T098. c Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2012) 2 / 26
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1 Conservation of Energy for a Closed System Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka-, Bangladesh zahurul@me.buet.ac.bd 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 δq = δw Mechanical equivalent of heat, J = { kj/kcal. in SI unit ME : Basic Thermodynamics T98 Example of a control mass undergoing a cycle. c Dr. Md. Zahurul Haq (BUET ME ( / 6 c Dr. Md. Zahurul Haq (BUET ME ( / 6 for a Change in State T4 δw = J δq δq = δw [J =. in SI unit] δq A + δq B = δw A + δw B δq C + δq B = δw C + δw B : δq A δq C = δw A δw C δq A δw A = δq C δw C (δq δw A = (δq δw C = As (δq δw is independent of the path involved 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. δq δw = de Q W = E c Dr. Md. Zahurul Haq (BUET ME ( 3 / 6 The physical significance of the property E is that 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. Potential Energy (PE: 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 & PE are external forms of energy as these are independent of the molecular structure of matter. These are associated with the coordinate frame that we select 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 ME ( 4 / 6
2 Internal Energy (U: A Thermodynamic Property E = U + KE + PE + δq δw = du + d(ke + d(pe + δq δw = du + d(ke + d(pe + = decm de cm = Q W du = du = U U = m(u u d(ke = mvdv = d(ke = m(v V d(pe = mgdz = d(pe = mg(z Z = mgh T33 Various forms of microscopic energies that make up sensible energy. T34 Internal energy of a system is the sum of all forms of the microscopic energies. Q W = [ (U U + m(v V + mg(z Z ] q w = [ (u u + (V V + g(z Z ] (u u c Dr. Md. Zahurul Haq (BUET ME ( 5 / 6 c Dr. Md. Zahurul Haq (BUET ME ( 6 / 6 Enthalpy (H: A Thermodynamic Property H U + PV h u + Pv Q W = E T9 Q W = U if KE, PE W = PdV = P(V V Q = U U + P(V V Q = (U + P V (U + P V T7 Q = H 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. T3 T3 T3 c Dr. Md. Zahurul Haq (BUET ME ( 7 / 6 c Dr. Md. Zahurul Haq (BUET ME ( 8 / 6
3 Example: Electric Heating of Gas at Constant Pressure. Joule s Free Expansion Experiment T8 Q net W net = E U Q net (W e + PdV = E U Q net W e P(V V = U U Q net = 3.7 kj W e = 7. kj h = h(3 kpa, x = h = 74.9 kj/kg h = h(3 kpa, T T =? Q net W e = P(V V + U U = H H = m(h h = h = kj/kg at 3 kpa & T = T = o C T39 Valve is opened and allowed to equilibrate. No change in water temperature. So, no heat transfer takes place. st Law: Q =, W =, = U =. P & V changed during this process, but without any change in U. So, U f (P,V = U = f (T for ideal gas. h = u + Pv = u + RT H = f (T for ideal gas. c Dr. Md. Zahurul Haq (BUET ME ( 9 / 6 Specific Heats C m δq C V m ( δq V = du + δw = du + Pdv ( c V = ( δu V + P dv = c V ( V = du dt = δu V = ( δu C P m + P dv V + ( δq P as u = f (T for ideal gas = du + δw = d(h Pv + Pdv = dh vdp ( = δh v dp c P = ( δh P v dp P = ( δh P = c P as h = f (T for ideal gas ( P = dh dt h = u+pdv = u+rt dh = du+rdt c P dt = c V dt+rdt Specific heat ratio, k c P c V = c P c V = R c V = R k c P = kr k c Dr. Md. Zahurul Haq (BUET ME ( / 6 c Dr. Md. Zahurul Haq (BUET ME ( / 6 Ideal gas models: R = Ru M c P (T c V (T = R c V (T = R k(t : k(t = c P(T c V (T du = c V dt u u = c V (TdT dh = c P dt h h = c P(TdT T8 T6 c T7 Dr. Md. Zahurul Haq (BUET ME ( / 6
4 Open (CV System Conservation of Energy for CV System T5 for closed system T99 Open (CV System Open (CV System c Dr. Md. Zahurul Haq (BUET ME ( 3 / 6 c Dr. Md. Zahurul Haq (BUET ME ( 4 / 6 T E cv = Q W + m i e i m e e e ( = Q W + m i (u i + V i + gz i m e u e + V e + gz e W consists of 3 parts: Flow work, W flow, associated with mass crossing the control surface, Shaft work, W shaft, which can be interchanged between the system and its surroundings, 3 If the CS is not rigid, boundary or PdV work may occur. Second and third components of work are lumped together as W net. c Dr. Md. Zahurul Haq (BUET ME ( 5 / 6 T Flow work, δw flow necessary to push a differential mass, δm into a CV is the product of force, F = PA and the distance,dx = V under uniform flow. δw flow = Fdx = PAV = Pv m W flow = Pv m W = W net + m i (Pv i m e (Pv e c Dr. Md. Zahurul Haq (BUET ME ( 6 / 6
![[ ( ] V = Q W net + m (h h + V + g(z z T δ m = ρv n da c T4 m = ρ V = ρav = AV v c Dr. Md.](/docs-images/72/66603478/images/5-1.jpg "Zahurul Haq (BUET ME ( 7 / 6 c Dr. Md. Zahurul Haq (BUET ME ( 8 / 6 Open (CV System Example: Feedwater Heater at Steady State. Determine m & V. Assume, v v f (T.")
5 Open (CV System Open (CV System (FLT for CV System Conservation of Mass = Q W net + i m i (h i + V i + gz i e ( m e h e + V e + gz e Closed System: m i = m e =. T3 = Q W net V avg = A c A c V n da c Steady-State-Steady Flow (SSSF System: dm cv = = i m i = e m e : = One-inlet, One-exit & Steady-state: m i = m e = m. [ ( ] V = Q W net + m (h h + V + g(z z T δ m = ρv n da c T4 m = ρ V = ρav = AV v c Dr. Md. Zahurul Haq (BUET ME ( 7 / 6 c Dr. Md. Zahurul Haq (BUET ME ( 8 / 6 Open (CV System Example: Feedwater Heater at Steady State. Determine m & V. Assume, v v f (T. Example: Steam Turbine Open (CV System dm cv = i dm cv / = m i e m e i m i = m + m e m e = m 3 m = ρav T5 m 3 = ρ 3 (AV 3 ρ = ρ(t = T,P = P ρ 3 = ρ(x =.,P = P 3 = m = 4.5 kg/s, V = 5.7 m/s. c Dr. Md. Zahurul Haq (BUET ME ( 9 / 6 T6 Steady-state Steady-flow: [ = Q W net + m (h h + ( V V h = h(mpa, 35 o C = J/kg h = h(.mpa,x =. = J/kg = W net = 678. kw ] + g(z z c Dr. Md. Zahurul Haq (BUET ME ( / 6
6 Mass Balance: t dm cv ( dmcv t = ( i = i m cv = m cv (t m cv ( = i m i e m e m i t ( e m e ( t m i m cv = m cv (t m cv ( = i e m i e ( t m e m e Energy Balance: t = Q W cv + i m i (h i + V i + gz i e t = Q t W cv + i t (h + V e + gz m e ( decv e ( m e h e + V e + gz e t (h + V + gz i m i If CV is fixed in space, E cv = U cv ( = ducv = d(mu = m u m u = U cv du cv = δq δw cv + ( i h + V i + gz dm i ( e h + V e + gz dm e U cv = Q W cv + ( h + V + gz i dm i ( h + V + gz e dm e du cv = d(mu = (mdu + udm: alternative form. c Dr. Md. Zahurul Haq (BUET ME ( / 6 c Dr. Md. Zahurul Haq (BUET ME ( / 6 Charging of Evacuated Vessel In the analysis of transient flow system, two models are widely used: Uniform State: the state within CV at any instant is uniform throughout the CV. However, the state within CV may change with time. It implies rapid or instantaneous approach to equilibrium at all times for the mass within CV. Uniform Flow: the state of mass crossing a CS is invariant with time. However, the mass flow rate across that particular CS may vary with time. This condition is frequently met when the flow into transient system is supplied from a very large reservoir of matter. c Dr. Md. Zahurul Haq (BUET ME ( 3 / 6 Mass enters only at one section of CS and no efflux of matter. T9 Assumptions: Adiabatic, rigid control volume Uniform state within the tank at any instant Negligible KE of inflowing gas Uniform inlet flow c Dr. Md. Zahurul Haq (BUET ME ( 4 / 6
7 Charging of Evacuated Vessel: CV Analysis Charging of Evacuated Vessel: CM Analysis T If tank is insulated or filled rapidly, Q W cv =, KE =, PE = Supply from a large reservoir uniform flow. m cv = m i m e m m = m i m e m =, m e = m = m i U cv = Q cv W cv + ( i dm i ( e dm e = h i m i U cv = m u m u = m u = h i = u T = kt i : for ideal gas. Example: Estimate T. h i = h(steam, P =. MPa, T = 3 o C = 35 kj/kg h i = u = u(steam, P =. MPa, T =? T = 456 o C T If tank is insulated or filled rapidly, Q cv W cv, KE =, PE = U cv = Q cv W cv U cv = m(u u Pressure in the line near the valve is constant. W = PdV = P(V V = P[V tank (V tank + V ] = mp v m(u u = mp v u = u + P v = h i = u : same result. c Dr. Md. Zahurul Haq (BUET ME ( 5 / 6 c Dr. Md. Zahurul Haq (BUET ME ( 6 / 6
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