CHEMICAL ENERGY BALANCE - EXAMPLE 11.3 : For liquid benzene, the CP constants are : a = 129440, b = - 169.5, c = 0.64781. Reference temperature is 298 K. The temperature of benzene is 60 degree Celsius. Calculate the enthalpy of benzene by using the formula H = a (DT) + (b/2) (T^2 - TREF^2) + (c/3) (T^3 - TREF^3) where ^ is power, DT is temperature difference with TREF = 298 K. H is in J / kmol. DT = T - TREF.

CHEMICAL ENERGY BALANCE - EXAMPLE 11.4 : Calculate the bubble temperature T at P = 85-kPa for a binary liquid with x(1) = 0.4. The liquid solution is ideal. The saturation pressures are Psat(1) = exp [ 14.3 - 2945 / (T + 224) ], Psat(2) = exp [ 14.2 - 2943 / (T + 209) ] where T is in degree Celsius. Please take note that x(1) + x(2) = 1. Please take note that y(1) + y(2) = 1, y(1) = [ x(1) * Psat(1) ] / P, y(2) = [ x(2) * Psat(2) ] / P, * is multiplication. P is in kPa.

REACTION ENGINEERING - EXAMPLE 13.2 : A batch reactor is designed for the system of the irreversible, elementary liquid-phase hydration of butylene oxide that produces butylene glycol. At the reaction temperature T = 323 K, the reaction rate constant is k = 0.00083 L / (mol - min). The initial concentration of butylene oxide is 0.25 mol / L = Ca. The reaction is conducted using water as the solvent, so that water is in large excess. (a) Let the molecular weight of water is 18 g / mol and the mass of 1 kg in 1 L of water, calculate the molar density of water, Cb in the unit of mol / L. (b) Determine the final conversion, X of butylene oxide in the batch reactor after t = 45 min of reaction time. Use the formula X = 1 - 1 / exp [ kt (Cb) ] derived from material balance. (c) Find the equation of t as a function of X.

BIOPROCESS ENGINEERING - EXAMPLE 14.2 : An aqueous solution with 2.5 g of a protein dissolved in 600 cubic centimeters of a solution at 20 degree Celsius was placed in a container that has a water-permeable membrane. Water permeated through the membrane until the h - level of the solution was 0.9 cm above the pure water. (a) Calculate the absolute temperature of the solution, T in Kelvin, where T (Kelvin) = T (degree Celsius) + 273.15. (b) Calculate the osmotic pressure, P of the solution by using the formula P = hrg where h is level of the solution, r is density of water with 1000 kg per cubic meter, g = 9.81 N / kg as gravitational acceleration. (c) Calculate the concentration of the protein solution, C in kg / cubic meter. (d) Calculate the molecular weight of the protein, (MW) = CRT / P where R = 8.314 Pa cubic meter / (mol K) as ideal gas constant.

ELECTRICAL TECHNOLOGY - EXAMPLE 16.1 : According to Shockley equation, the I - V characteristic of a diode is approximated by I = IS [ exp (nVD / VT) - 1 ]. For silicon, let the reverse bias saturation current IS as 0.000000000001. If n is ideality factor with value of 1.5, VT as thermal voltage drop of 0.026 V at room temperature, what is the value of current I that passes through the silicon diode in the heater of evaporator when the forward voltage drop VD = 0.026 V? Please take note that exp is the exponential function with e(1) = 2.718, e(2) = 7.389.

PROCESS DESIGN - EXAMPLE 21.1 : According to rules of thumb in chemical process design, consider the use of an expander for reducing the pressure of a gas when more than 20 horsepowers can be recovered. The theoretical adiabatic horsepower (THp) for expanding a gas could be estimated from the equation : THp = Q [ Ti / (8130a) ] [ 1 - (Po / Pi) ^ a ] where 3 ^ 3 is 3 power 3 or 27, Q is volumetric flowrate in standard cubic feet per minute, Ti is inlet temperature in degree Rankine, a = (k - 1) / k where k = Cp / Cv, Po and Pi are reference and systemic pressures respectively. (a) Assume Cp / Cv = 1.4, Po = 14.7 psia, (temperature in degree Rankine) = [ (temperature in degree Celsius) + 273.15 ] (9 / 5), nitrogen gas at Pi = 90 psia and 25 degree Celsius flowing at Q = 230 standard cubic feet per minute is to be vented to the atmosphere. According to rules of thumb, should an expander or a valve be used? (b) Find the outlet temperature To by using the equation To = Ti (Po / Pi) ^ a.

PROCESS DESIGN - EXAMPLE 21.2 : The names of the flow streams could be represented by : H1 for first hot stream, H2 for second hot stream, C1 for first cold stream, C2 for second cold stream. Data of supply temperature Ts in degree Celsius : 150 for H1, 170 for H2, 30 for C1, 30 for C2. Data of target temperature Tt in degree Celsius : 50 for H1, 169 for H2, 150 for C1, 40 for C2. Data of heat capacity Cp in kW / degree Celsius : 3 for H1, 360 for H2, 3 for C1, 30 for C2. (a) Find the enthalpy changes, dH for all streams of flow H1, H2, C1 and C2 in the unit of kW. Take note of the formula dH = (Cp) (Tt - Ts). (b) Match the hot streams H1 and H2 with the suitable cold streams C1 and C2 to achieve the maximum energy efficiency.

FOOD ENGINEERING - QUESTION 23.1 : (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calorie, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?

FOOD ENGINEERING - QUESTION 23.2 : (a) A dryer reduces the moisture content of 100 kg of a potato product from 80 % to 10 % moisture. Find the mass of the water removed in such drying process. (b) During the drying process, the air is cooled from 80 °C to 71 °C in passing through the dryer. If the latent heat of vaporization corresponding to a saturation temperature of 71 °C is 2331 kJ / kg for water, find the heat energy required to evaporate the water only. (c) Assume potato enters at 24 °C, which is also the ambient air temperature, and leaves at the same temperature as the exit air. The specific heat of potato is 3.43 kJ / (kg °C). Find the minimum heat energy required to raise the temperature of the potatoes. (d) 250 kg of steam at 70 kPa gauge is used to heat 49,800 cubic metre of air to 80 °C, and the air is cooled to 71 °C in passing through the dryer. If the latent heat of steam at 70 kPa gauge is 2283 kJ / kg, find the heat energy in steam. (e) Calculate the efficiency of the dryer based heat input and output, in drying air. Use the formula (Ti - To) / (Ti - Ta) where Ti is the inlet (high) air temperature into the dryer, To is the outlet air temperature from the dryer, and Ta is the ambient air temperature.

From 9 am to 2 pm the temperature rises at a constant rate from 14 degree F to 36 degree F . What was the temperature at noon? Option 1.)+16 2.)+26 3.)+31 4.) -4 +6

PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, 0.93 % argon, 0.039 % carbon dioxide, and small amounts of other gases by volume. Fact 2 : Volume occupied is directly proportional to the number of moles for ideal gases at constant temperature and pressure. Fact 3 : 12.5 moles of pure oxygen is required to completely burn 1 mole of pure octane. Fact 4 : Air-fuel ratio (AFR) is the mass ratio of dry air to fuel present in a combustion process such as in an internal combustion engine or industrial furnace. Fact 5 : Molecular weight of oxygen gas is 31.998 g / mole and molecular weight of nitrogen gas is 28.014 g / mole. (a) Find the molar ratio of nitrogen and oxygen, or (moles of nitrogen) / (moles of oxygen) in dry air, by assuming ideal features of nitrogen and oxygen gases. (b) How many moles of nitrogen are available if dry air is used to completely burn the 1 mole pure octane? (c) Find the mass of fuel of 1 mole of octane with molecular weight of 114.232 g / mole. (d) Find the mass of dry air with 12.5 moles of pure oxygen by assuming only oxygen and nitrogen gases exist in the air. (e) Find the air-fuel ratio (AFR) when octane is used as fuel. (f) Find the fuel-air ratio (FAR) when octane is used as fuel.

PETROLEUM ENGINEERING - QUESTION 25.3 : Liquid octane has a density of 703 kilograms per cubic metre and molar mass of 114.23 grams per mole. Its specific heat capacity is 255.68 J / (mol K). (a) Find the energy in J needed to increase the temperature of 1 cubic metre of octane for 1 Kelvin. (b) At 20 degree Celsius, the solubility of liquid octane in water is 0.007 mg / L as stated in a handbook. For a mixture of 1 L of liquid octane and 1 L of water, prove by calculations that liquid octane is almost insoluble in water.

NATURAL GAS ENGINEERING - QUESTION 26.2 : (a) The Hyperion sewage plant in Los Angeles burns 8 million cubic feet of natural gas per day to generate power in United States of America. If 1 metre = 3.28084 feet, then how many cubic metres of such gas is burnt per hour? (b) A reservoir of natural gas produces 50 mole % methane and 50 mole % ethane. At zero degree Celsius and one atmosphere, the density of methane gas is 0.716 g / L and the density of ethane gas is 1.3562 mg / (cubic cm). The molar mass of methane is 16.04 g / mol and molar mass of ethane is 30.07 g / mol. (i) Find the mass % of methane and ethane in the natural gas. (ii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the gases are ideal where final volume of the gas mixture is the sum of volume of the individual gases at constant temperature and pressure. (iii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the final mass of the gas mixture is the sum of mass of the individual gases. Assume the gases are ideal where mole % = volume % at constant pressure and temperature.

QUANTUM BIOLOGY - EXAMPLE 33.3 : In quantum biology, microtubule is used to store information in a cell. At temperature of T = 300 K, measured current I in a probe is directly proportional to the supplied voltage V, when passing through a microtubule with resistance R. (a) Form an equation of V as a function of I involving k as a constant. (b) If the microtubule has R = 1 ohm at such condition, find the value of V when I = 2 A. Hint : Ohm's law. (c) Find the relationship of k as a function of R.

QUANTUM BIOLOGY - EXAMPLE 33.6 : The time to maintain quantum coherence in a biological system is t. Let t = k / (MT) where k = constant, M = mass, T = temperature. For living farm animals, normal rectal temperatures ranges for pig and goat are 38.7 - 39.8 and 38.5 - 39.7 in degree Celsius respectively. Let a pig has a mass of 100 kg and a goat has a mass of 300 pounds. (a) Find the mass of goat in the unit of kilogram, when 1 pound = 0.4536 kg. (b) Find t (p) / t (g) for living animals when t (p) = t for pig and t (g) = t for goat. (c) State the assumption of your calculation in question (b).