Answer / kangchuentat

Answer 80 - (a) 1 cubic metre of liquid octane has 703 kilograms or 703 kilograms x mole / (0.11423 kilograms) = 6154.25 mole. Energy needed = [ 255.68 J / (mol K) ] x (6154.25 mol) K = 1573518.64 J where K is the unit symbol for Kelvin and mol is the unit symbol for mole. (b) Density of liquid octane = 703 kilograms per cubic metre = 0.703 grams per cubic centimetre = 703 g / L. There is 0.007 mg = 0.000007 g or 0.000007 g x (1 L / 703 g) = 0.0000000099573257 L of liquid octane in 1 L of water. With negligible amount of liquid octane in 1 L of water, octane is considered insoluble in water. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

How to TDS removed found in lab apparatus.

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Question 57 – 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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What is an effective means of removing silicone from aluminum?

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why do tangential entry require for a cyclone separator?

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dear sir/madam, I had applied for the post of officer trainee so plz if u have any papers both technical or aptitude plz inform me on sonalichavan01@gmail.com

0 Answers   HPCL,

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ENGINEERING NUMERICAL METHODS - EXAMPLE 19.2 : For a mixture of benzene (B), toluene (T) and xylene (X), the equation applies where x for B, T and X will sum up to 1. The equation of x for each component is x = (L / V + 1) (F) / (L / V + K). The data of F for each component are : 0.5 for B, 0.35 for T, 0.15 for X. The data of K for each component are : 1.98 for B, 0.76 for T, 0.24 for X. When x for B + x for T + x for X = 1, find the values of (a) L / V; (b) x for each component of B, T, X respectively. You may use Excel program - Data : What-If-Analysis for Goal Seek to perform the iterative calculations.

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Given, T1=600C, constant temperature of wall 800 , length of wit exchanger L=3m, dia=25 mm, Cp=14.91 KJ/kg.K, density ρ=914 kg/m3,mean velocity vm= 2 m/s, film heat transfer coefficient hf=72 Kcal/h.m2.K, what is the outlet temperature?

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can i drink deminaralised?

1 Answers   Dalkia,

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What is Draft? It releavent to furnace.

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Please convert 1 kg (mass) of Liquid LPG into Ltrs (volume) of LPG Vapor.

2 Answers   Wak gas,

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Question 49 - 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.

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could any one plzz give me the placement tech papers taken by iocl,and bpcl in chemical stream..... my id is.... mishra.gaurang@gmail.com

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