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# Engineering Analysis I PETR 1305

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GPA 3.7

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This 13 page Class Notes was uploaded by Davon Lueilwitz on Thursday October 22, 2015. The Class Notes belongs to PETR 1305 at Texas Tech University taught by Waylon House in Fall. Since its upload, it has received 65 views. For similar materials see /class/226429/petr-1305-texas-tech-university in Petroleum Engineering at Texas Tech University.

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Date Created: 10/22/15

Problem Set 3 PE1305 Due Sept 28 Problem 1 A 70 liter tank containing an ideal gas at a pressure of 40 atmospheres is connected by a valve to a 30 liter tank of an ideal gas at 70 atmospheres The valve is opened slowly and the gases in the two tanks mix Assuming the temperature throughout is constant what is the nal pressure in the tanks Problem 2 Three cylinders of an ideal gas are connected together though closed valves Cylinder A holds 20 liters at 60 atmospheres pressure Cylinder B holds 30 liters at 30 atmospheres Cylinder C holds 50 liters at 40 atmospheres The valves are opened and the cylinders contents allowed to mix Assuming the temperature is the same throughout what is the nal pressure in the cylinders Problem 3 A reference container of known volume VR is lled with Helium gas to an absolute pressure of P1 The reference container is then connected to an initially evacuated unknown volume Vs whereupon the pressure drops to an absolute value ong gt What is the ratio of VsVR in terms of P1 and P2 Ignore the volume of the connecting tubing What would the expression be if P1 and P2 were gauge pressures psig Problem 4 A piston pump containing 2 liters of nitrogen gas at 3000 psia absolute pressure empties its entire contents into a 10 liter pressure container Before the transfer of gas the 10 liter container already contained nitrogen gas at 100 psia absolute pressure Assuming nitrogen behaves as an ideal gas and assuming the temperature is the same throughout What is the nal pressure in the container T E0 9 Problem 5 A ship containing 704000 pounds of lique ed methane molecular weight 16 loses its refrigeration system just off the port of Elizabeth NJ and is forced to vent its entire cargo into the Winter air at T0 C and P21 atmosphere Assuming methane behaves as an ideal gas how many liters of gas are released How many cubic meters Suppose that released volume of gas occupies a hemispherical volume 12 of the volume of a sphere of radius r What would this radius be Note that lkilogram 22 lbs the gas constant R 082 atmlitersmole K You may remember that 1m3IO3 liters Problem 6 An idealized reservoir contains an ideal gas at a pressure of 2940 psia and a temperature of 91 C Under these conditions the volume ofgas in the reservoir is 109 What would be the volume of this gas at standard temperature and conditions 273 K and 1 atmos Assume 1 atmosphere equals 147 psia Problem 7 Employees ll canisters of butane 316 liters from a large 6000 liter storage tank lled with gaseous butane By What factor does the pressure in the tank fall every time a canister is lled Assume the canister pressure and the tank pressure are in equilibrium when the canister is removed after lling Assume butane behaves as an ideal gas and temperature is constant Problem 8 Using Excel do Problem 101 in Gottfried Print out you results and hand then in with the previous problems i A g 70 HS 30 H5 1 39 90 am 70 am WL 19V5M Fkv rpgv g PF VF 349 13 7v 7 70f373 f1 WIQM A W g C 2 2045 30 Mm 50 Mad 60 W6 30 cdm I LFOOJ W aAJ 2 VA 4 F3 V5 I V 5 inf J 20eo 20 30 4 rofwo 193175 2 P r am gx r 3 Z Passumc wag 602 SDWWE V gt vk 5 9 I M FIVK 1L V V5 gonad f OH 5 PZVLPLV W 7 quotP1252363 IQLVS 339 5 F Hg quot Pgl94 Vi CPw W j f fl zgwofgg V1 100 567 Fa 7 W 5MB W o 45 VFslo s 30675931 0005 PF 70mm Ho ch Ame gas bros QH 343 mazeth k 19 mm T507 P5 44 V rig v3 W W 000bsgt lg m g amp1033 22 a Lit 5 n 39100 5 gmmo m M16 7quot 5273le W 021 V RISK0 o lt 08317393 W I 5 m5 lisz QMS 05 1406 lm5 gt5 2mm 105 05111195 4 5 atworm L712 39 V loq 3 2 am 2223 ft 2 510 la fr qm ILf7fJM bmoc K W T5029 L oo o F sLHyo JZ m Fig HMO qu Iq r v T TL 5 L 1r v5 my rm PZVk V6 quot r 1 vprrnP T PPFVA 5 P15 Man W10 lkz rbq X Uu P i I V s 6000 9950 a Vn Wg 6316 630 3 96105 39 39 6631 03905 4 6039 HS glblb 7QLIq quot W0 6 S giEl 0915 66 Nltgtlt ltc IWDD39UOZ I 7lt Im39nrncn07gt 1187 ft 1187 ft 32 m2 158 mm 8 miles 400 g 171 oz 49 slugs 200 days 70 yrs 002 sec 648 N 383 Ibf 128 atm 07 atm 912 Pa 12 oz 85 L 1000 BTU 447 flb 102 kcal 800 Wh 200 HP 1000 MW 87 F 54 C 3617976 m 3617976 cm 496000992 in2 158000000 Angstroms 12374752 km 002740882 slugs 0106874984 Ibm 7150982735 kg 288000 min 2209032000 sec 555556E06 hr 1456761325 lbf 1703669026 N 1296959956 Pa 532 mmHg 0000900074 atm 0354959574 L 2244973958 gal 2521654885 cal thermo 0001785359 BTU 4270530743 1 6234358192 flb 1491402 w 134102006 HP 3037055556 K 1292 45de humm W w 1 In terms of LMt What is the dimension of the following physical variables a length L W b speed or velocitySEix c acceleration d area 1 3 e volume L 4 f density in 13 2 The uid in a cup sitting on the surface of a table exerts a force on the table given by the equation 39 i C F pgv Anya 6 quot cm A 4 exec Where p represents the uid density g the accelerationrdue to gravity and V the volume of the uid In terms of the fundamental dimensions LlLt What is the dimension of the right hand side of the equation What is the dimension of the force F F Ina 3 In the problem above the total force exerted by both uid and cup is given by F ng mg 39 Where in is the mass of the cup Write the dimensional equation for Ft in the following form x quot aquot k 4 3w er we loan lee lti 1 t 1 11 0 4 Inthe following 1 is distance F is force 11quot is heigh m is mass v is velocity V is volume P is pressure and g is the acceleration due to gravity Given the following de nitions of common variables Write their dimensions 1 L Work 2 force tunes distance traveled Work 2 N L EL zquot Potential Energy mgh RE 3 i l M 97 L 7 2 L Kinetic Energy I2 mv KB 7 d m L at 7 Torque force times leverarm Torque M L1 equot w J w quot LHS ideal gas law PV PV m L157 5 In the following expressions V is volume A is area t is time P is pressnre In is mass v is velocity p is density and g is acceleration due to gravity Indicate whether the expression is dimensionally correct or incorrect correct incorrect K 39 mt 32pVt PV 12mvZ mgVA Pv 32va IngV 52 4 03V m jam Hr 3N my 7 C Piaf I mail 2Q PJM 3 VIC Jim 3 L 15 WT mLtrl mm L3aa 213 ad 10 3mb39 31 4347 33 mtj 5 MIL yi 7 v0 2 MW 7 4 ijii 0 PE1305 Problem Set 2 Due Fridav Sept 14 2012 Perform the following conversions using the method demonstrated in class Use only the conversion factors shown below or shown in class Write out the entire conversion chain in detail The numerical answer is less important than showing the method by which you arrived at it 739quot 77 r 39 1 Convert 1187feetto meters 8 39klt M 39 we 2 Convert 1187 feet to centimeters quotreq q39 0 Him Z I 139 to 3 Convert 32 meters to inches 31 i L39m 39 39U8 Hit 4 Jv A 4 Convert 158 millimeters to angstroms iC BNQB ck 1m magma 3600 an WW 1 100m Q Wh lt 5 C 18 stat te 1 t lcil t rh39 M 009 me 7 onve u mtes o ome ers8 J K lgf qkm LHTO quot Hm I 9Mhb r I W K 6 Convert 400 grams to slugs 5 PRIME L W39J k l If 35 00 I J 7 Convert 171 square inches to square feet Q I 0 327111614 J K a 8 Convert 49 slugs to kilogramsli39l OCLe HVC it 150 E g 9 Convert 200 days to minutes 1335 ooom ukoi 1 7 7 7 10 Convert 70 years to seconds Q 90 S Q i w 7 11 Convert 02 seconds to hours1 53955 Xprquot Ci 12 Convert 648 Newtons to pounds force H 52 13 Convert 383 pounds force to newtons N J W7 7 39 86 l 7 14 Convert 128 atmospheres to newtons per square meter 11 q Fi gr 39 8 N m j 15 Convert 912 pascals to atmospheres I 8 01 13 2X 1 0 MWSf n t 61 1 3048 meters 1 mile 5280 feet 12 inches 1 foot 1 slug 9 32 lbf 1 kg 2201bf 1 angstrom 10quot0 meters 1 lbf 4448 Newtons 1 atmosphere 147 bfinz 1 lbfinz 68948 Newtonsm2 68948 Pascals Pa 9 WWu new NE 7 Ihgtv MOMS y gq v3 242M quot q AOL ow m 7 W r w ij M gt3 a 36 2 o 39 amp J 39 u Tag99woa H vU cgt lmx HM 7 lt7 y r A v 39 639OX0 osctjlt mmgw 3 IQBOM39PSN Hpr 124 Lay w 7 M 7 r 7 Sbamp M m M L a Iqqmm u gi jr W ahww Q Kiiw Egggw mvla9e a iwwi K fii i W Ra mm 115 MM i 39 7 watt PE 1305 Probl m Set 1 Due Jan 29 1 In terms of LMt what is the dimension of the following physical variables a length b speed or velocity 0 acceleration 1 area 6 volume f density 2 The uid sitting in a well exerts a force on the well oor given by the equation F ng Where p represents the uid mass density g the acceleration due to gravity and V the volume of the uid In terms of the fundamental dimensions LMt what is the dimension of the right hand side of the equation What is the dimension of the force F 3 If an additional mass is added to the uid column then the total force exerted by both uid and mass is given by Ft ng mg where m is the mass of the added object Write the dimensional equation for Ft in the following form l l l 4 In the following d is distance F is force h is heigh m is mass v is velocity V is volume P is pressure and g is the acceleration due to gravity Given the following de nitions of common variables write their dimensions Work force X distance traveled Work Potential Energy mgh PEl 7 Kinetic Energy 12 mv2 KE Torque force X leverarm Torque LHS ideal gas law PV PV W 5 In the following expressions V is volume A is area t is time P is pressure m is mass v is velocity p is density and g is acceleration due to gravity Indicate whether the expression is dimensionally correct or incorrect corr ct incorrect mt 32pVt PV 12mv2 mgVA PV 32mv2 mgV v rv F I 39ML39t397 7 ML JC39Z M a L39Jcquot ti 13 f f 3 e wa 3W Mmm Cquot owou LZ a W 3 LML3 CD 99D L39t z tlt L tnxf Ft W W V P E k Zj L ZK CMLl E Zj w f iE ML 4clt7i1EM L39JC 7 31LCML39 21 Workiht L EZZL w Aquot39 gt T 101211 i 39t 39 w CPWCMPI2L ljj i633 We 2 3aggtVt a EN lZBMvE rmqA 39 MLIW L3ML39t7 sz a w W BM mvHMgVquot E Lquotamp39ZEAL39wl ML39JcZ L3 y 39 M V quot I l 37 r I 1 i 1 1 E l

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