{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Test 1, Sp 08

# Test 1, Sp 08 - ChE 2114 First Test 1(15 points Automobile...

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ChE 2114 First Test February 15, 2008 1. (15 points) Automobile A costs \$17,000 and averages 24.0 miles per gallon of gasoline. Automobile B costs \$21,000 and averages 45.0 mpg. Assuming that the cost of gasoline is \$3.00 per gallon, how many miles would have to be driven by the two automobiles in order for the lower fuel consumption of Autornobile B to compensate for its higher purchase price? Il‘x) (15 points) A cylindrical wooden object has a height of 50.0 cm and a cross sectional area of 75.0 cmz. A vessel contains two immiscible liquid layers, the lower layer being water (p = 1.00 g/mL), and the top layer being a mixture of the two nonpolar hydrocarbons n-hexane (p = 0.659 g/mL) and benzene (p = 0.879 gme). This hydrocarbon mixture is 80.0 mass % n-hexane. The cylinder ﬂoats vertically in the liquids with the top 20.0 centimeters submerged in the hydrocarbon layer and the bottom 30.0 centimeters submerged in the water layer. Calculate the weight of the wooden object in Newtons. (g = 9.81 mfsz) Li.) (15 points) A vessel contains gas at a low pressure. Usng the sketch and informatiou provided below, calculate the gauge pressure (the pressure that will be read 011 the Bourdon gauge) for the gas 111 the vessel? Give your answer in units 011681112. Density 9 BOURBON GAUGE P (at the open end of the manometer) = atmospheric pressure p1 = 1.19gl’mL h = 49.0 cm p2=0.8?6 g/mL 01:30.5 cm p = 0.640 glmL FACTORS FOR UNIT CONVERSIONS Quality Mass Length Volume Force Pressure Energy Power Equivalent Values 1 kg 1 lbm 1000 g = 0.001 metric ton = 2.2046211)“, = 35.27392 02 16 oz = 5 X 10'4 ton = 453.593 g = 0.453593 kg ll || 1111 = 100 em =. 1000111111 = 106 microns (p.111) = 101D angstroms (A) z 39.31111. = 3.2808 ft = 1.0936 yd = 0.0006214 mile 111 = 12 in. = 1,43 yd = 0.304866 = 30.48 cm 1613 = 1000L = 105 01113 = 1051611 = 35.3145113 = 220.83 imperial gallons = 264.17 gal = 10506801 1113 = 1728013 = 7.4805 gal = 0028311613 = 28.3171. = 28.311 1:11:13 1N = lkgmlsz = 105 dynes = 105 g‘cmfsz = 0.2248110f 1le = 32.174 113,61sz2 = 44482 N = 4.4482 x 105 dynes 1 atm = 1.01325 x 105 Nlm2 (Pa) = 101.325 kPa = 1.01325 bar = 1101325 X 106 dyrleslcm2 = 760 mm Hg at 0°C (ton) = 10.333 1:: H20 at 4"C = 1469616916.2 (psi) = 33.9 ft H30 014°C = 29.921 in. Hg at (TC 11 = lN-m = 1.07 ergs ='1070yrle-crn = 2.178 x 10*? kah = 0.23901 ea] = 0.1376 11—le = 9.486 x 10“ Btu l W = 1st = 0.23901 calls = 0.7376 ft-Ibp's. = 9.486 X 10'4 Btul's = 1.341 x 10‘3 hp (20 points) A gaseous mixture of propane (CgHg) and 02 is 16.6 mol % C3H3 and 83.4 mol % 02 and flows at a rate of 680 grams per second. This stream is mixed with a stream of a_ir ﬂowing at a rate sufﬁcient to raise the ratio of mots of 03 to 111013 of C3H3 t010.0. (MW of C3H3 = 44.0; MW 0f02 = 32.0; MW ofair = 29.) Calculate the maSS ﬂow rate of the air stream. (15 points) A second-order reaction A —> products, with rate constant k = 8.4? (molesfliter)'1min" is occurring at steady-state in the liquid phase in a continuous stirred—tank reactor. The ﬂow rate of the inlet and outlet streams is 3.00 liters per minute. The concentration of reactant A in the inlet stream is 0.1?5 mols’L, and the concentration in the product stream leaving the reactor is 0.0350 mols’L. For a second-order reaction involving the single reactant A, the reaction rate is given by the equation: rate of reaction in moles per liter per minute = kc“2 where cA is mols per liter What is the volume of the reactor in liters? (20 points) There are four components (A, B, C, and D) in the feed to a distillation column, and the feed composition is 15.0 mole % A, 40.0% B 35.0% C, and 10.0% D. .. Only two streams leave the distillation column, the distillate and the bottoms. All of the A in the feed leaves in the distillate, and all of the D leaves in the bottoms. Components B and C appear both in the distillate and in the bottoms. The bottoms is 50.0 mole % D. The distillate is 37.5 mole % C. What is the mole % B in the distillate? ...
View Full Document

• Fall '08
• PLDurrill
• Kilogram, lower fuel consumption, wooden object, second-order reaction, higher purchase price, cylindrical wooden object

{[ snackBarMessage ]}

### Page1 / 6

Test 1, Sp 08 - ChE 2114 First Test 1(15 points Automobile...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online