Answer / eswar

# 1 it's generall 3 to 4 rpm....but rarely exceeds.

Answer / ashwin

IT GENERALLY 3-4 FOR ROTARY VACUUM DRIER.
& 4-5 FOR TUMBLER DRIER

Answer / shekhar

For rotocone type it ideally 6 to 7

Can condensate control in a reboiler cause water hammer problems?

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hi sir/madam i need placement paper of last years of DRDO in chemical sream...... send me the question paper as soon as possible on my email address praveen.zone@gmail.com

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ENGINEERING NUMERICAL METHODS - EXAMPLE 19.3 : There are 2 simultaneous equations : (A1) x + (B1) y = D1 and (A2) x + (B2) y = D2. (a) By using Excel program, find the values of x and y when A1 = 80, A2 = 150, B1 = 52, B2 = 100, D1 = 3.5 and D2 = 2.3. (b) Write the expression of Excel in the form of =MMULT(MINVERSE(W:X),Y:Z) in order to get the values of x and y. W, X, Y and Z may be A1, A2, B1, B2, D1 and D2.

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Question 107 - In N + 1 Rule in Quantum Chemistry, whenever a spin 1 / 2 nucleus is adjacent to N other nuclei, it is split into N + 1 distinct peaks. In 1 peak or singlet, there is only 1 magnitude. In 2 peaks or doublet, the ratio of magnitude of each peak is 1 : 1. In 3 peaks or triplet, the ratio of magnitude of each peak is 1 : 2 : 1. In 4 peaks or quartet, the ratio of magnitude of each peak is 1 : 3 : 3 : 1. In 5 peaks or quintet, the ratio of magnitude of each peak is 1 : 4 : 6 : 4 : 1. (a) By using binomial coefficients or Triangle of Pascal find the ratio of magnitude of each peak if 6 peaks exists. (b) How many adjacent nuclei are available in a spin 1 / 2 nucleus in such situation of 6 peaks?

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Question 72 - (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 calory, 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?

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NATURAL GAS ENGINEERING - QUESTION 26.1 : (a) In natural gas pipe sizing, the length of the pipe from the gas source metre to the farthest appliances is 60 feet. The maximum capacities for typical metallic pipes of 60 feet in length are : 66 cubic feet per hour for pipe size of 0.5 inches; 138 cubic feet per hour for pipe size of 0.75 inches; 260 cubic feet per hour for pipe size of 1 inch. By using the longest run method : (i) Find the best pipe size needed for the capacity of 75 cubic feet per hour. (ii) Estimate the suitable range of capacities for the pipe size of 1 inch. (b) The maximum capacities for typical metallic pipes of 50 feet in length are : 73 cubic feet per hour for pipe size of 0.5 inches; 151 cubic feet per hour for pipe size of 0.75 inches; 285 cubic feet per hour for pipe size of 1 inch. By using the branch method find the best pipe size needed for the capacity of 75 cubic feet per hour when the length of the pipe from the gas source metre to the appliance is 52 feet.

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Question 78 - 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.

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Explain what are the affinity laws associated with dynamics pumps?

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what is the PH of earth? does earth is neutral like ph value 7 for water indicating water is neutral?

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Question 68 – Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).

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Question - Chemical Engineering Material - In crystal material, hexagonal crystal system could form 4-digit index in certain direction of solid. For [1(-1)0] direction in the hexagonal crystal systems of particular catalyst applied in fume removal of incinerator, what is the four-digit index for this direction? Hint : The transformation equations between the 3-digit [h’k’l’] and the 4-digit [hkil] indices are : h = (1/3) (2h’ – k’); i = - (h + k); k = (1/3) (2k’ – h’); l = l’. A. [(-1)100] B. [1(-1)00] C. [(-1)000] D. [00(-1)(-1)] E. [(-1)0(-1)0]

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What kind of concerns is associated with temperature pinch points in condensers?

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