideal gas models: R = Ru M c T c V T = R c V T = R kt : kt = c T c V T du = c V dt u 2 u = c VTdT dh = c dt h 2 h = c TdT T28 T5 First Law of Thermodynamics for closed system T26 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics T27 ME 20 205 / 9 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 2 / 9
Open CV System Open CV System Mass Continuity Equation T23 V avg = A c A c V n da c T443 m cv t+m i = m cv t + t+m e T444 m cv t + t m cv t = m i m e mcvt+ t mcvt t if t 0 : dmcv = m i m e t = m i m e T22 δ m = ρv n da c T24 m = A ρv nda : m = ρav = AV v for d flow = i m i e m e c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 3 / 9 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 4 / 9 Open CV System [Moran Ex. 4.]: Feedwater heater at steady-state. Determine m 2 &. Assume, v 2 v f T 2. Open CV System Conservation of Energy for CV System = i / = 0 m i e m e i m i = m + m 2 e m e = m 3 m = ρav m 3 = ρ 3 AV 3 T25 ρ 2 = ρt = T 2, = 2 ρ 3 = ρx = 0.0, = 3 = m 2 = 4.5 kg/s, = 5.7 m/s. c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 5 / 9 T099 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 6 / 9
Open CV System Open CV System First Law of Thermodynamics FLT for CV System T00 = Q W cv + i m i h i + V2 i 2 + gz i e m e h e + V2 e = Q W + m i e i m e e e = Q W + m i u i + V2 i 2 + gz i = Q W cv + m i h i + V2 i 2 + gz i m e u e + V2 e m e h e + V2 e Closed System: m i = m e = 0. = Q W net Steady-State-Steady Flow SSSF System: = 0 = i m i = e m e : = 0 W = W s + W b + W f = W cv + W f W f = V i V e = m i v i m e v e h u + v One-inlet, One-exit & Steady-state: m i = m e = m. [ ] 0 = Q W cv + m h h 2 + 2 2 + gz z 2 Open CV System c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 7 / 9 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 8 / 9 Bernoulli s Equation h = u + v dh = du + dv + vd, so for isothermal process and incompressible fluid, dh = vd h h 2 = 2 ρ For a steady state flow device if E 0, KE 0, W cv = 0 and Q cv = 0: [ ] 0 = 0 0+ m h h 2 + 2 2 + gz z 2 ρg 2g z : pressure head : velocity head : elevation head ρg + V2 2g + z = 2 ρg + V2 2 2g + z 2 c Dr. Md. Zahurul Haq BUET First Law of Thermodynamics ME 20 205 9 / 9