CHEE 221: Chemical Processes and Systems Module 4. Energy Balances without Reaction Part c: Calculating Changes in Enthalpy (Felder & Rousseau Ch 8) Energy Balances. F&R Chapter 8 How do we calculate enthalpy (and internal energy) changes when we don’t have tabulated data (e.g., steam tables) for the process species? Basic procedures to calculate enthalpy (and internal energy changes) associated with the following processes are covered in Chapter 8 (no Reaction): Remember, 3 pieces of information set the thermodynamic state of matter: P, T and Phase. • Hˆ with change in P (at constant T and phase) (F&R 8.2) • Hˆ with change in T (at constant P and phase) (F&R 8.3) • Hˆ with change in phase (at constant T and P) (F&R 8.4) To keep track of our calculations, we will summarize enthalpies in an Inlet‐Outlet Enthalpy Table. Sections not covered in Ch 8 include 8.3c, 8.3e, 8.4b, 8.4d, 8.4e, 8.5 CHEE 221 2 Hypothetical Process Paths To calculate enthalpy changes, we need to construct a hypothetical process path, with: –A starting point: your defined reference state (phase, T and P) –An end point: the conditions of the stream of interest (inlet or outlet) SinceUˆ andHˆ are state properties (values are dependent only on the state of the species (phase, T and P) and not how they got there), any convenient process path from a reference state to a process state can be chosen The ideal process path will allow you to make use of: –sensible heats heat capacities (Table B2) – phase transition temperatures Tm, Tb (at which data are often listed, e.g. latent heats) (Table B1) –latent heats heat of vapourization, heat of melting (Table B1) CHEE 221 F&R Ch 8.1b 3 Hypothetical Process Paths Calculate the enthalpy change as 1 kg of ice at 0C is transformed to superheated steam at 400C at 10 bar. ˆ H2O(s) H ? H2O(v) 0C, 1 atm 400C, 10 atm We could use the steam tables and find (Remember: final‐initial): Hˆ 3264 (334) 3612 kJ/kg - 334 kJ/kg is the enthalpy of ice How would we calculate the enthalpy change if we didn’t have the steam tables? CHEE 221 4 Construction of the Hypothetical Process Path ˆ Our chosen H2O(s) H ? H2O(v) reference state (0C, 1 atm) true path (400C, 10 atm) Hˆ Hˆ at the 1 m ˆ normal melting point Hˆ Hˆ H5 0 1 5 ( A reasonable assumption (Tm ) of H2O (Table B.1) for small changes in P) H O H2O(l) 2 (v) (0C, 1 atm) (400C, 1 atm) 100 400 H 2O(l ) Hˆ C dT ˆ H 2O(v) 2 p Hˆ Hˆ H 4 C p dT 0 2 4 100 (C expressions are p ˆ (C p expressions are found in Table B.2) H3 found in Table B.2) H2O(l) H2O(v) (100C, 1 atm) (100C, 1 atm) ˆ ˆ H3 H v at the normal boiling point (Tb ) of H2O (Table B.1) ˆ ˆ ˆ ˆ ˆ ˆ ˆ H H final 0 H1 H 2 H3 H 4 H5 CHEE 221 5 Changes in H and U with P (constant T, phase) Ideal Gases: –By definition, Uˆ 0(molecules don’t interact, so changing P doesn’t change the internal energy) – H = U + PV but PV = nRT = 0 (constant T) Hˆ 0 Non‐Ideal Gases: – Changes in internal energy and enthalpy are small, provided P is small (< 5 atm) Uˆ Hˆ 0 –For steam, use tabulated values In a problem, state that changes in U and H with Liquids and Solids: respect to pressure are – Uˆ 0 small and will be –Hˆ PVˆ (but still very small) neglected (except for steam tables). CHEE 221 F&R Ch 8.2 6 Does H change with P? Steam Enthalpy Diagram Use the steam tables to determine the state (liquid, vapour or mixture of the two; saturated or supersaturated) and approximate temperature (no need to use extrapolate) of 1 kg of water at 1 bar with the following enthalpies (relative to liquid water at the triple point) a) 100 kJ b) 419 kJ c) 1500 kJ d) 2676 kJ e) 3000 kJ Conclusion: Enthalpy is not a strong function of pressure below 10 bar CHEE 221 7 Phase Changes (at constant T and P): Latent Heat Phase changes occur from the solid to the liquid phase, and from the liquid to the gas phase, and the reverse. The specific enthalpy change (heat) associated with the phase change at constant T and P is known as the latent heat of the phase change (i.e., latent heat of vapourization or simply heat of vapourization). ˆ H v (T, P) : liquid gas ˆ Hvapourization H m (T, P) : solid liquid Table B.1 reports these two latent heats for substances Hmelting at their normal melting and boiling points (i.e., at a pressure of 1 atm). CHEE 221 F&R Ch 8.4a 8 Changes in U and H with T (constant P, phase): Sensible Heat Sensible heat refers to heat that must be transferred to raise or lower the temperature of a substance without change in phase. 1) Sensible heat of solid, H (Tinitial Tmelting) Hvapourization 2) Sensible heat of liquid, H (Tmelting Tvapourization) Hmelting 3) Sensible heat of gas, H (Tvapourization Tfinal) Tinitial Tfinal CHEE 221 F&R Ch 8.3a 9 Changes with T (constant P, phase): Sensible Heat Closed System‐‐Find U . The quantity of sensible heat required to produce a temperature change in a system can be determined from the appropriate form of the first law of thermodynamics: Q = U (closed system; must be kept at constant volume) ˆ ˆ U U Cv (T ) lim T 0 T T V T2 ˆ U Cv (T)dT T1 Ideal gas: exact Solid or liquid: good approximation Slope = Cv = heat capacity at constant volume Nonideal gas: valid only if V constant CHEE 221 10 Changes with T (constant P, phase): Sensible Heat Open System‐‐Find H. Enthalpy, like internal energy, also depends strongly on temperature. Q˙ = H˙ (open system; calculate at constant pressure) Ideal gas: exact ˆ ˆ H H C p (T ) lim Solid or liquid: good approximation T 0 T T P Nonideal gas: exact only if P constant T2 Hˆ C (T )dT p Cp = heat capacity at constant pressure T1 Liquids and Solids: Cp Cv Ideal Gases: Cp = Cv + R CHEE 221 11 Heat Capacity Formulas Heat capacity – the amount of heat required to raise the temperature of one mole or one gram of a substance by one degree Celsius without change in phase units: J cal or mol K g C If Cp were constant, our job would be easy: H = CP (T2‐T1) But, heat capacities are functions of temperature and are expressed in polynomial form: 2 3 Cp = a + bT + cT + dT (Form “1”) or, ‐2 Cp = a + bT+ cT (Form “2”) Values of coefficients a, b, c, and d are given in Table B.2. CHEE 221 F&R Ch 8.3b 12 Heat Capacity Calculations –Integration kJ 2 3 C p mol C a bT cT dT T2 ˆ kJ 2 3 H a bT cT dT dT mol T1 b c d a(T T ) (T 2 T 2 ) (T 3 T 3 ) (T 4 T 4 ) 2 1 2 2 1 3 2 1 4 2 1 CHEE 221 13 Notes Regarding Table B.2 Be sure you use the correct functional form 2 3 ‐2 – Cp = a + bT + cT + dT (Form 1) or Cp = a + bT+ cT (Form 2) Temperature units are sometimes K and sometimes C Positive exponent in table heading means you use negative exponent in the expression – E.g., if a x 103 = 123.0 a = 123.0 x 10‐3 2 2 2 Be careful when you integrate! (T2 –T1 ) (T2 –T1) CHEE 221 14 Specific Enthalpies of Gases –Table B.8 Table B.8 lists specific enthalpies (kJ/mol) of selected gases (mainly combustion products, i.e. this table might be useful when solving energy balances for combustion problems) as a function of temperature. •Can be used to estimate H changes as an alternative to integrating the Cp equation. • Interpolation may be required. •The reference state of these gases is: 1 atm and 25C. •Use this table as you would for the steam tables, however, note that for H2O, the units and reference state are different than the steam tables. CHEE 221 F&R Ch 8.3b 15 Putting it all together: Hypothetical Process Path ˆ reference H2O(s) H ? H2O(v) state (0C, 1 atm) true path (400C, 10 atm) Hˆ Hˆ at the 1 m ˆ normal melting point Hˆ Hˆ H5 0 1 5 ( A reasonable assumption (Tm ) of H2O (Table B.1) for small changes in P) H O H2O(l) 2 (v) (0C, 1 atm) (400C, 1 atm) 100 400 H 2O(l ) Hˆ C dT ˆ H 2O(v) 2 p Hˆ Hˆ H 4 C p dT 0 2 4 100 (C expressions are p ˆ (C p expressions are found in Table B.2) H3 found in Table B.2) H2O(l) H2O(v) (100C, 1 atm) (100C, 1 atm) ˆ ˆ H3 H v at the normal boiling point (Tb ) of H2O (Table B.1) ˆ ˆ ˆ ˆ ˆ ˆ ˆ H H final 0 H1 H 2 H3 H 4 H5 CHEE 221 16 Calculating enthalpy changes: Examples 1.