Heat, Internal Heat, Internal Energy and the First Law

Unit 3 – First Law September 9, 2010 Outline • Quiz solution • Unit three – heat, internal energy and the first law of thermodynamics Unit Three – Heat, Internal Energy and the First Law – Heat, Q, is energy in transit due only to a temperature gradient – Internal energy, U, is a property found from tables or relations for ideal gases – Q = ΔU + W or Qin – Qout = ΔU + Wout – Win – Specific internal energy: ΔU = mΔu 2 Unit Three Goals Unit Three Goals Continued – understand the meaning of the internal energy as a property giving the amount of energy stored in a body. – use the first law as Q = ΔU + W = mΔu + W to solve problems in closed (fixed mass) systems – use the sign convention for heat and work • As a result of studying this unit you should be able to – find properties more easily than you were able to do after units one and two – describe the path for a process and determine the work with more confidence – understand that heat is energy in transit due only to a temperature difference • Heat added to a system is positive; heat removed from a system is negative: Q = Qin – Qout • Work done by a system is positive; work done on a system is negative: W = Wout – Win 3 Unit Three Goals Continued Unit Three Goals Concluded – find the internal energy from the enthalpy as u = h – Pv (with correct units) – this is needed for quiz tables – use the enthalpy in place of the internal energy when enthalpy is given in a table but internal energy is not – work problems, possible involving trial-anderror solutions, using the first law, property relations and a path equation – we do not have to worry about the sign convention when we use W = ∫PdV; we automatically get the correct sign – use the first law as q = Δu + w where q = Q/m and w = W/m, in problem solving – find the internal energy in tables from any definition of the state in one-phase or mixed regions 5 ME 370 – Thermodynamics 4 6 1 Unit 3 – First Law September 9, 2010 First Law as Energy Balance First-law Problems • Change in Energy = energy added as heat – energy lost by doing work • Heat and work are processes • We can store energy as kinetic energy, potential energy or thermodynamic internal energy, U = mu • Start with closed (fixed mass) systems • Here ΔU = mΔu, and we find u from tables or ideal gas relationships • In tables, u acts like volume – Find u(T,P) in gas (superheat) region (or compressed liquid tables for water) – u = uf + x(ug – uf) in mixed region – Compressed liquid: u(T,P) ≈ uf(T) – Can find u given v and (T or P) by same rules used to find state previously – u is a property (depends on state) • Most processes have negligible KE and PE changes; consider only ΔU here 7 Solving First-law Problems 8 First Law for a Complex Path • Data usually given: partial information about state points along a path • Usually want to find heat transfer • Have to do same calculations done previously to find work: W = ∫pathPdV • Here we have an initial point (1) and a P final point (4) 2 3 • Work = area under path = (P1 + P2)•(V2 – 4 1 V1)/2 + P2-3(V3 – V2) + V (P3 + P4)(V4 – V3)/2 • Q = (Ufinal – Uinitial) + W = m(u4 – u1) + W • Do not need intermediate u values • Use property data for u (and possibly v) – May have other types of work such as electrical work or applied torque • Use property data to get u at initial and final states of overall process • Q = m(ufinal – uinitial) + W 9 10 Enthalpy Is Electricity Heat or Work? • Defined property H ≡ U + PV • h = H/m = U/m + P(V/m) = u + Pv = u + P/ρ • Constant pressure path has solution that Q = m(u2 – u1) + P1-2 m(v2 – v1) = m[(u2 + P2v2) – (u1 + P1v1)] = m(h2 – h1) • For constant pressure process Q = ΔH • H used in open systems and quiz tables require calculation of u = h – Pv • We know that “resistance heating” is I2R • If the system boundary is defined such that electricity crosses boundary then we do I2R work on system 11 ME 370 – Thermodynamics – For Q = 0, ΔU = –W = –(–I2R) = I2R • If resistor is outside system boundary then I2R heat is added to system – For W = 0, ΔU = Q = I2R • Either approach gives ΔU = I2R 12 2 Unit 3 – First Law September 9, 2010 Example Calculation Example Calculation II • Given data: • Refrigerant 134a, initially at 100oF and 20 psia fills a container with an initial volume of 20 ft3. It is cooled in two steps: (1) at constant volume to 40oF and (2) at constant pressure to 0oF. • What is the heat transfer? – Initial state: p1 = 20 psia, T1 = 100oF, V1 = 20 ft3 – Path: Constant volume to intermediate state then constant pressure to final state – Intermediate state: v2 = v1 and T2 = 40oF – Final state: P3 = P2 and T3 = 0oF • 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 1 first (constant-volume) part of the path (1 to 2) P 2 3 • Total work comes from second (constantpressure) part (2 to 3) V • W = P2-3(V3 – V2) where • Q = ΔU + W P2-3 = P2 = P3 = m(u3 – u1) • V3 = mv3 and V2 = mv2 + mP2-3(v3 – v2) • Q = m(u3 – u1) + mP2-3(v3 – v2) • m = V1 / v1 • We have to find u1 and v1, P2 = P3 = P2-3, v2, v3 and u3. • State 1 has P1 = 20 psia, T1 = 100oF • State 2 has v2 = v1 and T2 = 40oF • State 3 has P3 = P2 and T3 = 0oF • Find properties from R-134a tables 15 16 States 1 and 2 and Mass Interpolate for Final State • We can find P1 = 20 psia and T1 = 100oF in superheat tables giving v1 = 2.8726 ft3/lb and u1 = 112.66 Btu/lbm 3 m V1 20 ft = = 6.962 lbm v1 2.8726 ft 3 lbm State 2 has v2 = v1 and m= Properties from Table A-13E, page 978 • T2 = 40oF • This is between 15 and 20 psia 20 psia − 15 psia P2 = 15 psia + 2.5306 ft 3 3.4074 ft 3 − lbm lbm ⎛ 2.8726 ft 3 3.4074 ft 3 ⎞ ⎜⎜ ⎟⎟ = 18.086 psia − lbm lbm ⎝ ⎠ 17 ME 370 – Thermodynamics • Final state has P3 = P2 = 18.086 psia and T3 = 0oF – interpolate between 15 and 20 psia to find u3 and v3 v3 = 3.1001 ft 3 18.086 psia − 15 psia ⎛ 2.2922 ft 3 3.1001 ft 3 ⎞ 2.549 ft 3 ⎜ ⎟⎟ = + − 20 psia − 15 psia ⎜⎝ lbm lbm lbm lbm ⎠ u3 = 95.08 Btu 18.086 psia − 15 psia ⎛ 94.72 Btu 95.08 Btu ⎞ 94.86 Btu ⎜ ⎟⎟ = + − lbm 20 psia − 15 psia ⎜⎝ lbm lbm lbm ⎠ • Common interpolation factor for u and v 18 3 Unit 3 – First Law September 9, 2010 Compute Heat Transfer Another Example Calculation • Q = m(u3 – u1) + mP2-3(v3 – v2) • Apply previous results to obtain ⎛ 94.86 Btu 112.66 Btu ⎞ ⎟⎟ + − Q = (6.962 lbm )⎜⎜ lbm ⎝ lbm ⎠ 3 3 2 18.086 lb f ⎛ 2.549 ft 2.8726 ft ⎞ in ⋅ Btu ⎜ ⎟ (6.962 lbm ) − ⎜ lb ⎟ 5.40395 lb ⋅ ft 3 = −145 Btu in 2 lbm m f ⎝ ⎠ • Negative Q means heat is rejected from system • Given: 10 kg of H2O at 200oC and 50% quality is expanded to 400oC at constant pressure • Find: Heat Transfer • Given data: – Initial state: T1 = 200oC, x1 = 0.5 – Path: Constant pressure P1 = P2 = P1-2 – Final state: T2 = 400oC – Substance: Water, use property tables – Mass: m = 10 kg 19 Another Example Calculation II • Equations: – First Law: Q = ΔU + W = m(u2 – u1) + W – W = ∫path PdV – W = P1-2 (V2 – V1) for constant pressure, P1-2 = P1 = P2 • Combine first law and work equation to get Q = m(u2 – u1) + P1-2 (V2 – V1) – V2 = mv2 and V1 = mv1 • We have to find v1, v2, P1 = P2 = P1-2, u1, and u2 • Use property tables for water 20 Find Initial State Properties • Initial state is x1 = 50% and T1 = 200oC • x1 = 50% quality implies mixed region at T1 = 200oC • P1 = Psat(T1 = 200oC) = 1554.9 kPa • v1 = (1 – x1) vf(T1 = 200oC) + x1 vg(T1 = 200oC) = 0.06426 m3/kg • u1 = (1 – x1) uf(T1 = 200oC) + x1 ug(T1 = 200oC) = 1723.0 kJ/kg Saturation properties from Table A-4, page 916 21 Find Final State Properties Interpolation Details • Point 2 has T2 = 400oC and P2 = P1 = 1554.9 kPa • Interpolation in superheat table between 1.40 MPa and 1.60 MPa at T2 = 400oC – At P = 1.40 MPa, u = 2953.1 and v = 0.21782 m3/kg – At P = 1.60 MPa, u = 2950.8 kJ/kg and v = 0.19007 m3/kg • Interpolation (details next page) gives v2 = 0.19646 m3/kg and u2 = 2951.3 kJ/kg Properties from Table A-6, page 921 ME 370 – Thermodynamics 22 23 • Interpolation calculations for both u and v have common factor for pressure u = 2953.1 kJ kg + 1.5549 MPa − 1.40 MPa • 1.60 MPa − 1.40 MPa ⎛⎜ 2950.8 kJ − 2953.1 kJ ⎞⎟ = 2951.4 kJ kg kg ⎠ kg ⎝ v = 0.21782 m 3 kg + 1.5549 MPa − 1.40 MPa • 1.60 MPa − 1.40 MPa ⎛ 0.19007 m3 − 0.21782 m3 ⎞ = 0.19646 m3 ⎜ kg kg ⎟⎠ kg ⎝ 24 4 Unit 3 – First Law September 9, 2010 First Law Gives Answer Third Example Calculation • Q = ΔU + W = ΔU + P1-2 (V2 – V1) = m(u2 – u1) + P1-2 m(v2 – v1) 1 MJ Q = (10 kg )⎛⎜ 2951.4 kJ − 1723.0 kJ ⎞⎟ + kg kg ⎠ 1000 kJ ⎝ 3 3 ⎞ 1 MJ (10 kg )(1.5549 MPa)⎛⎜ 0.19646 m − 0.06426 m kg kg ⎟⎠ MPa ⋅ m 3 ⎝ • Q = 14.33 MJ (added to system) 25 Third Example Calculation II – Initial state: T1 = 200oC, x1 = 0.5 – Path: Straight line, P = a + bV – Final state: T2 = 400oC, P2 =10 MPa – Substance: Water, use property tables – Mass: 110 kg 26 Find Initial State Properties • Equations: – First Law: Q = ΔU + W = m(u2 – u1) + W – W = ∫path PdV – W = (P1 + P2)(V2 – V1)/2 for straight line path (area of a trapezoid) – Combine first law and work equation to get Q = m(u2 – u1) + (P1 + P2)(V2 – V1)/2 – V2 = mv2 and V1 = mv1 • We have to find v1, v2, P1, u1, and u2 • Use property tables for water 27 Final State, Work and Heat • Initial state: x1 = 50% and T1 = 200oC • x1 = 50% quality implies mixed region at T1 = 200oC • P1 = Psat(T1 = 200oC) = 1554.9 kPa • v1 = (1 – x1) vf(T1 = 200oC) + x1 vg(T1 = 200oC) = 0.06426 m3/kg • u1 = (1 – x1) uf(T1 = 200oC) + x1 ug(T1 = 200oC) = 1723.0 kJ/kg Saturation properties from Table A-4, page 916 28 Solving First Law Problems • At point 2, T2 = 400oC and P2 = 10 MPa • In superheat table, v2 = 0.026436 m3/kg and u2 = 2833.1 kJ/kg (page 922) P1 + P2 (V2 − V1 ) = P1 + P2 m(v2 − v1 ) = 1.5549 MPa + 10 MPa (10 kg ) • 2 2 2 ⎛⎜ 0.026436 m 3 − 0.06426 m3 ⎞⎟ 1 MJ = −0.4373 MJ kg kg ⎠ 1 MPa ⋅ m3 ⎝ W= 1 MJ Q = (10 kg )⎛⎜ 2833.1 kJ − 1723.0 kJ ⎞⎟ − 0.4373 MJ kg kg ⎠ 1000 kJ ⎝ • Q = 10.67 MJ (heat added) 29 ME 370 – Thermodynamics • Given: 110 kg of H2O at 200oC and 50% quality is compressed along a linear path, P = a + bV to 400oC and 10 MPa • Find: Heat Transfer • Given data: • Get moving boundary work as done previously: W = ∫pathPdV • May have other work terms • Find internal energy from tables • May find m = V/v from given data • Q = ΔU + W = m(ufinal – uinitial) + W • If m not known, q = Δu + w • Observe that u depends more on temperature, T, than pressure, P 30 5