CHE 205 – Chemical Process PrinciplesSection 3: F&R, Chapter 5 EQUATIONS OF STATE FOR GASES Questions A gas enters a reactor at a rate of 255 SCMH. What does that mean? An orifice meter mounted in a process gas line indicates a flow rate of 24 ft3/min. The gas temperature is 195oF and the pressure is 62 psig. The gas is a mixture containing 70 mole% CO and the balance H2. What is the mass flow rate of the hydrogen in the gas? A reactor feed stream consists of O2 flowing at 32 kg/s. The gas is to be compressed from 37oC and 2.8 atm absolute to 54oC and 284 atm. What are the volumetric flow rates at the inlet and outlet (needed to rate the compressor)? A pitot tube indicates that the velocity of a stack gas is 5.0 m/s at 175oC. The stack diameter is 4.0 m. A continuous stack analyzer indicates an SO2 level of 2500 ppm (2500 moles SO2/106 moles gas). At what rate in kg/s is SO2 being discharged into the atmosphere? A 70.0 m3 tank is rated at 2000 kPa. If 150 kg of helium is charged into the tank, what will the pressure be? How much more helium can be added before the rated pressure is attained? Answers: Need an equation of state: relationship between temperature (T), pressure (P), volume (V), and number of moles (n) of a gas. In Chapter 4, streams on flow charts labeled like this: 100 mol/s 0.600 mol A/mol 0.400 mol B/mol In this chapter, stream data just as likely to look like this: 250 L/s @ 37oC, 800 mm Hg pA = 420 mm Hg (partial pressure of A) For material balances, however, we still need moles and mole fractions. The job now becomes one of converting volumetric flow rates (or volumes) to molar flow rates (or moles), and (for gases) partial pressures to mole fractions. (Latter is easy: yA = pA/P ) Convert volumes to moles – Solids & liquids: use tabulated densities (volume to mass) & molecular weights (mass to moles). Mixtures—either look up mixture density data or assume volume additivity & calculate density from Eq. (5.1-1). – Gases, can’t use tabulated densities. (Why not?) Instead, need an equation of state (EOS) — a formula relating V, n, T, and P. Simplest is the ideal gas EOS. Read Section 5.1. We won’t lecture on it, but you need to know it. 3-1 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 Ideal Gas Equation of State (Sections 5.2a & 5.2b) (continuous) PV nRT (batch) or PV nRT or PVˆ RT where Vˆ V / n (or V / n ) is the specific molar volume of the gas T and P must be absolute temperature (K, oR) and absolute pressure (not gauge). R is the gas constant—values given on inside back cover of text. Convenient—applies regardless of what the gas is, & whether the gas has a single component or is a mixture. If you can’t assume ideal gas behavior (i.e., if gas is nonideal or real), Approximate—greatest validity at low gas densities (high T, low P), when gas molecules are far enough apart for intermolecular forces to be negligible (behave like billiard balls). Usually ok for temperatures at or above 0oC & pressures at or below 1 atm. Rule of thumb for when to use it given in Eqs. (5.2-3) on p. 192. Three will get you four. Given any three of the variables P,V (or V ), T , and n (or n ) , calculate the fourth one. Example The volumetric flow rate of a stream of propane at 150oC and 70.0 atm being fed to a combustion furnace is measured and found to be 29.0 m3/h. (a) Determine its molar flow rate in kmol/h. Solution. We are given three of the four gas law variables (_____, ______, and ______) and so can determine the fourth one (________). From the inside back cover, R 0.08206 (L atm)/(mol K) . 70.0 atm kmol C3 H8 PV n h RT = 58.5 29.0 m3 h mol K 0.08206 L atm kmol h (b) Now, suppose an analysis of the combustion chamber products shows that the molar flow rate of the propane was 101 kmol/h. Think of four possible reasons for the discrepancy between the two stated values of the molar flow rate. 1. _______________________________________________________________________ 2. _______________________________________________________________________ 3. _______________________________________________________________________ 4. _______________________________________________________________________ 3-2 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 (c) Check the assumption of ideal gas behavior. Solution: RT 0.08206 L atm mol K P __________ L , therefore from Eq. ____________, mol ideal gas behavior is a ___________ assumption. For another example, work through Problem 5.10 in the workbook. Standard temperature and pressure (STP): 0oC (273.16K, 491.67oR), 1 atm. At STP, 1 mol occupies 22.415 L, 1 kmol occupies 22.415 m3, & 1 lb-mole occupies 359.05 ft3. (Memorize) While our text (and many other thermodynamic texts) use 0oC and 1 atm as STP, there are some other ‘standards’. In this course, please use the values on p. 194. Suppose you are told that a gas flows at a rate of, say, 1280 SCFH [standard cubic feet per hour, or ft3(STP)/h] (a) It does not mean that the gas is at standard temperature and pressure. It does mean that if you brought it from whatever its temperature and pressure really are to 0oC and 1 atm, its volumetric flow rate would be 1280 ft3/h. (See Example 5.2-4.) (b) You can calculate the molar flow rate of the gas as n 1280 ft 3 (STP) h 1 lb-mole 3 359.05 ft (STP) 3.56 lb-mole h (c) If you know that the actual temperature of the gas is 120oC (393K) and 0.800 atm, you can calculate its actual flow rate as 1280 ft 3 (STP) 393 K 1 atm ft 3 V 2300 h 273 K 0.800 atm h (See Example 5.2-3.) 3-3 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 Ideal Gas Mixtures, Partial Pressures, and Volume Percentages (Section 5.2c) Suppose yA = mole fraction of a component A in a mixture of gases at pressure P and volume V. (Example: A mixture of gases at P = 1000 mm Hg with a volume of 200 liters contains 30 mole% CH4, 50 mole% C2H6, and 20 mole% C2H4.) Partial pressure of a component of a gas: pA = yAP Example: pCH4 0.30 1000 mm Hg = 300 mm Hg The partial pressures of all components of a mixture add up to the total pressure (prove it). Pure component volume and percentage by volume: The pure component volume of A is the volume A would occupy if it were by itself at the mixture temperature and pressure Pv A nA RT Divide v A nA yA PV nRT V n (volume fraction = mole fraction) The percentage by volume (% v/v) is 100 times the volume fraction. Thus, % v/v = mole% for an ideal gas mixture % v/v has no practical significance for a nonideal gas mixture Three alternative ways of telling you the value of a mole fraction MOLE FRACTIONS yA PARTIAL PRESSURE pA = yAP VOLUME FRACTION % v/v = mole% for an ideal gas mixture From now on, any of these specifications may be given in material balance problems. You should immediately convert the first two to mole fractions when you label the flow chart, and if you label the partial pressure on the flow chart also label the mole fraction and count pA = yAP as another equation in the degree-of-freedom analysis. 3-4 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 Exercise. Liquid acetone (C3H6O) is fed at a rate of 400 L/min into a heated chamber where it evaporates into a nitrogen stream which enters at 27oC and 475 mm Hg (gauge). The gas leaving the heater is diluted by another nitrogen stream flowing at a measured rate of 419 m3(STP) at 25oC and 2.5 atm. The combined gases are then compressed to a total pressure P = 6.3 atm (gauge) at a temperature of 325oC. The partial pressure of acetone in this stream is pa = 501 torr (501 mm Hg). Atmospheric pressure is 763 torr. (a) Write the complete set of equations you would solve to determine the molar composition of the product gas stream and the volumetric flow rate of the nitrogen entering the evaporator. (b) How is it possible for the second nitrogen stream to be at standard temperature and pressure and at 25oC and 2.5 atm? Solution. Verify that the flow chart is completely labeled. V1 (m 3 / min) n1 (mol N 2 / min) 27 o C, 475 torr Evaporator Compressor 400 L/min C3H6O (l) n2 (mol/min) n 4 (mol/ min) y4 (mol C3H6O(v)/mol) (1–y4) (mol N2/mol) 325oC, 6.3 atm (gauge) pa = 501 torr 419 m3/min (STP) N2/min n3 (mol N 2 / min) 25oC, 2.5 atm (a) Degree-of-freedom analysis: System equations: (b) _________________________________________________________________________________ 3-5 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 Nonideal (Real) Gases (Section 5.3) How would you measure P, V, T relationships in a laboratory? Would the relationships be different for each gas species? When you can’t assume __________ behavior (low ___, high ____, outside criteria of Eqs. 5.2-3), rather than create massive tables of P, V, T data for each gas species, need to find a universal ‘equation of state’ that incorporates properties of the gas species. Read p. 5-4 of the workbook. It turns out that different gases behave similarly at the same _______________________ Section 5.3a. Each gas has a unique value for Tc and Pc (Table B.1). Let’s review supercritical behavior. We will discuss three approaches for non-ideal equations of state: 1. Virial equations of state (Section 5.3b). 2. Cubic equations of state and the SRK equation (Section 5.3c). Commonly used for single species (for mixtures, use compressibility factor EOS). The SRK equation of state (Eq. 5.3-7) may be the most commonly used EOS other than the ideal gas equation. Easy or hard to use, depending on which of the three variables ( P,Vˆ , T ) is unknown. PVˆ BP (see p. 201 for how to calculate B) 1 RT RT Procedure: For given species, look up Tc , Pc , and the Pitzer acentric factor (Table 5.3-1 for selected species). Calculate a, b, and m from Eqs. 5.3-8, 5.3-9, & 5.3-10. If T and Vˆ are known, evaluate Tr from Eq. 5.3-11 and from Eq. 5.3-12, solve Eq. 5.3-7 for P. If T and P are known, enter Eq. 5.3-7 in Solver, enter all known values, and solve for Vˆ . (Alternatively, use Solver in Excel, as in Example 5.3-3.) If P and Vˆ are known, enter Eqs. 5.3-7, 5.3-11 for Tr, and 5.3-12 for into Excel’s Solver, enter all known values, and solve for T. 3-6 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 3. Compressibility factor equation of state (Section 5.4): PV = znRT , where z is the compressibility factor (a fudge factor — the farther it is from 1, the farther the gas is from ideal). Either you have tables where you can look up z for a given T and P (Section 5.4a, single species only), or you use the law of corresponding states (Section 5.4b) to estimate z. Procedure: Given two of the variables T, P, and Vˆ (or V and n or V and n ) 1. Look up Tc and Pc (e.g. in Table B.1). Apply Newton’s corrections for H2, He (p. 208). 2. Calculate two of the quantities reduced temperature. Tr = T/Tc , reduced pressure Pr = P/Pc , and Vˆ , depending on which two of the variables T, P, and Vˆ are ideal reduced volume, Vˆrideal RTc / Pc known. 3. Look up z on one of the generalized compressibility charts, Figs. 5.4-1 – 5.4-4. 4. Substitute known variables and z into the compressibility factor equation of state to determine the unknown variable. Example: Example 5.4-2, p. 209. Note: If the gas were anything other than nitrogen, then Z will change because the critical constants are different. For example, Z = 1.4 for hydrogen. Go through Test Yourself on p. 210. Kay’s rule: PVT calculations for nonideal gas mixtures using compressibility charts (Section 5.4c): Calculate pseudocritical temperature and pseudocritical pressure by weighting Tci and Pci by mole fractions of ith component (Eqs. 5.4-9 and 5.4-10), then proceed as for single component. This is the only method we will present in this book for doing PVT calculations on nonideal gas mixtures. 3-7 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 Exercise. A natural gas (85 mole% CH4, 15% C2H6) at 20oC and 80 atm is burned completely with 30% excess air. Vnatural gas 285 L/s. The stack gas emerges at 280oC and 1 atm. What is Vstack gas ? Solution. (a) Draw and label the flow chart. Furnace CH4 + 2O2 CO2 + 2H2O C2H6 + 7 2 O2 2CO2 + 3H2O (b) What equations of state should you use to relate the volumetric flow rates of the fuel and stack gases to their molar flow rates? Fuel gas: __________________________________________________ Stack gas: _________________________________________________ (c) Do the degree-of-freedom analysis (use atomic balances). 3-8 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 (d) Write out the seven equations required to solve for the unknowns, letting (Tcm , Pcm) and (Tce , Pce) = critical temperatures of methane and ethane, respectively. (You could find their values in Table B1 but don’t bother.) 3-9 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014) CHE 205 – Chemical Process Principles Section 3: F&R, Chapter 5 3-10 Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014)