Answer / kang chuen tat (malaysia - pen

Answer 17 : Let V / t = 80 L / s = 0.08 cubic metres / s. V / t = Av : (a) pai = 3.142 A = (pai)(radius square). (3.142)(0.16)(0.16)v = 0.08, v = 0.995 m / s. (b) (3.142)(0.08)(0.08)v = 0.08, v = 3.98 m / s. (c) (18000) + (1000)(9.81)(0) + 0.5(1000)(0.995)(0.995) – (1000)(9.81)(2) – 0.5(1000)(3.98) = 158885 Pa at point 2. 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.

We are having proplem with the persperation smell in ETT/ECHO room in our cardiac hospital.The 2 Ton of AC is not conditioning the environment. We have to use exhaust fan to through out of the room bad air. Could anyone suggest how to coup with this effectively using chemical engineering. Any permanent solution to neutralise the bad smell.The Air freshners are not doing enough instead making a bad odour.

1 Answers  

---

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.

1 Answers  

---

CHEMICAL ENERGY BALANCE - EXAMPLE 11.5 : According to Margules Equation, P = x(1) p(1) g(1) + x(2) p(2) g(2) for a two-component mixture where P is bubble pressure, x is mole fraction, p is saturation pressure, g is constant given by ln g(1) = x(2) A x(2). Find the value of A as a constant when P = 1.08 bar, p(1) = 0.82 bar, p(2) = 1.93 bar in a 50 : 50 mole fraction mixture. Estimate the pressure required to completely liquefy the 30 : 70 mixture using the same equation, by proving P = 1.39 bar. Take note that ln g(2) = x(1) A x(1), ln g(1) = x(2) A x(2).

1 Answers  

---

which thing is responsible for making petroleum?

6 Answers   OGDCL,

---

plz send me hpcl and reliance entrance test papers on my id s,kumar_bk@yahoo.co.in

1 Answers   Reliance,

---

Explain what is the method involved in analyzing powders for composition?

0 Answers  

---

Why is respiration considered as an exothermic reaction? Explain?

0 Answers  

---

What are some common precipitating agents used to remove metals from aqueous waste streams?

0 Answers  

---

A distillation column separates 10000 kg / hr of a mixture containing equal mass of benzene and toluene. The product D recovered from the condenser at the top of the column contains 95 % benzene, and the bottom W from the column contains 96 % toluene. The vapor V entering the condenser from the top of the column is 8000 kg / hr. A portion of the product from the condenser is returned to the column as reflux R, and the rest is withdrawn as the final product D. Assume that V, R, and D are identical in composition since V is condensed completely. Find the ratio of the amount refluxed R to the product withdrawn D. Hint : Solve the simultaneous equations as follow in order to find the answer (R / D) : 10000 = D + W; 10000 (0.5) = D (0.95) + W (0.04); 8000 = R + D.

2 Answers  

---

How can one determine the particle size distribution for a given bulk solid?

0 Answers  

---

UNIT OPERATION - EXAMPLE 9.4 : Acetone and ethanol are separated using a distillation column with a partial condenser and partial reboiler. An equimolar, sub-cooled liquid feed enters at 100 kmol / hr and condenses 1 mole of vapor for every 6 moles of feed. The separation requires a distillate vapor that is 95 mol % acetone and bottoms liquid that is 5 mol % acetone. The reflux is returned from the condenser to the column as a saturated liquid and the operation is run at (L / V) = 1.4 * (L / V) min. Assume constant overflow conditions. (a) Feed operating line is y = [ q / (q - 1) ] x - z / (q - 1) where z = 0.5 for equimolar liquid mixture of 2 components, q = (L'- L) / F where L' = L + F + (F / 6) for condensation of 1 mole of vapor / 6 moles of feed. What is y = f(x)? (b) The rectifying operating line is y = (L / V) x + (D / V) (xd) where (L / V) min goes through the points A (0.95, 0.95) and B (0.53, 0.69), V = L + D. What is y = f(x)? Let xd = 0.95. (L / V) min is the slope of the 2 points A and B.

1 Answers  

---

In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).

1 Answers  

---