Unit 3 – First Law September 9, 2010 Outline Unit Three – Heat, Internal • Quiz solution Energy and the First Law • Unit three – heat, internal energy and the first law of thermodynamics Mechanical Engineering 370 – Heat, Q, is energy in transit due only to a temperature gradient Thermodynamics – Internal energy, U, is a property found from tables or relations for ideal gases Larry Caretto –Q = ΔU + W or Q –Q = ΔU + W –W September 9, 2010 in out out in – Specific internal energy: ΔU = mΔu 2 Unit Three Goals Unit Three Goals Continued – understand the meaning of the internal • As a result of studying this unit you energy as a property giving the amount of should be able to energy stored in a body. – find properties more easily than you were – use the first law as Q = ΔU + W = mΔu+ W able to do after units one and two to solve problems in closed (fixed mass) – describe the path for a process and systems determine the work with more confidence – use the sign convention for heat and work – understand that heat is energy in transit • Heat added to a system is positive; heat removed from a system is negative: Q = Q – due only to a temperature difference in Qout • Work done by a system is positive; work done on a system is negative: W = Wout –Win 3 4 Unit Three Goals Continued Unit Three Goals Concluded – we do not have to worry about the sign – find the internal energy from the enthalpy convention when we use W = ∫PdV; we as u = h – Pv (with correct units) – this is automatically get the correct sign needed for quiz tables – use the first law as q = Δu + w where q – use the enthalpy in place of the internal = Q/m and w = W/m, in problem solving energy when enthalpy is given in a table but internal energy is not – find the internal energy in tables from – work problems, possible involving trial-and- any definition of the state in one-phase error solutions, using the first law, property or mixed regions relations and a path equation 5 6 ME 370 – Thermodynamics 1 Unit 3 – First Law September 9, 2010 First Law as Energy Balance First-law Problems • Change in Energy = energy added as • Start with closed (fixed mass) systems heat – energy lost by doing work • Here ΔU = mΔu, and we find u from • Heat and work are processes tables or ideal gas relationships • We can store energy as kinetic energy, • In tables, u acts like volume potential energy or thermodynamic – Find u(T,P) in gas (superheat) region (or internal energy, U = mu compressed liquid tables for water) – u is a property (depends on state) –u = uf + x(ug –uf) in mixed region • Most processes have negligible KE and – Compressed liquid: u(T,P) ≈ uf(T) PE changes; consider only ΔU here – Can find u given v and (T or P) by same rules used to find state previously 7 8 Solving First-law Problems First Law for a Complex Path • Data usually given: partial information • Here we have an about state points along a path initial point (1) and a P • Usually want to find heat transfer final point (4) 23 • Have to do same calculations done • Work = area under path = (P + P )•(V – previously to find work: W = ∫pathPdV 1 2 2 1 4 – May have other types of work such as V1)/2 + P2-3(V3 –V2) + V electrical work or applied torque (P3 + P4)(V4 –V3)/2 • Use property data to get u at initial and •Q = (Ufinal –Uinitial) + W = m(u4 –u1) + W final states of overall process • Do not need intermediate u values • Q = m(ufinal –uinitial) + W • Use property data for u (and possibly v) 9 10 Enthalpy Is Electricity Heat or Work? • Defined property H ≡ U + PV • We know that “resistance heating” is I2R • h = H/m = U/m + P(V/m) = u + Pv = u + • If the system boundary is defined such P/ρ that electricity crosses boundary then • Constant pressure path has solution we do I2R work on system – For Q = 0, U = –W = –(–I2R) = I2R that Q = m(u2 –u1) + P1-2 m(v2 –v1) = Δ m[(u2 + P2v2) – (u1 + P1v1)] = m(h2 –h1) • If resistor is outside system boundary • For constant pressure process Q = ΔH then I2R heat is added to system • H used in open systems and quiz tables – For W = 0, ΔU = Q = I2R require calculation of u = h – Pv • Either approach gives ΔU = I2R 11 12 ME 370 – Thermodynamics 2 Unit 3 – First Law September 9, 2010 Example Calculation Example Calculation II • Given data: o • Refrigerant 134a, initially at 100 F and o 3 – Initial state: p1 = 20 psia, T1 = 100 F, V1 = 20 ft 20 psia fills a container with an initial – Path: Constant volume to intermediate state 3 volume of 20 ft . It is cooled in two then constant pressure to final state o steps: (1) at constant volume to 40 F – Intermediate state: v = v and T = 40oF o 2 1 2 and (2) at constant pressure to 0 F. o – Final state: P3 = P2 and T3 = 0 F • What is the heat transfer? • Equations: – First law: Q = ΔU + W = m(u3 –u1) + W – Work: W = ∫pathPdV – Mass: m = V1 / v1 13 14 Finding the Work Solving the Problem • No work done during the • Q = m(u –u) + mP (v –v) 1 3 1 2-3 3 2 first (constant-volume) •m = V/ v part of the path (1 to 2) P 1 1 2 • We have to find u and v , P = P = • Total work comes from 3 1 1 2 3 second (constant- P2-3, v2, v3 and u3. o pressure) part (2 to 3) V • State 1 has P1 = 20 psia, T1 = 100 F •W = P (V –V) where o 2-3 3 2 •Q = ΔU + W • State 2 has v2 = v1 and T2 = 40 F P2-3 = P2 = P3 o = m(u3 –u1) • State 3 has P3 = P2 and T3 = 0 F •V = mv and V = mv 3 3 2 2 + mP2-3(v3 – • Find properties from R-134a tables v2) 15 16 States 1 and 2 and Mass Interpolate for Final State • We can find P = 20 psia and T = 1 1 • Final state has P = P = 18.086 psia 100oF in superheat tables giving v = 3 2 1 and T = 0oF – interpolate between 15 2.8726 ft3/lb and u = 112.66 Btu/lb 3 3 m 1 m V1 20 ft and 20 psia to find u3 and v3 m = = 3 = 6.962 lbm Properties from Table v1 2.8726 ft A-13E, page 978 3 3 3 3 lb 3.1001 ft 18.086 psia −15 psia ⎛ 2.2922 ft 3.1001 ft ⎞ 2.549 ft m v = + ⎜ − ⎟ = • State 2 has v = v and T = 40oF 3 ⎜ ⎟ 2 1 2 lbm 20 psia −15 psia ⎝ lbm lbm ⎠ lbm • This is between 15 and 20 psia 95.08 Btu 18.086 psia −15 psia ⎛ 94.72 Btu 95.08 Btu ⎞ 94.86 Btu ⎜ ⎟ u3 = + ⎜ − ⎟ = 3 3 lb 20 psia −15 psia lb lb lb 20 psia −15 psia ⎛ 2.8726 ft 3.4074 ft ⎞ m ⎝ m m ⎠ m P =15 psia + ⎜ − ⎟ =18.086 psia 2 2.5306 ft 3 3.4074 ft 3 ⎜ lb lb ⎟ − ⎝ m m ⎠ • Common interpolation factor for u and v lbm lbm 17 18 ME 370 – Thermodynamics 3 Unit 3 – First Law September 9, 2010 Compute Heat Transfer Another Example Calculation o • Given: 10 kg of H2O at 200 C and 50% • Q = m(u3 –u1) + mP2-3(v3 –v2) quality is expanded to 400oC at • Apply previous results to obtain constant pressure ⎛ 94.86 Btu 112.66 Btu ⎞ • Find: Heat Transfer ⎜ ⎟ Q = ()6.962 lbm ⎜ − ⎟ + ⎝ lbm lbm ⎠ • Given data: 18.086 lb ⎛ 2.549 ft 3 2.8726 ft 3 ⎞ in2 ⋅ Btu o 6.962 lb f ⎜ − ⎟ = −145 Btu – Initial state: T1 = 200 C, x1 = 0.5 ()m 2 ⎜ ⎟ 3 in ⎝ lbm lbm ⎠ 5.40395 lbf ⋅ ft – Path: Constant pressure P1 = P2 = P1-2 o – Final state: T2 = 400 C • Negative Q means heat is rejected from – Substance: Water, use property tables system – Mass: m = 10 kg 19 20 Another Example Calculation II Find Initial State Properties o • Equations: • Initial state is x1 = 50% and T1 = 200 C – First Law: Q = ΔU + W = m(u2 –u1) + W •x1 = 50% quality implies mixed region at –W = ∫path PdV o T1 = 200 C –W = P1-2 (V2 –V1) for constant pressure, P1-2 = o P1 = P2 •P1 = Psat(T1 = 200 C) = 1554.9 kPa • Combine first law and work equation to get Q = o •v1 = (1 – x1) vf(T1 = 200 C) + x1 vg(T1 = m(u2 –u1) + P1-2 (V2 –V1) 200oC) = 0.06426 m3/kg –V = mv and V = mv 2 2 1 1 •u = (1 – x ) u (T = 200oC) + x u (T = • We have to find v , v , P = P = P , u , and u 1 1 f 1 1 g 1 1 2 1 2 1-2 1 2 200oC) = 1723.0 kJ/kg • Use property tables for water Saturation properties from 21 Table A-4, page 916 22 Find Final State Properties Interpolation Details • Interpolation calculations for both u and • Point 2 has T = 400oC and P = P = 2 2 1 v have common factor for pressure 1554.9 kPa 1.5549 MPa −1.40 MPa u = 2953.1kJ + • • Interpolation in superheat table between kg o 1.60 MPa −1.40 MPa 1.40 MPa and 1.60 MPa at T2 = 400 C – At P = 1.40 MPa, u = 2953.1 and v = ⎜⎛2950.8kJ − 2953.1kJ ⎟⎞ = 2951.4kJ 0.21782 m3/kg ⎝ kg kg ⎠ kg – At P = 1.60 MPa, u = 2950.8 kJ/kg and v = 3 1.5549 MPa −1.40 MPa 0.19007 m3/kg v = 0.21782m + • kg 1.60 MPa −1.40 MPa • Interpolation (details next page) gives v 2 ⎛ 3 3 ⎞ 3 = 0.19646 m3/kg and u = 2951.3 kJ/kg ⎜0.19007m − 0.21782m ⎟ = 0.19646m 2 ⎝ kg kg ⎠ kg Properties from Table A-6, page 921 23 24 ME 370 – Thermodynamics 4 Unit 3 – First Law September 9, 2010 First Law Gives Answer Third Example Calculation o • Given: 110 kg of H2O at 200 C and •Q = ΔU + W = ΔU + P1-2 (V2 –V1) = 50% quality is compressed along a o m(u2 –u1) + P1-2 m(v2 –v1) linear path, P = a + bV to 400 C and 10 MPa 1 MJ Q = (10 kg)⎛2951.4kJ −1723.0kJ ⎞ + • Find: Heat Transfer ⎜ kg kg ⎟ ⎝ ⎠1000 kJ • Given data: ⎛ 3 3 ⎞ 1 MJ o (10 kg)(1.5549 MPa) 0.19646 m − 0.06426 m – Initial state: T1 = 200 C, x1 = 0.5 ⎜ kg kg ⎟ 3 ⎝ ⎠ MPa ⋅m – Path: Straight line, P = a + bV o • Q = 14.33 MJ (added to system) – Final state: T2 = 400 C, P2 =10 MPa – Substance: Water, use property tables – Mass: 110 kg 25 26 Third Example Calculation II Find Initial State Properties • Equations: o • Initial state: x1 = 50% and T1 = 200 C – First Law: Q = ΔU + W = m(u2 –u1) + W •x = 50% quality implies mixed region at –W = ∫path PdV 1 T = 200oC –W = (P1 + P2)(V2 –V1)/2 for straight line path 1 (area of a trapezoid) o •P1 = Psat(T1 = 200 C) = 1554.9 kPa – Combine first law and work equation to get Q o •v1 = (1 – x1) vf(T1 = 200 C) + x1 vg(T1 = = m(u2 –u1) + (P1 + P2)(V2 –V1)/2 200oC) = 0.06426 m3/kg –V2 = mv2 and V1 = mv1