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Energy

Energy. Kinetic Energy (E k ). Nomenclature: “ ˙ ” above means “rate” “ ˆ ” above means “per mole or per unit mass”. Potential Energy ( E p ). Energy due to position in a potential field (gravity / magnetic) or confirmation relative to equilibrium (eg, spring). Examples.

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Energy

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  1. Energy

  2. Kinetic Energy (Ek) Nomenclature: “ ˙ ” above means “rate” “ ˆ ” above means “per mole or per unit mass”

  3. Potential Energy (Ep) Energy due to position in a potential field (gravity / magnetic) or confirmation relative to equilibrium (eg, spring)

  4. Examples • Water flows into a process unit through a 2-cm inner diameter(ID) pipe at a rate of 2 m3/hr. Calculate Ek for this stream (J/s). • Crude oil is pumped at a rate of 15.0 kg/s from a well 220 m deep to a storage tank 20 m above ground level. Calculate the rate at which potential energy increases (J/s).

  5. Energy Transfer • Exchange of fluid across system boundary • Heat (Q) across system boundary due to DT from Thigh to T low (positive Q from surroundings to system) • Work (W) across system boundary due to force, torque, voltage, etc. (positive W from surroundings to system)

  6. Dimensions and Units Energy [=] force * distance [=] M/L / t2 *L = M*L2/ t2 Energy [=] N*m = J, dyne * cm = erg, ft lbf

  7. Extra Practice Problems Handout Problem Set: IV-1 – IV-8

  8. Physical Properties • Extensive Properties = ƒ(amount of material) examples: mass, volume, Ek, Ep • Intensive Properties ≠ ƒ(amount of material) examples: T, p,r • Specific Property: an intensive property obtained by dividing an extensive property by an amount of material example: specific volume is volume per unit mass example: Ek / m = Êk example: Ĥ  = Û + p V, where  Ĥ is enthalpy Û accounts for thermal effects on molecular motion, accounts for both thermal and pressure effects. Ĥ accounts for both thermal and pressure effects

  9. State Variables vs Path Functions Consider a property l that is a function of x and y. l(x,y) is a point function if & only if the finite Dl (or differential dl) is independent of path. • that is, Dl = lf – li ... change in l is obtained by difference • in this case, l is a point function or state variable If l is not a point function or state variable - it is a path function or path variable Consider the figure below showing a person starting at initial Point 1 and later at final Point 2 Consider path A and path B position does not depend on the path taken; that is, position is apoint function

  10. • position change is calculated by Dl = lf – li • for a round trip ∫dl = lf – li = 0 • the work required to move from 1 to 2 depends on the path (A or B); that is, work is apath function • for a round trip ∫dl = lf – li ≠ 0and depends on path • all intensive properties are state variables (point functions) • two important energy balance path functions are Q & W 2 Y B A 1 X

  11. Internal Energy (U) Û = Û(T,V) ... Specific Internal Energy ... molecular motion • Û is a point function because it only depends on the amount of energy at time of determination • DÛ can be calculated using DÛ = ∫Cv(T)dT where Cv is the heat capacity at constant volume • For ideal gas: this equation is accurate because Û is not a function of V • For solids & liquids: this equation is a good approximation because Û is not a strong function of V • For real gases: this equation is valid only if V is constant • Note: only differences in Û can be calculated

  12. Enthalpy (Ĥ)  Ĥ= Û + pV , where p is absolute pressure • Ĥ is a point function = combination of two point functions - Ĥ =  Ĥ (T,p) • Let Cp .... heat capacity at constant pressure • DĤ  can be calculated using the following - DĤ = ∫Cp(T)dT • for ideal gas: this equation is accurate because H is not a function of p • for real gases: this equation is valid only if p is constant • for solids & liquids: add D(V p) to RHS of equation • Note: only differences in  Ĥ can be calculated, must be a reference state where Ĥ = 0

  13. Heat Capacities Heat capacity definition - amount of heat /energy required to raise a unit of mass by one degree • Cp & Cv [=] J /mole °C [=] Btu / lbm °F ... functions of Temperature • ideal gas: Cp = Cv + R, where R = Gas Constant. • real gas: see Figure 23.4 • liquids and solids: Cv ~ Cp • Cp0 (J / gmol DK) → Cp0 = a + b(T) + c(T)2 + d(T)3 where a, b, c, d given in Appendix E • for mixtures: Cp' = ∑ yi Cpi & Cv' = ∑ yi Cvi • to obtain DH  for any process use DH  = ∫Cp(T) dT

  14. Example Determine the enthalpy change for CO2 gas cooled from 300°C to 40°C, Compare the calculated value to the value obtained using Figure 23.4 at the average temperature of 170 °C. a = 36.11 X 10-3 b = 4.233 X 10-5 c = -2.887 X10-8 d = 7.464X10-12

  15. Enthalpy Tables Since only DĤ  can be calculated, define reference condition (T, p, state of aggregation) at this reference condition set   Ĥ = 0 then tabulated value is H  from this condition. This is, establishing a relative scale just like Celsius and Fahrenheit did for T Different texts & different handbooks use different reference conditions so the tabulated values differ, but DĤ  is the same regardless of source of data but you can not mix data from sources with different reference points.

  16. Steam Tables

  17. P-H Diagram

  18. Example Steam at 150 psia (absolute) with 350°F of superheat is fed to a turbine at a rate of 2000 lbm/hr. The turbine operation is adiabatic, and the effluent is saturated vapor at 15 psia. Calculate the rate of enthalpy change in the turbine (kilowatts).

  19. Extra Practice Problems Problem Handout Set: IV-15 – IV-49

  20. Phase Change Operations When changing phases, bonds are broken in the first phase and reformed in the second phase • if breaking old bonds requires more energy than is released by forming new bonds, then energy must be transferred into the system • energy transfer in the form of heat • for a closed system: DU + DEk + DEp = Q + W • adding heat to system causes internal energy to increase but does not cause the temperature to increase • use DĤ  rather than DÛ • since DĤ  = DÛ + D(pV ) balance is now Q = DĤ  – D(pV)

  21. Phase Change Operations • if the phase change is liquid ↔ vapor / gas. • assumed the gas is ideal, then Q = DĤ  – RDT • if phase change is liquid to solid at constant T & p • then Q = DĤ  ... heat causes a change in specific enthalpy Consider three important phase changes • for each change, DĤ  = strong ƒ(T) & weak ƒ(p) • Specific enthalpy change of VAPORIZATION: DĤvap • liquid →vaporat constant T & p • Specific enthalpy change of CONDENSATION: DĤcond • vapor → liquidat constant T & p • DĤcond = – DĤ vap ...... usually tabulate as DĤvap

  22. Phase Change Operations • Specific enthalpy change of MELTING: DĤm • solid → liquid at constant T & p • Specific enthalpy change of FREEZING: DĤfreeze • liquid →solid at constant T & p • DĤfreeze = – DĤm • Specific enthalpy change of SUBLIMATION: DHs • solid → vapor at constant T & p • each of these three quantities is the amount of energy required to change the phase of one unit mass or mole at constant T & p • Also known as latent heat of .... from the Latin for “hidden” • heat is added but T does not increase • each of the DĤ change = ƒ(T) • we usually refer to the normal boiling point or normal melting / freezing point .which occurs at p = 1 atm

  23. Phase Change Data

  24. Phase Change Data

  25. Phase Change Data

  26. State Properties and Hypothetical Paths Even when you cannot find tabulated enthalpies for a substance, you will often find data that allows you to calculate enthalpy changes associated with the following: • change in pressure at constant T • change in T at constant p • change in phase at a specific T & p (usually 1 atm) • to determine DH  for a process, use a hypothetical process path for which you can calculate DH  for each step.

  27. Example If you are given DĤm (T1,1 atm) and you wish to calculate DĤm ( T2 , P atm), make use of hypothetical process path: (solid, T2, p atm) (liquid, T2,p atm) DĤm (T2) ∫ CpdT ∫ CpdT (solid, T1,1 atm) DĤm (T1) (liquid, T1,1 atm)

  28. 10 Minute Problem Calculate DĤ  (J / gmol) for the process in which ice at –5°C and 1 atm is converted to steam at 300°C and 5 atm. Super Heated Steam T= 300 °C P = 5 atm Saturated Water T= 100 °C P = 1 atm DĤ Steam Tables DĤ  = ∫ Cp dT Ice T= 0 °C P = 1 atm Subcooled Water T= 0 °C P = 1 atm DĤfusion Data: Cp(Ice) = 23.7 J/gmol DK Cp (water) = 75.4 J/gmol DK DĤ  = ∫ Cp dT Ice T= -5 °C P = 1 atm

  29. Closed Systems Integral Energy Balance Summation of energy changes between two points in time (initial and final) Accumulation = final – initital final = Uf + Ekf +Epf initial = Ui + Eki + Epi transfer = Q + W First Law of Thermodynamics for Closed System: DU + DEk + DEp = Q + W • If T surroundings = T system, Q = 0 • If well insulated or adiabatic, Q = 0 • If no change in T, phase, or composition, DU = 0 • W is done against a resistance • W = 0 if no moving parts (shaft work)

  30. Internal Energy (U) Û = Û(T,V) ... Specific Internal Energy ... molecular motion • Û is a point function because it only depends on the amount of energy at time of determination • DÛ can be calculated using DÛ = ∫Cv(T)dT where Cv is the heat capacity at constant volume • For ideal gas: this equation is accurate because Û is not a function of V • For solids & liquids: this equation is a good approximation because Û is not a strong function of V • For real gases: this equation is valid only if V is constant • Note: only differences in Û can be calculated

  31. Example A gas is contained in a cylinder fitted with a movable piston. The initial gas temperature is 25°C. The cylinder is placed in boiling water with the piston held in a fixed position. Heat in the amount of 2 kcal is absorbed by the gas, which equilibrates at 100°C (and a higher pressure). The piston is then released, and the gas does 100 J of work in moving the piston to its new equilibrium position. The final gas temperature is 100°C. Write an energy balance equation for each of the two stages of this process, and in each case solve for the unknown energy term in the equation. In solving this problem, consider the gas in the cylinder to be the system, neglect the change in potential energy of the gas as the piston moves vertically, and assume the gas behaves ideally. Express all energies in joules.

  32. Extra Practice Problems Handout Problem Set IV-9 - IV-14

  33. Open (Continuous Flow) Systems

  34. Balances On Open Systems • Continuous Process (heating water flowing in pipe) – continuous flowing streams; thus, use flow rates – Differential balance = analysis at an instant in time • Consider W as the sum of two types of work: Ws + Wflow • Ws = Shaft Work or Mechanical Work • Due to moving parts within system (turbine, propeller) • Due to movement of system boundary (piston)

  35. • Wflow = Flow Work = pV Work due to fluid acting on adjacent fluid: in – out • Consider pipe full of flowing fluid due to Dp ... p1 > p2 • Volume of system = length (L) l area (A) • force new fluid volume (dashed box) into the system L • likewise Wout = poutVout • thus, Wf = piVi – poVo • for multiple inlets and outlets Wf = ∑ piVi – ∑ poVo • since • multiple inlets and outlets A

  36. • Wtotal = Wshaft + Wflow = Wshaft + ∑ pi mi Vi – ∑ pomoVo Do energy balance on the system • Accumulation = In – Out + Generated – Consumed • at steady state balance yields: In = Out • isolating “in”/“positive” on LHS ... ∑Ei + Q + W = ∑Eo • rearranging gives: Q + W = ∑Eo – ∑Ei • substituting for W gives: Q + Wshaft + ∑pimiVi – ∑pomoVo = ∑m(Ûo + Êko + Êpo ) – ∑m (Ûi + Êki + Êpi ) • rearranging gives Q + Wshaft = ∑mo (Ûo+ poVo + Êko + Êpo ) –∑mi (Ûi+ piVi + Êki + Êpi) • recall   Ĥ = Û + pV ; • therefore, Q + Wshaft = ∑mo ( Ĥo + Êko + Êpo ) - ∑mi ( Ĥi + Êki + Êpi)

  37. First Law of Thermodynamics Open Systems DH + DEk + DEp = Q + Wshaft • if Tsurround = Tsystem, Q = 0 • if well insulated or adiabatic, Q = 0 • Wshaft = zero if there are no moving parts

  38. Enthalpy (Ĥ)  Ĥ= Û + pV , where p is absolute pressure • Ĥ is a point function = combination of two point functions - Ĥ =  Ĥ (T,p) • Let Cp .... heat capacity at constant pressure • DĤ  can be calculated using the following - DĤ = ∫Cp(T)dT • for ideal gas: this equation is accurate because H is not a function of p • for real gases: this equation is valid only if p is constant • for solids & liquids: add D(V p) to RHS of equation • Note: only differences in  Ĥ can be calculated, must be a reference state where Ĥ = 0

  39. Examples 500 kg of water vapor per hour drive a turbine. The steam enters at 44 atm and 450°C at a velocity of 60 m/s. It leaves 5 m lower at atmospheric pressure and 360 m/s. The turbine delivers W = 700 kW and there is a heat loss of 10,000 kcal/hr. Calculate the change in specific enthalpy in kJ/kg. Two streams of liquid water (one 120 kg/min @ 30°C and the other 175 kg/min @ 65°C) are mixed to form the feed to a boiler operating at 17 bars absolute. The exiting steam emerges from the boiler through a 6-cm ID pipe. Calculate the required heat input to the boiler in kilojoules/minute if the emerging steam is saturated at the boiler pressure. Neglect the kinetic energies of the liquid inlet streams.

  40. 10 Minute Problem Saturated steam at 1 atm is discharged from a turbine at a rate of 1000 kg/hr. Superheated steam at 300°C and 1 atm is needed as a feed to a heat exchanger; to produce it, the turbine discharge stream is mixed with superheated steam available from a second source at 400°C and 1 atm. The mixing unit operates adiabatically. Calculate the rate of steam (kg H2O(v)/hr) at 300°C produced, and the required volumetric flow rate (m3 H2O(v)/hr) of the 400°C steam

  41. Extra Practice Problems Problem Set Handout: IV-50 – IV-104

  42. Energy Balances and Chemical Reactions

  43. Enthalpy (Heat) of Reaction Define: DHrxn (T,p) = ∑Hproducts – ∑Hreactants • true if and only if (i) feed is in stoichiometric quantities (ii) the reaction is complete ... review (iii) feed and products are at the same T,p At low to moderate pressure, DHrxn ≠ƒ(p) • DHrxn < 0 .... means product has less enthalpy than reactants (referred to as exothermic reaction at constant T) • DHrxn > 0 .... product has more enthalpy than reactants (referred to as endothermic reaction at constant T)

  44. Enthalpy (Heat) of Reaction Consider the following reaction: a0R0 + a1R1 + a2R2 + .... → b1P1 + b2P2 + b3P3 + ... Divide by a0 so that one coefficient is unity yields: R0 + c1R1 + c2R2 + .... → g1P1 + g2P2 + g3P3 + Specific enthalpy of reaction (in terms of R0 reacted) • DHrxn (kJ or kcal/ mole R0) = ∑(gjHP,j) – ∑ (cHR,i +  HR,o) • specific enthalpy of rxn in terms of Rior Pj DH’rxn (kJ or kcal/ mole Rior Pj ) = DH rxn (kJ or kcal/ mole R0) ci or gi

  45. Examples Example The reaction 2A + 3B → 4C is carried out in a continuous reactor. Heat flowing out of the reactor is 50 kJ. Open System Energy Balance: DH + DEk + DEp = Q + Wshaft Q = – 50 kJ DHrxn (kJ/mole A reacted) = –50/2 = –25 DH’rxn (kJ/mole B reacted) = –50/3 DH’’rxn (kJ/mole C produced) = –50/4 Example: 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O .... DH = –5756 kJ DHrxn (kJ/mole C4H10 reacted) = –2878 DH’rxn (kJ/mole O2 reacted) = –442.8 DH’’rxn (kJ/mole CO2 produced) = –719.5 DH’’’rxn  (kJ/mole H2O produced) = –575.6

  46. Standard Heats of Reaction Rather than tabulate DHrxn at every possible T,p combination we use : Standard Heat of Reaction(DH°rxn  )  @ 25°C & 1 atm • superscript ° indicates “standard T and p” If a reaction is at some other pressure: • assume at low to moderate pressure, DHrxn  ≠ƒ(p) If a reaction is at some other temperature, • use hypothetical path because H is a state variable • DHrxn = ƒ(state of aggregation) Example: A(g) + B(g) → C(g) + D(g) ...... DHrxn,1 A(g) + B(g) → C(g) + D(liq) ...... DHrxn,2 DHrxn,1 ≠ DHrxn,2they differ by the amount of heat required to evaporate D

  47. Hess’s Law If a series of reactions (1, 2, 3) can be rearranged/adjusted by a series of algebraic operations to yield the desired reaction equation, then the heat (enthalpy) of the desired reaction can be obtained by performing the same algebraic operations to DH°rxn,1, DH°rxn,2,DH°rxn,3

  48. Example 2 C(s) + O2 (g) → 2 CO(g) Determine DH°rxn ? .... can we do this experimentally ? No cannot be carried out at 25°C & 1 atm At higher temperatures: C + O2 → CO2 or a mix of CO2 & CO, but never pure CO Hypothetical path with reactions for which we know or can experimentally measure DH°rxn,1 From tables of DH°rxn or experiments we can get: 1) C(s) + O2 (g) → CO2 (g) ..... DH°rxn,1  = –393.51 kJ/mole C 2) CO(g) + ½ O2→CO2 (g) ..... DH°rxn,2  = –282.99 kJ/mole CO

  49. DH°rxn C + ½ O2 CO DH°rxn,1 -DH°rxn,2 CO2 • Following the hypothetical path gives: • DH°rxn= DH°rxn,1 + (- DH°rxn,2 )= –393.51 + 282.99 = – 110.52 • Same result by subtracting equation 2 from equation 1 to get 3 • C + O2 – CO – ½ O2 → CO2 – CO2 • • which gives C + ½ O2 → CO and DH°rxn= DH°rxn,1 + (- DH°rxn,2 )

  50. Example The standard heats of the following combustion reactions have been determined: (a.) C2H6 + 7/2 O2 → 2 CO2 + 3H2O : DH°rxn,a= –1559.8 kJ/mole (b.) C + O2 → CO2 : DH°rxn,b= –393.5 kJ/mole (c.) H2+ ½ O2 → H2O : DH°rxn,c= –285.8 kJ/mole Use Hess’s law and the given heats of reaction to determine the standard heat of the reaction: (d.) 2C + 3 H2 → C2H6 : DH°rxn,d= ?

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