Day 1 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12)
13) 14) 15)
16) 17)
18) 19)
20)
Number of carbon atoms in olvanil Who discovered x-ray? Amino acids in human protein Definition of ozonolysis Given dimensions of a peso bill (length & width). Assuming ½ billion of gold atoms can be spread across its length, determine total mass. Freezing point depression Boiling point rise Evaluate given structural formulas (paraffins) Center of an atom Balancing chemical reactions Given density of Sulfuric acid determine its molarity Given a high-rating octane fuel -mileage (mi/gal) -density -distance covered Calculate CO2 produced Blanket of pollutants Pollutant produced from coal and petroleum Nearest layer of atmosphere from earth’s surface a) Troposphere b) Stratosphere c) Mesosphere d) Thermosphere pH problems (buffer) pH problem: Given NH4OH soln initially at 30 mL diluted to 400 mL then titrated with HCl. Compute pH a) at the start b) at the equiv point c) after addition of 40 mL of HCl Ksp problems (2) Metabolic reactions For 12 mold of ATP produced from every mole CH3COO- there are 7.4 kcal. For 1 gram of CH3COO- determine Edman Degradation (sequencing of amino acids)
Day 2 Problem 1.7 Coulson and Richardson A cyclone separator, 0.3 m in diameter and 1.2 m long, has a circular inlet 75 mm in diameter and an outlet of the same size. If the gas enters at a velocity of 1.5 m/s, at what particle size will the theoretical cut occur? The viscosity of air is 0.018 mN s/m2, the density of air is 1.3 kg/m3 and the density of the particles is 2700 kg/m3. Using the data provided: cross-sectional area at the gas inlet, Ai = (π/4)(0.075)2 = 4.42 × 10−3 m2 gas outlet diameter, d0 = 0.075 m gas density, ρ = 1.30 kg/m3 height of separator, Z = 1.2 m, separator diameter, dt = 0.3 m. Thus: mass flow of gas, G = (1.5 × 4.42 × 10 −3 × 1.30) = 8.62 × 10−3 kg/s The terminal velocity of the smallest particle retained by the separator, u0 = 0.2Ai2d0ρg/(πZdtG)
(equation 1.54)
or: u0 = [0.2 × (4.42 × 10−3)2 × 0.075 × 1.3 × 9.81]/[π × 1.2 × 0.3 × 8.62 × 10 −3] = 3.83 × 10−4 m/s Use is now made of Stokes’ law (Chapter 3) to find the particle diameter, as follows: u0 = d2g(ρs − ρ)/18μ
(equation 3.24)
or: d = [u0 × 18μ/g(ρs − ρ)]0.5 = [(3.83 × 10−4 × 18 × 0.018 × 10−3)/(9.81(2700 − 1.30))]0.5 = 2.17 × 10−6 m or 2.17 μm Note tas may tanong pa na ano daw yung particle size na papasok sa cylone at yung mareretain sa bottom something parang ganun.
PROBLEM 6.14 Coulson and Richardson Glass spheres are fluidised by water at a velocity equal to one half of their terminal falling velocities. Calculate:
(a) voidage (b) the density of the fluidised bed, (c) the pressure gradient in the bed attributable to the presence of the particles. The particles are 2 mm in diameter and have a density of 2500 kg/m3. The density and viscosity of water are 1000 kg/m3 and 1 mNs/m2 respectively. (sa original na tanong given naman yung galileo no.) Solution The Galileo number is given by: Ga = d3ρ(ρs − ρ)g/μ2 = [(2 × 10−3)3 × 1000(2500 − 1000) × 9.81]/(1 × 10−3)2 = 117,720 From equation 5.79: (4.8 − n)/(n − 2.4) = 0.043Ga0.57 = 0.043 × 117,7200.57 = 33.4 and: n = 2.47 u/u0 = 0.5 = e2.47 and hence: e = 0.755 The bed density is given by: (1 − e)ρs + eρ = (1 − 0.755) × 2500 + (0.755 × 1000) = 1367 kg/m3 The pressure gradient due to the solids is given by: {[(1 − e)ρs + eρ] − ρ}g = (1 − e)(ρs − ρ)g = (1 − 0.755)(2500 − 1000)9.81 = 3605 (N/m2)/m
PROBLEM 14.13 (Coulson and Richardson) A liquid with no appreciable elevation of boiling-point is concentrated in a triple-effect evaporator. If the temperature of the steam to the first effect is 395 K and vacuum is applied to the third effect so that the boiling-point is 325 K, what are the approximate boiling-points in the three effects? The overall transfer coefficients may be taken as 3.1, 2.3, and 1.1 kW/m2 K in the three effects respectively. Solution For equal thermal loads in each effect, that is Q1 = Q2 = Q3, then: U1A1ΔT1 = U2A2ΔT2 = U3A3ΔT3
(equation 14.7)
or, for equal areas in each effect: U1ΔT1 = U2ΔT2 = U3ΔT3
(equation 14.8)
In this case: 3.1ΔT1 = 2.3ΔT2 = 1.1ΔT3 Thus: Δ T1 = 0.742ΔT2 and ΔT3 = 2.091ΔT2 ΣΔT = ΔT1 +Δ T2 + ΔT3 = (395 − 325) = 70 deg K Thus: 0.742ΔT2 + ΔT2 + 2.091ΔT2 = 70 deg K and ΔT2 = 18.3 deg K and: ΔT1 = 13.5 deg K, ΔT3 = 38.2 deg K The temperatures in each effect are therefore: T1 = (395 − 13.5) = 381.5 K T2 = (381.5 − 18.3) = 363.2 K, and T3 = (363.2 − 38.2) = 325 K Example 13.5 (Coulson and Richardson) In order to extract acetic acid from dilute aqueous solution with isopropyl ether, the two immiscible phases are passed countercurrently through a packed column 3 m in length and 75 mm in diameter. It is found that, if 0.5 kg/m2s of the pure ether is used to extract 0.25 kg/m2s of 4.0 per cent acid bymass, then the ether phase leaves the column with a concentration of 1.0 per cent acid by mass. Calculate: (a) the number of overall transfer units based on the raffinate phase, and (b) the overall extraction coefficient based on the raffinate phase.
(pero ang tanong sa min eh, what daw ung conc. Ng acid sa raffinate, logarithmic driving force and no. of transfer unit) The equilibrium relationship is given by: (walang given na ganito) (kg acid/kg isopropyl ether) = 0.3 (kg acid/kg water).
Note: Sa board exam di sinabi kung ilan yung retention na 2 kg insoluble matter/1kg solution kenneth eto lang yun naaala ko pero may mga ganito pa, 1.
super heated steam daw tas papasok sa nozzle, tapos may given na P1 P2 at T1, ang tinatanong eh enthalpy at temp T2. (Over-all Energy Balance Problem)
2.
may lumabas ding fluid flow, sabi ni hero nasa mccabe daw xa, kaso nagulat xa kc iba ung sagot sa mccabe eh same na same naman daw ang tanong pati given. Example 4.6 (McCabe and Smith) A pump draws a solution of specific gravity 1.84 from a storage tank through a 3 in. (75-mm) Schedule 40 steel pipe. The efficiency of the pump is 60 percent. The velocity in the suction line is 3 ft/s (0.914 m/s). The pump discharges through a 2-in. (50-mm) Schedule 40 pipe to an overhead tank. The end of the discharge pipe is 50 ft (15.2 m) above the level of the solution in the feed tank. Friction losses in the entire piping system are 10 ft lbf/lbm (29.9 J/kg). What pressure must the pump develop? What is the power delivered to the fluid by the pump?
3.
Yung TIo2 na tinuro mo Kenneth TiCl4 can be formed by reacting titanium dioxide (TiO2) with hydrochloric acid. TiO2 is available as an ore containing 78 % TiO2 and 22 % inerts. The HCl is available as 45 wt% solution (the balance is water). The per pass conversion of TiO2 is 75 %. The HCl is fed into the reactor in 20 % excess based on the reaction. Pure unreacted TiO2 is recycled back to mix with the TiO2 feed. TiO2 + 4 HCl → TiCl4 + 2 H2O For 1 kg of TiCl4 produced, determine: a. the kg of TiO2 ore fed. b. the kg of 45 wt % HCl solution fed. c. the ratio of recycle stream to fresh TiO2 ore (in kg). (MW : TiO2 79.9; HCl 36.47; TiCl4 189.7)
4.
Yung production ng ammonia, ung sinabi ko sau Kenneth, may given given na mass nila, tas tinatanong mga conversion something etc. The synthesis of ammonia proceeds according to the following reaction N2 + 3 H2 -----> 2 NH3 In a given plant, 4202 lb of nitrogen and 1046 lb of hydrogen are fed to the synthesis reactor per hour. Production of pure ammonia from this reactor is 3060 lb per hour. a. What is the limiting reactant. b. What is the percent excess reactant. c. What is the percent conversion obtained (based on the limiting reactant).
5.
Liquid fuel combustion, may given na 20% excess air, tas sa fuel eh binigay ang % w/w ng carbon, hydrogen, oxygen at nitrogen. Tinatanong ang moles ng co2 sa fuel gas. 02 na issupply given na 20% x’s air. May given din mga calorific value something, ung pangatlong tanong gagamitin ung mga data ng calorific value something
6.
Ang isa pang multi effect evap. Given naman xa maxado, kukunin mo lang yung V1 na mag eevaporate sa unang evaporator. Tas amount na pasok sa evaporator 2 tas ang pangatlo eh area ang hinahanap
7.
Absorption ng sulfur something, may given na recovery. Water ang absorpption medium. Tapos ung water with sulfur eh magrereact sa NaOH para marecover yung sulfur. May amount na given sa sulfur sa product, tas ayun stoic stoic ata gagamitin ditto. (MATERIAL BALANCE)
8.
Thermo. 2 given na tanks, tas magkakonek cla sa pamamagitan ng isang tube something, tas initially ung tank1 eh may given na pressure na given at temp. tapos ayun tanong eh wat temp at pressure pag equal na volume dun sa tank 1 at 2. Parang ganun. (haha ang gulo)
9.
Simpleng PV=nRT (buti na lang meron neto,)
10. Conversion ng units, pero may halong pagkachecal xa Conversion of units ng kinetic energy
11. 3 Heat of reaction na heat of formation yung given 12. Given eh 57.8 mole % hexane, 8.9 heptane, tas 33.3 steam. Tas may given na Antoine sa hexane at heptane. Pero kakaiba ba xa kc A and B lang yung constant. Tas ayun ang tanong temp ng first liquid na magcocondensed, tas temp ng 2nd liquid na mag cocondensed tas ano daw yung conc. Product after nun
