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by: Peter Idenu

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# Thermodynamics Study Guide II ENGR 222

Marketplace > Louisiana Tech University > Applied Science > ENGR 222 > Thermodynamics Study Guide II
Peter Idenu
LA Tech
GPA 3.5
Thermodynamics
Dr.Moore

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This is a sample exam with solutions included
COURSE
Thermodynamics
PROF.
Dr.Moore
TYPE
Study Guide
PAGES
5
WORDS
KARMA
50 ?

## Popular in Applied Science

This 5 page Study Guide was uploaded by Peter Idenu on Wednesday September 30, 2015. The Study Guide belongs to ENGR 222 at Louisiana Tech University taught by Dr.Moore in Summer 2015. Since its upload, it has received 28 views. For similar materials see Thermodynamics in Applied Science at Louisiana Tech University.

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Date Created: 09/30/15
EN GR 222 THERMODYNAMICS Exam 1 1 5 points Identify each of the following system properties as being either intensive or extensive i volume extensive ii specific volume intensive iii temperature intensive iv pressure intensive v mass extensive 2 5 points Define the following terms i isobaric process a process that occurs at constant pressure ii closed system a system for which there is no mass transfer across the boundaries iii open system a system for which there is mass transfer across the boundaries iv work any energy exchange with a system that is not heat OR energy transfer associated with a force acting through a distance v isothermal process process that occurs at constant temperature 3 5 points A manufacturing process involves having a rigid container maintain a vacuum of 98 kPa If the local atmospheric pressure is 103 kPa what is the absolute pressure within the container If instead of a vacuum the container were pressurized to 80 kPa above the local atmospheric pressure what would be the gage pressure of the container a Pvac Patm Pabs gt Pabs Patm Pvac 103 kPa 98 kPa 5 kPa b Because the problem states that the tank is pressured to 80 kPa ABOVE Patm Pgage PabsPatm is not needed as Pgage 80 kPa is essentially given This just tests your understanding of the real meaning of gage pressure 4 5 points Can gaseous air typically be considered a pure substance Why or why not Yes Gaseous air is a mixture of a variety of gases but so long as it can be said to have a uniform homogeneous composition it can be considered a pure substance 5 10 points A womout motor that produces 150hp of shaft output has an efficiency of 88 It is replaced With a higher efficiency motor of the same output With 96 efficiency Find the reduction in heat being transferred to the room by installing the new motor at full load Pmotor lsmp noldmotor 88 nnewmotor 96 Qoldmotor Pmotor391 noldmotor Qoldmotor lshp Qnewmotor Pmotor391 nnewmotor Qnewmotor 611p 6 20 points A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 deg C A valve is opened and half of the mass is allowed to escape If the final pressure in the tank is 22 atm What is the final temperature of the tank m1 2 2kg P1 2 4atm T1 2 40 C In1 1112 7 n2 21kg P2 2 22atm Ideal gas from problem statement so PvRT and P1v1T1 P2v2T2 Rearranging give T2P2v2v1P1T1 For a rigid container if m205m1 then v22v1 and v2v1 2 This indicates that even though it wasn39t stated there was likely some additional heat added to the container in order for the pressure to be at 22 atm 7 20 points A 4 m3 rigid vessel contains H20 at 5 MPa and 1000 deg C Describe the phase of the substance and determine the mass and internal energy contained Within the vessel 3 Vol 2 4m PHZO 2 5MPa THZO 2 1000 C From Table A5 at 5 MPa 5000 kPa Tsat26394 deg G Since our temperature of 1000 C is greater than this the H20 is superheated vapor You could have also determined this using Table A6 From Table A6 at 5 MPA and 1000 deg C the specific volume of the H20 is 3 111 VHZO 2 kg mHZO VHZO 8 30 points 10 kg of Rl34a at 300 kPa fills a rigid container Whose volume is 14 L Determine the temperature and total enthalpy in the container The container is now heated until the pressure is 600 kPa Determine the temperature and total enthalpy When the heating is complete Vol 11134313 199 PR134a3 300dquot VR134a3 141 V VR134a 3 m From Table A12 at 300 kPa and the specific volume just determined the substance is a mixture of liquid and vapor but interpolation will be needed to use the table data 246 C 125 C 32kPa 28kPa T sat l 25 C 30kPa 28kPa Similarly interpolating for the specific volume values of saturated fluid and vapor 3 3 3 00007771 0000769711i vf 00007697n1 g g 30kPa 28kPa kg 32kPa 28kPa 3 4 m Vf 39 kg 3 0063681 0072434 m kg kg v 0072434 301ltPa 28kPa g kg 320kPa 28kPa 3 v O 11 g kg VR134a Vf x Vg Vf 3 X 93X 10 very small nearly all liquid k k 5514 J 5016 J k k k hf 5016 g g 300kPa 28kPa kg 32kPa 28kPa kJ hf 526 kg k k U 25193k 24977k h 24977 g g 30kPa 28kPa g kg 32kPa 28kPa k h 2508 g kg hR134az hf Xhg hf k h 545 R134a kg HR134a le34ahR134r Now the container is heated until the pressure is 600 kPa However there is no change in the mass within the container or its volume so the specific volume remains 1 4 10 3 m3 and based on the data in Table A12 at 600kPa the substance VR134a 39 X kg is still a saturated mixture However here there IS an entry for 600 kPa and no interpolation is necessary T2 Tsat at 600 kPa 2155 deg C 3 m v 00008198 stM kg 3 v 0034335Ii M8 kg k h 8150 VR134a Vf 2 vg vf x2 17gtlt 10 hRmahf XZhg hf kJ h 846 R134a kg 33 1111349111344 HRl34a 8464

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