1. A hydrocarbon is burnt with excess air. The Orsat analysis of the flue gas shows 10.81% CO2, 3.78% O2 and 85.40 N2. Calculate the atomic ratio of C:H in the hydrocarbon and the % excess air.

Calculations:

100 moles of dry flue gas (i.e., excepting H2O)The Orsat analysis shows the compositions of the flue gases by not taking into account of H2O.From the composition of air(mole %),

Here, nitrogen is the tie component.79 mole of N2 h 21 mole of O2Therefore, O2 that is entering the burner = 85.4 x 21/79 = 22.7C + O2 à CO21 mole of CO2 h 1 mole of O2 h 1 atom of C ( i.e.,1 mole of O2 reacts with 1 atom of C to produce 1 mole of CO2)

Therefore, O2 used up for reacting with carbon = 10.81 mole and,Carbon in the hydrocarbon = 10.81 atomsO2 reacted with Hydrogen in the hydrocarbon = 22.7 - (10.81 + 3.78) = 8.114H + O2 à 2H2O1 mole of O2 reacts with 4 atoms of hydrogen.Therefore, hydrogen in the hydrocarbon = 8.11 x 4 = 32.44 atoms.C:H ratio in the hydrocarbon = 1 : 32.44/10.81 = 1 : 3Theoretical air demand = air needed for complete conversion of carbon to carbon dioxide and hydrogen to water vapor% excess air = 100 x (actual air used - theoretical air demand) / theoretical air demand= 100 x (22.7 - (10.81 + 8.11))/( 10.81 + 8.11) = 20%

2. Natural gas containing 80% CH4, 15% C2H6 and 5% C3H8 is burnt with 50% excess air. Assuming that 90% of the carbon in the hydrocarbons are converted to CO2 and the rest to CO, determine

Flue gas analysisOrsat analysisCalculations:

Basis: 1 mole of natural gas

Component Mole Atoms of C Atoms of HCH4 0.80 0.8 x 1 = 0.8 0.8 x 4 = 3.2C2H6 0.15 0.15 x 2 = 0.3 0.15 x 6 = 0.9C3H8 0.05 0.05 x 3 = 0.15 0.05 x 8 = 0.4Total 1.0 1.25 4.5Reactions:

C + O2 à CO2 -- I

C + 1/2 O2 à CO -- II

H + 1/4 O2 à 1/2 H2O -- III

90% of Carbon is converted by reaction I, and 10% of carbon is converted by II.Amount of CO2 produced = 1.25 x 0.9 = 1.125 moleAmount of CO produced = 1.25 x 0.1 = 0.125 mole

Amount of H2O produced = 4.5 / 2 = 2.25 moleAmount of O2 used by hydrocarbon = O2 used by reactions I, II and III.= 1.125 + 0.125 x (1/2) + 4.5 x (1/4) = 2.3125 moleTheoretical O2 needed = Oxygen for complete conversion of C to CO2 and H to H2O.= 1.25 + 4.5 x (1/4) = 2.375 moleOxygen entering = 150% of theoretical = 1.5 x 2.375 = 3.5625 moleTherefore, nitrogen entering = 3.5625 x 79/21 = 13.4018 mole = N2 in the flue gasO2 in the flue gases = O2 entering - O2 used = 3.5625 - 2.3125 = 1.25 moleFlue gas analysis:

Component Moles Mole %CO2 1.125 6.20CO 0.125 0.69H2O 2.25 12.39O2 1.25 6.89N2 13.4018 73.83Total 18.1518 100Orsat analysis (Water free):

Component Moles Mole %CO2 1.125 7.07CO 0.125 0.79O2 1.25 7.86N2 13.4018 84.28Total 15.9018 100

Balance of Individual Reactions:Basis: 1 mole of Natural gas

CH4 0.80C2H6 0.15C3H8 0.05Total 1.0CH4 + 2 O2 à CO2 + 2 H2O -- 1

CH4 + 3/2 O2 à CO + 2 H2O -- 2

C2H6 + 7/2 O2 à 2 CO2 + 3 H2O -- 3

C2H6 + 5/2 O2 à 2 CO + 3 H2O -- 4

C3H8 + 5 O2 à 3 CO2 + 4 H2O -- 5

C3H8 + 7/2 O2 à 3 CO + 4 H2O -- 6

In the above reactions CO2 is produced from reactions 1, 3 and 5.Since 90% of Carbon is converted to CO2 and 10% to CO,CO2 produced = (1 x 0.8 + 2 x 0.15 + 3 x 0.05) x 0.9 = 1.125 moleSimilarly CO is obtained from reactions 2, 4 and 6.

CO produced = (1 x 0.8 + 2 x 0.15 + 3 x 0.05) x 0.1 = 0.125 moleH2O produced = (2 x 0.8 + 3 x 0.15 + 4 x 0.05) x 0.9 + (2 x 0.8 + 3 x 0.15 + 4 x 0.05) x 0.1 = 2.25 moleO2 used up in these reactions = (2 x 0.8 + 3.5 x 0.15 + 5 x 0.05) x 0.9 + (1.5 x 0.8 + 2.5 x 0.15 + 3.5 x 0.05) x 0.1= 2.3125 mole

Theoretical O2 needed = moles of O2 needed for Conversion of C to CO2 and H to H2O.= 2 x 0.8 + 3.5 x 0.15 + 5 x 0.05 = 2.375 moleO2 entering = 50 % excess = 150% of theoretical = 2.375 x 1.5 = 3.5625 moleN2 entering along with O2 in the air = 3.5625 x 79/21 = 13.4018 mole (sine air is 21% O2 and 79% N2 by volume).

O2 in the flue gas = O2 entering - O2 used up = 3.5625 - 2.3125 = 1.25 moleN2 in the flue gas = N2 entering = 13.4018 moleTherefore, for 1 mole of Natural gas entering, the flue gas coming out are:

Component MolesCO2 1.125CO 0.125H2O 2.25O2 1.25N2 13.4018Total 18.1518On comparing with the data obtained from atomic balance and balance of individual reactions, it can be seen that the results are same from either method. But instead of lengthy calculations for individual reactions, we can very well make use of atomic balances.

3. Air at 30oC and 150 kPa in a closed container is compressed and cooled. It is found that the first droplet of water condenses at 200 kPa and 15oC. Calculate the percent relative humidity of the original air. The vapor pressures of water at 15oC and 30oC are 1.7051 kPa and 4.246 kPa respectively.

Calculations:

At 15oC, and 200 kPa water is at 100% humidity.i.e., Percentage humidity or percentage absolute humidity = 100 x [pA(p - pS)]/[ pS(p - pA)] = 100

Therefore,pA/(p - pA) = pS/(p - pS)where pA = partial pressure of water vapor;pS = vapor pressure of water vaporp = total pressure of systemi.e., no of moles of water vapor per mole of dry air = 1.7051/(200 - 1.7051) = 0.0086

This ratio (moles of water vapor / mole of dry air) is not going to change for a closed system. Therefore, partial pressure of A at 30oC and 150 kPa is found as follows:

pA/(p - pA) = 0.0086pA/(150 - pA) = 0.0086pA + 0.0086 pA = 1.29pA = 1.29/1.0086 = 1.279 kPa

Percentage relative humidity of the original air:percentage relative humidity = 100 x (pA/pS) = 100 x 1.279/4.246 = 30.12%

4. Nitrogen from a cylinder is bubbled through acetone at 1.1 bar and 323 K at the rate of 2 x 10-4 m3/min. The nitrogen, saturated with acetone vapor leaves at 1.013 bar, 308 K at the rate of 3.83 x 10-4 m3/min. What is the vapor pressure of acetone at 308 K?

Calculations:

Using Ideal gas law, molal flow rate of nitrogen is calculated as follows:

Molal density of nitrogen rm = n/V = P/RT = 1.1 x 105/(8314 x 323) = 0.04096 kmol/m3Therefore, molal flow rate = 2 x 10-4 x 0.04096 = 8.1934 x 10-6 kmol/minSimilarly the molal density of leaving gases = 1.013 x 105/(8314 x 308) = 0.03956 kmol/m3and the molal flow rate of leaving gases = 3.83 x 10-4 x 0.03956 = 1.5151 x 10-5 kmol/minSince the molal rate of nitrogen is not going to change,molal flow rate of acetone in the leaving gases = 1.5151 x 10-5 - 8.1934 x 10-6 = 6.9578 x 10-6 kmol/minMole fraction of acetone in the leaving gases = 6.9578 x 10-6 / 1.5151 x 10-5 = 0.4592

And partial pressure of acetone in the leaving gases = 0.4592 x 1.013 = 0.4652 BarAt saturation conditions, partial pressure of vapor is equal to its vapor pressure. Therefore, vapor pressure of acetone at 308 K = 0.4652 Bar.

5. An evaporator is fed continuously with 25 kg/hr of a solution which contains 10% NaCl, 10% NaOH and 80% H2O. During evaporation, H2O is removed from the solution and NaCl precipitates as crystals which is settled and removed. The concentrated liquor leaving the evaporator contains 50% NaOH, 2% NaCl and 48% H2O. Calculate

(i)Weight of salt precipitated per hour. (6)(ii) Weight of concentrated liquor leaving per hour. (6)

Calculations:

Data:Flow rate of feed = 25 kg/hr

Composition of feed (F):NaCl 10%NaOH 10%H2O 80%

Composition of concentrated liquor (P):NaCl 2%NaOH 48%H2O 50%Balance for NaOH:NaOH entering the in feed = NaOH leaving in concentrated liquori.e., 0.1 x 25 = 0.50 PTherefore, the mass flow rate of concentrated liquor (P) = 0.1 x 25 / 0.50 = 5 kg/hrBalance for NaCl:NaCl entering in the feed = 25 x 0.1 = 2.5 kg/hrThe entering NaCl is coming out as crystals and in the concentrated liquor.NaCl leaving in the concentrated liquor = 0.02 P = 0.02 x 5 = 0.1 kg/hrTherefore,NaCl precipitated = 2.5 - 0.1 = 2.4 kg/hr

6. A solution of ethyl alcohol containing 8.6% alcohol by weight is fed at the rate of 5000 kg/hr to a continuous fractionating column operating at atmospheric pressure. The distillate which is the desired product contains 95.4% alcohol by weight and the residue from the bottom of the column contains 0.1% alcohol by weight. Calculate the following:

i. the mass flow rates of the distillate and residue in kg/hr, andii. the percentage loss of alcohol.

Calculations:

Overall material balance:Feed = Distillate + Residuei.e., F = D + RD + R = 5000 à 1Component balance for Alcohol:FxF = DxD + RxR

Where x 's are the mole fraction of alcohol in various streams.0.954 D + 0.001 R = 5000 x 0.086 = 430 à 2Equations 1 and 2 contains 2 unknowns D and R, and that can be solved as follows:Multiplying equation 1 by 0.001,0.001D + 0.001 R = 5 à 3subtracting equation 3 from equation 2,0.953 D = 425D = 446 kg/hri.e., Distillate flow rate = 446 kg/hrTherefore, R = 5000 - 446 = 4554 kg/hri.e., Residue flow rate = 4554 kg/hrEthanol leaving in the residue (i.e., ethanol lost) = 4554 x 0.001 = 4.554 kg/hrEthanol entering in the feed = 5000 x 0.086 = 430 kg/hrPercentage loss of ethanol = 100 x (4.554 / 430) = 1.06

7. 220 kg of CO2 gas at 27oC and 1 atm is compressed adiabatically to 1/5th of its volume. It is then cooled to its original temperature at constant volume. Find Q, DU and W for each step and for the entire process.

Formula:For an ideal gas, PV = nRTFor an adiabatic process, PVg = constant,V2/V1 = (P1/P2)(1/g )T2/T1 = (V1/V2)(g -1)Adiabatic work of compression, W = (P2V2 - P1V1) / (g - 1) è 1Q + W = DUDU = mCV(T2 - T1) è 2Calculations:Here there are two steps taking place.

1. Adiabatic compression, and2. Constant volume heat removal.

n = 220 / Molecular weight of CO2 = 220/44 = 5 kmolV1 = 5 x 8314 x (273 + 27) / (1.01325 x 105) = 123.08 m3.V2 = (1/5) x 123.08 = 24.616 m3.g = 1.3 (for CO2: data)T2 = 300 x (5)0.3 = 486.2 KP1/P2 = (V2/V1)gP2 = 1 / (1/5) g

i.e., P2 = 8.103 atm.Adiabatic compression work W can be calculated from equn.1 or equn.2 (since DU = Q + W; and Q = 0 for an adiabatic process).W = (8.103 x 1.01325 x 105 x 24.616 - 1.01325 x 105 x 123.08) / 0.3= 25.7985 MJSince the temperature of the system is returned to its original state, the internal energy change of the total process is zero. Therefore, DU for the heat removal step is = -25.7985 MJ.For the constant volume cooling step, W = 0, and DU = -25.7985 MJ. Therefore, Q for this step = -25.7985 MJ.Summary:

  Q W DU

Step 1 0 25.7985 MJ 25.7985 MJ

Step 2 -25.7985 MJ 0 -25.7985 MJ

Over all-25.7985 MJ (heat removed from the system)

25.7985 MJ (work added to the system)

0

8. A vertical cylinder with a freely floating piston contains 0.1 kg air at 1.2 bar and a small electric resistor. The resistor is wired to an external 12 Volt battery. When a current of 1.5 Amps is passed through the resistor for 90 sec, the piston sweeps a volume of 0.01 m 3. Assume (i) piston and the cylinder are insulated and (ii) air behaves as an ideal gas with Cv = 700 J/(kg.K). Find the rise in temperature of air.

Calculations:Energy added by electric resistor = V I t = 12 x 1.5 x 90 = 1620 JWork done = -ò PdV = -P(V2 - V1) = -1.2 x 105 x DV = 1.2 x 105 x 0.01 = 1200 JDU = Q + W = 1620 - 1200 = 420 J (energy added to the system and work added to the system are taken as positive.)DU = mCVDTDT = 420/(0.1 x 700) = 6 K.

9. A 28 liter rigid enclosure contains air at 140 kPa and 20oC. Heat is added to the container until the pressure reaches 345 kPa. Calculate the heat added.

Formulae:For an ideal gas, PV = nRTFrom first law of thermodynamics,Q + W = DUFor an ideal gas DU is a function of temperature only, and work done at constant volume = 0Therefore, Q = DU.And, DU = mCV(T2 - T1)m = n x Molecular weightCalculations:n = PV / (RT) = 140 x 1000 x 0.028 / (8314 x 293) = 1.609 x 10-3 kmolm = 1.609 x 10-3 x 29 = 0.0467 kg.For constant volume system, P a T.Therefore, P1/T1 = P2/T2

140/293 = 345/T2

T2 = 722 K.CV = 0.718 kJ/kg.oC (data, for air at 20oC)Q = DU = 0.0467 x 0.718 x (722 - 293) = 14.385 kJi.e., Heat added = 14.385 kJ

10. A thermodynamic system undergoes a cycle composed of a series of three processes for which Q1 = +10 kJ, Q2 = +30 kJ, Q3 = -5 kJ. For the first process, DE = +20 kJ, and for the third process, DE = -20 kJ. What is the work in the second process, and the net work output of the cycle?

Calculations:Q + W = DEWork done in the first process = 20 - 10 = 10 kJ (i.e., work is done on the system)Work done in the third process = -20 - (-5) = -15 kJFor a cyclic process, the overall internal energy change is zero. (i.e., DE = 0)Therefore, DE in the second process = (0 - (20 - 20)) = 0 kJTherefore, work done in the second process = 0 - 30 kJ = -30 kJ.Total work done during the cycle = 10 + (-15) + (-30) = -35 kJ (i.e., 35 kJ of work is done by the system).

Total work done can also be calculated as follows:Total Q = Q1 + Q2 + Q3 = 10 + 30 - 5 = 35 kJTherefore, net work done during the cycle = 0 - 35 kJ = -35 kJ (the negative sign indicates that work is done by the system).

11. Air is compressed from 2 atm absolute and 28oC to 6 atm absolute and 28oC by heating at constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and DE and DH of the air.

Calculations:PV = nRTWe have to rise the pressure to 6 atm by heating at constant volume.At constant volume, P a T.Therefore, T2/T1 = P2/P1

T2 = (6/2) x (273 +28) = 903 KFor a constant volume process, W = 0 and DE = QCV = 0.718 kJ/kg.oC (for air; data)Q = mCV(T2 - T1) = 0.718 x (903 - 301) = 432.24 kJ/kg. DE = 432.24 kJ/kgFor cooling at constant pressure, heat removed = mCP(T2 - T1)= 1.005 x (903 - 301) = 605.01 kJ/kg (Cp - Cv = R for ideal gas)Internal energy change for the total process consisting the above two steps is zero (since internal energy is a function of temperature alone).Work done on the gas during constant pressure cooling = 605.01 -  432.24 = 172.77 kJ/kg. Summary:

  Heat required Work required DE DH

Constant Volume heat addition 432.24 kJ/kg   432.24 kJ/kg 605.01 kJ/kg

Constant Pressure heat removal -605.01 kJ/kg 172.77 kJ/kg -432.24 kJ/kg -605.01 kJ/kg

Over All -172.77 kJ/kg 172.77 kJ/kg 0 0

12. A tank having a volume of 0.1 m3 contains air at 14 MPa and 50oC. It is connected through a valve to a larger tank having a volume of 15 m3, which is completely evacuated. The entire assembly is completely insulated. The valve is opened and the gas allowed to come to equilibrium in both tanks. Calculate the final pressure.

Calculations:For this process, Q = 0 and W = 0. Therefore, DU = 0.DU = mCvDT = 0.Therefore, DT = 0.For an ideal gas, PV = nRTSince T is constant, P1V1 = P2V2

14 x 0.1 = P2 x (0.1 + 15)i.e., final pressureP2 = 92.715 KPa.

13. Liquid water at 25oC flows in a straight horizontal pipe, in which there is no exchange of either heat or work with the surroundings. Its velocity is 12 m/s in a pipe with an i.d. of 2.5 cm until it flows into a section where the pipe diameter increases to 7.5 cm. What is the temperature change?

Calculations:For the steady flow process, the first law is written asDH + Du2/2 + gDz = Q + Ws

since there is no shaft work, Ws = 0

and flow is horizontal, Dz = 0Therefore,DH + Du2/2 = Qand since there is no heat transfer, Q = 0Therefore, DH + Du2/2 = 0Applying continuity equations,u1A1r1 = u2A2r2

for water r1 = r2

Therefore,u2/u1 = A1/A2 = d1

2/d22 = 2.52/7.52 = 0.1111

u2 = 12 x 0.1111 = 1.333 m/s(u2

2 - u12) / 2 = (1.3332 - 122) / 2 = -71.11

Therefore, DH = 71.11 J/kgEnthalpy change per kg mass = 71.11 JWe know, DH = mCpDT and, m = 1000 g; Cp = 4.184 J/g.oCTherefore,DT = 71.11/(1000 x 4.184) = 0.017oCTemperature change = 0.017oC

14. An existing process consists of two steps:(a) One mole of air T1 = 900 K and P1 = 3 bar is cooled at constant volume to T2 = 300 K.(b) The air is then heated at constant pressure until its temperature reaches 900 K.It is proposed to replace this two-step process by a single isothermal expansion of the air from 900 K and 3 bar to some final pressure P. What is the value of P that makes the work of the proposed process equal to that of the existing process? Assume mechanical reversibility and treat air as an ideal gas with Cp = 7R/2 and Cv = 5R/2Calculations:Work in the original process consisting of two steps:(a) For the constant volume process:W = 0Pressure at the end of the constant volume process (P2):P1V1 = RT1

V1 = R x 900 / 3 = 300RP2V2 = RT2

Since V1 = V2,P2 = R x 300 / (300R) = 1 bar(b) For the constant pressure process:P2 = 1 bar; T2 = 300 K; V2 = 300RP3 = 1 bar; T3 = 900 KP3V3 = RT3

V3 = R x 900 / 1 = 900RW = ò PdV = P2(V3 - V2) = 1 x (900R - 300R) = 600RTotal work done in the two step process = 600RFor the single step isothermal process:W = 600R = RT ln (P1/P)i.e., 600R = R x 900 x ln (3/P)ln(3/P2) = 600/900 = 2/3e(2/3) = 3/P1.9477 = 3/PP = 3/1.9477 = 1.54 bar

15. One kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1 bar until its volume triples. Calculate W, Q, DU, and DH for the process. Assume air obeys the relation PV/T = 83.14 bar cm3 mol-1 K-1 and that Cp = 29 J mol-1 K-1.

Calculations:From the relation, PV/T = 83.14 bar cm3 mol-1 K-1

Initial molal volume = 83.14 x 300 /1 = 24942 cm3/mol= 24.942 m3/kmoli.e. Initial volume of 1 kg of air (V1) = 24.942 m3 / 29 = 0.86 m3

Final volume (V2)= 3 x V1 = 3 x 0.86 = 2.58 m3 = 2.58 x 106 cm3.Final specific volume = 24942 x 3 = 74826 cm3/molFinal temperature:P2 x 74826 = T2 x 83.141 x 74826 = T2 x 83.14 T2 = 900 KFor the constant pressure heating process,W = -P(V2 - V1) = -1 x (2.58 x 106 - 0.86 x 106) bar cm3 = -1 x 105 x (2.58 - 0.86) Nm = -172000 J = -172 kJ. (work is done by the system by expansion)Moles of air in 1 kg mass = 1000/29 = 34.483 molQ = mCpDT = 34.483 x 29 x (900 - 300) = 600000 J = 600 kJ.DU = Q + W = 600 - 172 = 428 kJ (heat given to the system and work done on the system are positive quantities)DH = Q = 600 kJSummary:W -172 kJQ 600 kJDU 428 kJDH 600 kJ

16. Liquid water at 100oC and 1 bar has an internal energy (on an arbitrary scale) of 419 kJ/kg and a specific volume of 1.044 cm3/g.

(a) What is its enthalpy?(b) The water is brought to the vapor state at 200oC and 800 kPa, where its enthalpy is 2838.6 kJ/kg and its specific volume is 260.79 cm3/g. Calculate DU and DH for the process.Calculations:(a) Enthalpy = U + PV= 419 + 100 kPa x 1.044 x 10-3 m3/kg = 419 + 0.1044 = 419.1044 kJ/kg(b)Internal energy of water vapor at 200oC and 800 kPa = H - PV= 2838.6 - PV = 2838.6 - 800 x 260.79 x 10-3 = 2838.6 - 208.632 = 2629.968 kJ/kgDU = 2629.968 - 419 = 2211.968 kJ/kgDH = 2838.6 - 419.1044 = 2419.4956 kJ/kg

17. The decomposition of NO2 follows a second order rate equation. Data at different temperatures are as follows:

T (K) 592 603 627 651.5 656k (cm3/gmol.sec) 522 755 1700 4020 5030Compute the energy of activation Energy from the data. The reaction is 2NO2 à 2NO + O2

Calculations:Activation energy is found from the Arrhenius' relationk = ko e-E/RT

i.e.,ln k vs. 1/T is a straight line with a slope of -E/Rwhere, E is the activation energy; andR is the universal gas constant.T (K) 592 603 627 651.5 656

k (cm3/gmol.sec) 522 755 1700 4020 5030

1/T (oK-1) 0.001689 0.001658 0.001595 0.001535 0.001524

ln k 6.257668 6.626718 7.438384 8.299037 8.523175From the above data, the following graph is drawn, and the slope is = -13622

Therefore,E/R = 13622E = 13622 x 8314 = 113253.3 kJ/kmol

18. The thermal decomposition of nitrous oxide (N2O) in the gas phase at 1030oK is studied in a constant volume vessel at various initial pressures of N2O. The half-life data so obtained are as follows:

po (mm Hg) 52.5 139 290 360

t1/2 (sec) 860 470 255 212

2N2O à 2N2 + O2

Determine the rate equation that fits the data.Calculations:t1/2 Vs. CAO in a log-log plot is a straight line with a slope of 1-nCAo = pAo/RT = po/RT (for an ideal gas)ln(CAo) Vs. ln(t1/2) in a linear XY plot gives a straight line, and the slope of the line is 1 - n

from the above figure,1-n = -0.7469therefore, n = 1 + 0.7469 = 1.7469 = 1.75i.e., Order of the reaction (n) = 1.75t1/2 and CAo are related by the formula:

for a value of po = 290, and t1/2 = 255,CAo = (290/760) x 1.01325 x 10-5/(8314 x 1030) = 4.515 x 10-3 gmol/lit.therefore,k = (21.75 - 1 - 1) x (4.515 x 10-3)(1-1.75) / (255 x (1.75 - 1)) = 0.2046 sec-1.(gmol/lit) -0.75

the rate expression for the given reaction is,-rN2O = 0.2046 (C N2O)1.75

19. Establish the kinetics of the thermal decomposition of Nitrous oxide from the following data and find the reaction rate at 990oC and an initial pressure of 200 mm Hg.

t, min 20 53 100

% decomposition 23 50 73Calculations:Conversion (XA)of Nitrous oxide (A) for various time values are given are given.We shall test the date for first order, second order kinetics by integral method of analysis.For a first order reaction, -ln(1 - XA) vs. t is a straight line.t, min 20 53 100

XA 0.23 0.50 0.73 t, min 20 53 100

XA 0.23 0.5 0.73

1-XA 0.77 0.5 0.27

-LN(1 - XA) 0.261365 0.693147 1.309333From the above data the following graph is drawn:

 For a second order reaction,XA/(CAo (1- XA) ) = kti.e., XA/(1 - XA) vs. t is a straight linet, min 20 53 100

XA/(1 - XA) 0.298701 1 2.703704From the above data the following graph is drawn: By comparing the two graphs, it can be seen that First order kinetics is well fiiting the given data. Fisrt order rate constant k is obtained from the slope of the first order data graph, and it is = 0.0131 sec-1.Therefore, -rA = kCA = kCAo(1 - XA) = kpAo(1 - XA)/RT = (1 - XA) x 0.0131 x (200/760) x 101325 /(8314 x (990 + 273)) = 3.327 x 10-5(1 - XA)i.e., -rA = 3.327 x 10-5(1 - XA)where XA is fractional decomposition of Nitrous oxide, and -rA is the rate of decomposition of Nitrous oxide.

20. The gas phase decomposition of A takes place according to the irreversible reaction, A à 3P. The kinetics of the reaction was studied by measuring the increase in pressure in a constant volume reaction vessel. At 504oC and an initial pressure of 312 mm Hg, the following data were obtained:

Time (Sec) 390 777 1195 3155 aTotal pressure (mm Hg) 408 488 562 779 931

a. Test for a first order reaction.b. Calculate the value of the specific reaction rate at 504oC

Calculations:Total pressure of the system (p) and partial pressure of A (pA)are related as:pA = pAo - (a/Dn)(p - po)where, po is the initial pressure of the system;pAo is the initial pressure of A;a is the stoichiometric coefficient of A;and Dn = stoichiometric coefficient of products minus reactants.For the given problem, po = pAo = 312 mm Hg (Assuming feed is pure A)a = 1Dn = 3-1 = 2pA = 312 - (1/2)(p - 312)i.e.,pA = 312 - 0.5 (p - 312)pA for various times are calculated as follows:Time (Sec) 390 777 1195 3155 aTotal pressure (mm Hg) 408 488 562 779 931pA (mm Hg) 264 224 187 78.5 2.5Fractional conversion of A is given by:XA = (pAo - pA) / pAo

For various time values, XA and -ln(1- XA) is calculated and tabulated as follows:Time (Sec) 390 777 1195 3155 apA (mm Hg) 264 224 187 78.5 2.5XA 0.154 0.282 0.401 0.748 0.992-ln(1 - XA) 0.167 0.331 0.512 1.378 4.828For a first order reaction, a plot of t Vs. -ln(1 - XA) in a linear XY graph is a straight line with a slope of k.From the following plot, it is seen that the given data is for a first order reaction.

21. Acetic acid is hydrolysed in three stirred tank reactors operated in series. The feed flows to the first reactor (V = 1 lit) at a rate of 400 cm3/min. The second and third reactors have volumes of 2 and 1.5 litres respectively. The first order irreversible rate constant is 0.158 min-1. Calculate the fraction hydrolysed in the effluent from the third reactor.

Calculations:The design equation for series, steady flow mixed reactor isVi =FAo (XA,i - XA,i-1) / (-rA)i

Where Vi = volume of reactor iFAo = molal flow rate of A into the first reactorXA,i = fractional conversion of A in the reactor iXA,i-1 = fractional conversion of A in the reactor i-1For first order reaction, -rA,i = kCA,i = kCAo(1 - XA,i)v = volumetric flow rate of A = 400 cm3/min = 0.4 lit/minFor the first reactor: (V = 1 lit)(-rA)1 = (kCA)1 = k CA,1 = k CAo ( 1- XA,1)CAo= FAo / vi.e., FAo = v CAo

XA,i-1 = XA,0 = 0Therefore,Vi =FAo (XA,i - XA,i-1) / (-rA)i

1 = 0.4 (XA,1 - 0) / (0.158 x ( 1 - XA,1 ) )XA,1 = 0.283For the second reactor: (V = 2 lit)(-rA)2 = (kCA)2 = k CA,2 = k CAo ( 1- XA,2)Therefore,(-rA)2 = k CAo ( 1- XA,2)XA,1 = 0.283FAo = v CAo

Vi =FAo (XA,i - XA,i-1) / (-rA)i

2 = 0.4 (XA,2 - 0.283) / ( k ( 1- XA,2) )5 k = (XA,2 - 0.283) / ( 1- XA,2)0.79 - 0.79 XA,2 = XA,2 - 0.2831.073 = 1.79 XA,2

XA,2 = 0.60For the third reactor: (V = 1.5 lit)(-rA)3= (kCA)3 = k CA,3 = k CAo ( 1- XA,3)XA,2 = 0.6FAo = v CAo

Vi =FAo (XA,i - XA,i-1) / (-rA)i

1.5 = 0.4 (XA,3 - 0.60) / ( k ( 1- XA,3) )0.5925 = (XA,3 - 0.60) / ( 1- XA,3)0.5925 - 0.5925 XA,3 = XA,3 - 0.601.1925 = 1.5925 XA,3

XA,3 = 0.749The fraction hydrolyzed in the effluent from the third reactor = 0.749

22. The decomposition of phosphine is irreversible and first order at 650oC.4PH3(g) à P4(g) + 6H2(g)The rate constant (sec-1) is reported aslog k = (-18993/T ) + 2 log T + 12.13     (log to the base 10)where T is in oK.In a closed vessel (constant volume) initially containing phosphine at 1 atm pressure, what will be the pressure after 50, 100 and 500 sec. The temperature is maintained at 650oC.Calculations:Total pressure of the system (p) and partial pressure of A (pA)are related as:

pA = pAo - (a/Dn)(p - po)If phosphine is taken as A, thenpAo = po = 1 atma = stoichiometric coefficient of A = 4Dn = stoichiometric coefficient of products - stoichiometric coefficient of reactants= 1 + 6 - 4 = 3Therefore,pA = 1 - 4/3 (p - 1),4/3 (p - 1) = 1 - pA

i.e.,p = 0.75 (1 - pA) + 1Rate constant k, at 650oC (= 650 + 273 = 923 K)is found from the relationship for k and Tlog k = (-18993/T ) + 2 log T + 12.13log k = -18993/923 + 2 log 923 + 12.13log k = -2.517k = 10-2.517

Therefore, k = 3.04 x 10-3 sec-1.For the first order equation,-ln (CA/CAo) = k tor-ln (pA/pAo) = ktpA/pAo = e-kt

pA = pAo e-kt = 1 x exp(-3.04 x 10-3 t) = exp(-3.04 x 10-3 t)for various time values, pA is calculated and tabulated as follows:t, sec 50 100 500pA, atm 0.860 0.738 0.219 From the relationship of total pressure(p) and partial pressure of A (pA),p = 0.75 (1 - pA) + 1total pressure for various time values are calculated as:t, sec 50 100 500pA, atm 0.860 0.738 0.219p, atm 1.105 1.197 1.586

23. The kinetics of an aqueous phase decomposition of A is investigated in two CSTR's in series, the first having half the volume of the second. At steady state with a feed concentration of 4 gmol/lit., and mean residence time of 65 sec in the second reactor, the concentration of feed to second from the first is found to be 2 gmol/lit., while that of the stream leaving the second is 1 gmol/lit. Find the suitable kinetic rate expression.

Calculations:Data:Residence time of second reactor = 65 sect1 = V1/v = (C0 - C1)/ (-r)1

t2 = V2/v = (C1 - C2)/ (-r)2

Conversions based on the feed to the reactors = 50% = 0.5(4 to 2 gmol/lit in first reactor, and 2 to 1 gmol/lit in second reactor)Test for 1st order:for first order reaction, -rA = k CAo(1 - XA) andwhere, CAo = concentration of A in the feed to the reactor; andXA = fractional conversion of A in the reactorfor the second reactor (-r)2 = k 2 x 0.5 = ksince t2 = 65 sec, and C1 - C2 = 2 - 1 = 1 gmol/liti.e., 65 = 1/k à 1for the first reactor, V1= V2/2and t1 = 65/2 = 32.5 sec.for the first reactor (-r)1 = k 4 x 0.5 = 2k

i.e., 32.5 = (4 - 2)/2k = 1/k à 2from equn.1 and 2 it is seen that 1st order mechanism is not suitable.Test for 2nd order:for 2nd order reaction, -rA = k CAo

2(1 - XA)2 andfor the second reactor (-r)2 = k x 22 x 0.52 = kt2 = V2/v = (C1 - C2)/ (-r)2

i.e.,65 = (2 - 1)/k = 1/k65 = 1/k à 3for the first reactor (-r)1 = k x 42 x 0.52 = 4kt1 = V1/v = (C0 - C1)/ (-r)1

32.5 = (4 - 2)/4k = 1/2ki.e.,65 = 1/k à 4On comparing equn.3 and 4 it could be seen that the given data is for a second order reaction.k = 1/65 lit/(gmol.sec) = 0.0154 lit/(gmol.sec)and, -rA = 0.0154 CA

2 gmol/(lit.sec)

24. Water at 20oC (viscosity = 1 cp) flows through a smooth straight pipe A of inside diameter 4 cm at an average velocity of 50 cm/sec. Oil flows through another pipe B of inside diameter 10 cm. Assuming similarities, calculate the velocity of oil through pipe B. Specific gravity of oil is 0.8 and its viscosity is 2 cp.

Data:A:Dia of A = 4 cmVelocity of fluid in A = 50 cm/secDensity of fluid in A = 1 g/ccViscosity of fluid in A = 1 cpB:Dia of B = 10 cmDensity of fluid in B = 0.8 g/ccViscosity of fluid in A = 2 cpFormulae:For dynamic similarity, ratio of forces to be equalInertial forceA / Inertial forceB = Viscous forceA / Viscous forceB

i.e. NReA = NReB

Calculations:NReA = NReB

Therefore,4 x 50 x 1 / 1 = 10 x VB x 0.8 / 2200 = 4 VB

VB = 200/4 = 50 cm/secVelocity of oil through B = 50 cm/sec

25. A geometrically similar model of an air duct is built 1:30 scale and tested with water which is 50 times more viscous and 800 times more dense than air. When tested under dynamically similar conditions, the pressure drop is 2.25 atm in the model. Find the corresponding pressure drop in the full-scale prototype.

Data:Dimensions of the model = 1/30 times the dimensions of prototypeViscosity of fluid in the model = 50 times the viscosity of fluid in prototypeDensity of fluid in the model = 800 times the density of the fluid in prototypeFormulae:For dynamic similarity, ratio of forces to be equalInertial forcem / Inertial forcep = Pressure forcem / Pressure focep

= Viscous forcem / Viscous forcep à 1Pressure force = (2frLv2 / D)D2 (i.e. = frictional pressure drop x area) à 2From 1,NRem = NRep

 Calculations:NRem = NRep

1 x vm x 800 / 50 = 30 x vp x 1 / 1vm = 1.875 vp

Pressure force ratio = (2fm x 800 x 1 x vm2 / 1) x 12 / ((2fp x 1 x 30 x (vm / 1.875)2 / 30) x 302)

= 1600fmvm2 / 512fpvm

2

=3.125fm/fp

f = 0.079 (Dvr/m)-0.25

fm/fp = (1 x vm x 800 / 50)-0.25 / (30 x (vm/1.875) x 1 / 1)-0.25 = 16-0.25/16-0.25 = 1pressure ratio = 3.125 x 302 / 12 = 2812.5 ( i.e. pressure = pressure force/area)Pressure drop of the prototype = 2.25/2812.5 atm = 0.0008 atm = 81.06 N/m2

26. A U-tube manometer filled with mercury is connected between two points in a pipeline. If the manometer reading is 26 mm of Hg, calculate the pressure difference between the points when (a) water is flowing through the pipe (b) air at atmospheric pressure and 20oC is flowing in the pipe. Density of mercury = 13.6 gm/cc Density of water = 1 gm/cc Molecular weight of air = 28.8

Data:Manometer reading(h) = 26 mm Hg = 0.026 m HgDensity of mercury(rm) = 13.6 gm/ccDensity of water = 1 gm/ccMolecular weight of air(MW) = 28.8Temperature of air = 20 o C = 293 KR = 8314 J/(kmol.K)Formulae:For simple U - tube manometer,P1 - P2 = Dp = (rm - r)gh.Ideal Gas lawPV = n RTMolal density = n/V = P/(RT)Mass density(r) = Molecular weight x molal densityCalculations:(a) Water is flowing through the pipe:Dp = (rm - r)gh = (13600 - 1000) x 9.812 x 0.026 = 3214.4 N/m2

(b) Air at atmospheric pressure and 20oC is flowing in the pipe:r = 28.8 x 101325/(8314 x 293) = 1.2 kg/m3

Dp = (rm - r)gh = (13600 - 1.2) x 9.812 x 0.026 = 3469.2 N/m2

27. 2.16 m3/h water at 320 K is pumped through a 40 mm I.D. pipe through a length of 150 m in a horizontal direction and up through a vertical height of 12 m. In the pipe there are fittings equivalent to 260 pipe diameters. What power must be supplied to the pump if it is 60% efficient? Take the value of fanning friction factor as 0.008. Water viscosity is 0.65 cp, and density = 1 gm/cc.

Data:Flow rate(Q) = 2.16 m3/h = 2.16/3600 m3/sec = 0.0006 m3/secDia of pipe(D) = 40 mm = 0.04 mLength of pipe in horizontal direction = 150 mLength of pipe in vertical direction(Dz) = 12 mEquivalent length of fittings = 260 pipe diametersFriction factor(f) = 0.008Efficiency of pump(h ) = 0.6Viscosity of fluid(m) = 0.65 cp = 0.00065 kg/(m.sec)

Density of fluid(r) = 1 gm/cc = 1000 kg/m3

Formulae:1. Bernoulli's equation

2. Frictional losses per unit mass of flowing fluid

3. Power required for pumping = (mass flow rate x head developed by pump)/ h= (volumetric flow rate x pressure developed by pump)/ hCalculations:Length of pipe with fittings = 150 + 12 + 260 x 0.04 m = 172.4 mVelocity = volumetric flow rate/flow cross sectional area= 0.0006/((p/4) x 0.042) = 0.477 m/sechf = 2 x 0.008 x 172.4 x 0.4772 / 0.04 = 15.69 m2/sec2

frictional losses per unit weight of fluid(h) = hf / g = 15.69/9.812 = 1.6 mpump head(q) = Dz + h = 12 + 1.6 = 13.6 mpressure developed by pump = 13.6 x 1000 x 9.812 = 133443.2 N/m2

power required for pumping = 0.0006 x 133443.2 / 0.6 = 133.4 watt = 133.4/736 HP = 0.181 HP

28. Brine of specific gravity 1.2 is flowing through a 10 cm I.D. pipeline at a maximum flow rate of 1200 liters/min. A sharp edged orifice connected to a simple U-tube mercury manometer is to be installed for the purpose of measurements. The maximum reading of the manometer is limited to 40 cm. Assuming the orifice coefficient to be 0.62, calculate the size of the orifice required.

Data:Density of brine (r) = 1.2 x 1000 kg/m3 = 1200 kg/m3

Dia of pipe (Da) = 10 cm = 0.1 mMaximum flow rate (Q) = 1200 liters/min = 1.2 m3/min = 0.02 m3/secMaximum manometer reading (hm) = 40 cm = 0.4 m of HgDensity of manometric fluid (rm) = 13600 kg/m3

Orifice coefficient (Co) = 0.62Formulae:Velocity at the orifice

Where b = Db/Da

Db = dia of orificeDa = dia of pipeFor the U-tube manometer,Pa - Pb = (rm - r)ghm

Calculations:Pa - Pb = (13600 - 1200) x 9.812 x 0.4 = 48667.5 N/m2

The quantity (1 - b4) will be approximately 1.Therefore,Vb = 0.62 x ( 2 x 48667.5 / 1200 )0.5 = 5.584 m/secCross sectional area of orifice = volumetric flow rate / velocity at the orifice = 0.02 / 5.584 = 0.00358 m2

Dia of orifice (Db) = (0.00358 x 4/p)0.5 = 0.0675 m = 6.75 cm

29. A Newtonian fluid having a viscosity of 1.23 poise, and a density of 0.893 gm/cm3, is flowing through a straight, circular pipe having an inside diameter of 5 cm. A pitot tube is installed on the pipeline with its impact tube located at the center of the pipe cross section. At a certain flow rate, the pitot tube indicates a reading of 8 cm of mercury. Determine the volumetric flow rate of the fluid.

Data:Viscosity of fluid (m) = 1.23 poise = 0.123 kg/(m.sec)Density of fluid (r) = 0.893 gm/cm3 = 893 kg/m3

Dia of pipe (D) = 5 cm = 0.05 mManometer reading (hm) = 8 cm of Hg = 0.08 m of HgFormulae:Velocity at the center of pipeVo = ( 2 x (Ps - Po ) / r )0.5 = ( 2 x hm x (rm - r) x g / r )0.5

Calculations:Vo = ( 2 x 0.08 x (13.6 - 0.893) x 9.812 / 0.893 )0.5 = 4.73 m/secNRe = D x Vo x r / m = 0.05 x 4.73 x 893 / 0.123 = 1717Since the flow is laminar (NRe < 2100), the average velocity is given byVavg = Vo / 2 = 4.73 / 2 = 2.365 m/sec (velocity at the centerline is the maximum velocity)Volumetric flow rate = Vavg x p x D2 / 4 = 2.365 x p x 0.052 / 4 = 0.00464 m3/sec = 4.64 lit/sec

30. A rotameter calibrated for metering has a scale ranging from 0.014 m3/min to 0.14 m3/min. It is intended to use this meter for metering a gas of density 1.3 kg/m3 with in a flow range of 0.028 m3/min to 0.28 m3/min. What should be the density of the new float if the original one has a density of 1900 kg/m3? Both the floats can be assumed to have the same volume and shape.

Data:Old:Density of float (rf)= 1900 kg/m3

Density of gas (r) = 1.3 kg/m3

Qmin = 0.014 m3/minNew:Density of gas (r) = 1.3 kg/m3

Qmin = 0.028 m3/minFormula:

Where Q = volumetric flow rateA2 = area of annulus (area between the pipe and the float)A1 = area of pipeAf = area of floatVf = volume of floatCD = rotameter coefficientCalculations:From the equation, for the same float area and float volume and the pipe geometry,Q = k (rf - r)0.5

where k is a proportionality constantQnew / Qold = ( (rfnew - r) / (rfold - r) )0.5

i.e. 0.028 / 0 .014 = ( (rfnew - 1.3) / (1900 - 1.3) )0.5

2 = (rfnew - 1.3)0.5 / 43.574(rfnew - 1.3)0.5 = 87.15(rfnew - 1.3) = 7595.1rfnew = 7596 kg/m3