DESIGN CONSIDERATIONS * System: Ethanol – water * Feed rate: 225kmol/h * Feed composition: 28 mol% ethanol * Feed condition: 50% saturated liquid & 50% saturated vapor * 97% of ethanol recovery is required * Operating pressure: 1bar * Distillate composition: 81 mol% ethanol * Column type: Sieve tray column * Operating condition: 70% of flooding Applying material balance to the rectifying section (Eqn 01); V=L+D Applying material balance for the more volatile
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OBJECTIVE The main purpose of this experiment is:- i. To demonstrate the working principles of industrial heat exchangers ii. To investigate the efficiency of the heat exchanger in parallel and counter flow arrangements 1.0 INTRODUCTION A heat exchanger is equipment in which heat exchange takes place between 2 fluids that enter and exit at different temperatures. The main function of heat exchanger is to either remove heat from a hot fluid or to add heat to the cold fluid. The direction
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Units and tasks DAVID HILTON Unit 7 Management Accounting
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References: Liang‚ F.‚ Xu‚ B.G.‚ Shi‚ X.H.‚ Ming‚ S.R.‚ “Advances in desulfurization with wet oxidation process”‚ Modern Chem. Ind.‚ 23 (5)‚ 21-24 (2003) McManus‚ D.‚ Martell‚ A.E.‚ “The evolution‚ chemistry and applications of chelated iron hydrogen sulfide removal and oxidation processes”‚ J. Mol. Cat. A.‚ 117 (1)‚ 289-297 (1997). their iron complexes”‚ Inorg. Chim. Acta.‚ 278 (2)
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------------------------------------------------- ASSIGNMENT 1 1. A rigid tank with a volume of 1.8 m3 contains 15 kg of saturated liquid vapor mixture of water at 90oC. Now the water is slowly heated. Determine the temperature at which the liquid in the tank is completely vaporized. Also‚ show the process on a T-v diagram with respect to saturation lines. (5 marks) 2. A 0.5 m3 vessel contains 10 kg of refrigerant-134a at -20oC. Determine (a) the pressure‚ (b) the total internal energy
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incorrect. This fee payment is valid only for your particular course of study or program. If you fall out of status‚ apply for a new F-1‚ F-3‚ M-1‚ M3 or J-1 Non-immigrant visa‚ or if you want to change your Non-immigrant category to an F-1‚ F-3‚ M-1‚ M-3 or J-1‚ you may be required to pay another fee. APPLICANT STATUS: J-1 DATE OF BIRTH: Jun 23‚ 1991 PROGRAM NUMBER: P305708 EXCHANGE VISITOR CATEGORY: SUMMER WORK/TRAVEL AMOUNT RECEIVED: $35.00 SEVIS IDENTIFICATION NUMBER: N0011118101 THIS
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PROPERTIES Unit Value (1) ASTM Method Resin Properties Melt Flow Rate @ 190°C & 2.16 kg load Density @ 23°C g/10 min. kg/m3 5 935 D 1238 D 1505 Tensile Strength @ Yield Tensile Strength @ Break Tensile Elongation @ Break 1% Secant Modulus Flexural Strength Hardness (Shore D) ESCR (100% Igepal)‚ F50 MPa MPa % MPa MPa hrs 16 17 590 420 13 66 >150 D 638 D 638 D 638 D 790 D 790 D 2240 D 1693B THERMAL PROPERTIES Vicat Softening Point Brittleness
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The net force F = ΔP.A that acts on the ball is from the high pressure area(surrounding) to the low pressure area(above the fan) which makes the ball return to its initial position. Recorded air velocity = 5.4 +/- 0.5 m/s. Air density = 1.3 kg/m3 Diameter= (5.040 +/- 0.005) ×10-2m [pic]=[pic] [pic]= [pic]=[pic]= 1.6×10-5 m2 Ball area= (7.980 +/- 0.016)[pic]m2 Ball mass = (3.30 +/- 0.05)[pic]kg [pic] Now F of air resistance equals the ball’s weight since it is just lifted above
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Ideal Gas Law Packet Name ______________________________ 12.3 Date __________________ Period _______ Given: Ideal Gas Law = then P = n = V = T = R = 1. What pressure is required to contain 0.023 moles of nitrogen gas in a 4.2 L container at a temperature of 20.(C? 2. Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0 L. What temperature must be maintained on 0.500 moles of this gas in order
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the net force is zero in both cases. 1-7 A plastic tank is filled with water. The weight of the combined system is to be determined. Assumptions The density of water is constant throughout. Properties The density of water is given to be ρ = 1000 kg/m3. Analysis The mass of the water in the tank and the total mass are V =0.2 m 3 mtank = 3 kg mw =ρV =(1000 kg/m )(0.2 m ) = 200 kg 3 3 mtotal = mw + mtank = 200 + 3 = 203 kg Thus‚ 1N W = mg = (203 kg)(9.81 m/s 2 ) 2 1 kg ⋅ m/s
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