Test 1, F07 - First Test ChE 2114 September 14, 2007 A...

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Unformatted text preview: First Test ChE 2114 September 14, 2007 A Table of Conversion Factors is at the bottom of page 2 of the test. 1. (20 points) A cylindrical sample of wood is 50.0 centimeters in height and has a cross sectional area of 300 square centimeters. The specific gravity of the sample is 0820 20/4. If the wood cylinder is floating vertically in a liquid whose density is 1.60 gme, what is the minimum force, in Newtons, that must be applied to the top of the wood sample in order to submerge it 100% in the liquid? FACTORS FOR UNIT CONVERSIONS Length Volume Force Pressure Energy Power Ex) (20 points) A open end mercury manometer (density = 13.6 g/c-rnE) is attached to a tank as shown in the sketch below. The tank contains air at very low pressure. The mercury in the higher pressure side of the manometer is covered with 2.00 meters of water that lies between the Hg and the atmosphere. What is the gauge pressure in the tank in Pa? Equivalent Values . 1 kg = 1000 g = 0.001 metrioton = 2.2046211:m = 35.27392 02 #1113.1 = 16 oz = 5 >4 1074 ton = 453.593 g = 0.453593 kg 1 m = 100 em = 1000 mm = 106 microns (p.111) =- 1019 angstroms (A) = 39.37 'm. = 3.2808ft = 1.0936 yd : 0.0006214 13:61.6 1 ft = 12 in. = 1X3 yd = 0.3048 111 = 30.48 cm 1613 = 1000L = 10° 66:3 = 10° mL = 35.3145 113 = 220.83 imperial gallan = 264.17 ga1 = 1056.68 qt _ 1 £13 = 1728 111.3 = 7.4805 gal = 0.028317 613 = 28.317L = 28,317 61113 . IN = 1 kg-mfsz = 105 dynes = 105 g-cmr’sz = 0.22481 116p 111:.f = 32.1741bm-ftfsz = 4.4482N = 4.4482 x 105 dynes 1 arm = 1.01325 x105 14me (Pa) = 101.325 kPa =_1.01325 bar = 1.01325 >410r3 636165761112 - . . = 760mmHg at 0°C (torr) = 10.333 1311—120 at 4°C = 146961138111.2 (psi) = 33.9 8 H20 at 4°C = 29.921 in. Hg at 0°C 1 J = 1N-m = 107 ergs = 107 dyne-Cm = 2.778 X 10—7 kW-h = 0.23901 cal = 0.7376 0:48p = 9.486 >< 10'4 Btu 1 w = 1st = 0.23901 cars = 0.7376 ft-lbpts = 9.486 x 10—43606 = 1.341 x 10—3 hp g = 9.81 mas2 density of water = 1.00 gme Air: 79 moi % N2 (MW = 28.0) 21 m0] % 02 (MW = 32.0) La) (10 points) As shown in the sketch below, two streams (F and W) enter a mixing vessel and one product stream (P) leaves the vessel. Components A, B, and C are involved and the x values are mole fractions. Given: x,“ = 0.100 KW = 0.600 ch : 0-400 1-“ = 250 moles per minute. We need to calculate values for W and P in moles per minute and all of the unspecified compositions. How man}r degrees of freedom does this system, as described above, have? F W Composition Composition HP ‘4»! "HF 1:19 Xe: P Composition Kg; ‘3? W (10 points) Information regarding saturated aqueous solutions of sodium bicarbonate is given below. “Solubility” at a given temperature is the number of grams of NaHC03 that will dissolve in 100 grams of water to give a saturated solution. An unsaturated solution, initially at 50.0 0C, is 10.0 mass % NaHCO3. To what temperature must this solution be cooled in order for 20.0% of the NaHCO3 in the initial solution to erystaliize out? solubility temp. (°C) g NaHC03/100 3 H10 60 16.4 so 14.45 40 123 30 llJ 20 Qfi 10 8.15 UI (20 points) A gas mixture is 75.0 mole % C4H10 (MW = 58.0) and 25.0% N; (MW = 28.0). This stream mixes with a stream of air whose mass flow rate is two times the mass flow rate of the C4H10/N2 stream. What is the average molecular weight of the combined gas stream produced by the mixing described above. (20 points) A liquid mixture is T00 mass % alcohol (A), 20.0 mass % water (W), and 10.0 mass % non-volatile solute (S). The mixture is fed to a continuous distillation column operating at steady-state. The overhead product is 90.0 mass % alcohol and contains no S. If the mass % S in the bottoms product is 40.0%, what is the mass % water in this bottoms product? ...
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This note was uploaded on 09/28/2008 for the course CHE 2114 taught by Professor Pldurrill during the Fall '08 term at Virginia Tech.

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Test 1, F07 - First Test ChE 2114 September 14, 2007 A...

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