TAMU CHEN 205 - extra-sol234 (6 pages)

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extra-sol234

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6
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Texas A&M University
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Chen 205 - Chem Engr Thermo I
Chem Engr Thermo I Documents
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ME 24 221 THERMODYNAMICS I Solutions to extra problems in Chapters 2 3 and 4 December 5 2000 J Murthy 2 25 Two reservoirs A and B open to the atmosphere are connected with a mercury manometer Reservoir A is moved up down so the two top surfaces are level at h3 as shown in Fig P2 25 Assuming that you know A Hg and measure the heights h1 h2 and h3 find the density B Solution Balance forces on each side P0 Ag h3 h2 Hggh2 P0 Bg h3 h1 Hggh1 h 3 h2 h 2 h1 B A Hg h 3 h1 h 3 h1 2 32 Two piston cylinder arrangements A and B have their gas chambers connected by a pipe Cross sectional areas are AA 75 cm2 and AB 25 cm2 with the piston mass in A being mA 25 kg Outside pressure is 100 kPa and standard gravitation Find the mass mB so that none of the pistons have to rest on the bottom Solution P0 P0 Force balance for both pistons A mPAg P0AA PAA B A B F F mPBg P0AB PAB Same P in A and B gives no flow between them mPBg mPAg P0 P0 AA AB mPB mPA AA AB 25 25 75 8 33 kg 2 36 Two cylinders are connected by a piston as shown in Fig P2 36 Cylinder A is used as a hydraulic lift and pumped up to 500 kPa The piston mass is 25 kg and there is standard gravity What is the gas pressure in cylinder B Solution Force balance for the piston PBAB mpg P0 AA AB PAAA AA 4 0 12 0 00785 m2 AB 4 0 0252 0 000491 m2 PBAB PAAA mpg P0 AA AB 500 0 00785 25 9 807 1000 100 0 00785 0 000491 2 944 kN PB 2 944 0 000491 5996 kPa 6 0 MPa 4 21 A cylinder having an initial volume of 3 m3 contains 0 1 kg of water at 40 C The water is then compressed in an isothermal quasi equilibrium process until it has a quality of 50 Calculate the work done in the process Assume the water vapor is an ideal gas Solution C V Water v1 V1 m T PG 40 oC 3 P1 Tbl B 1 1 PG 7 384 kPa very low so H2O ideal gas from 1 2 1 2 P 1 PG 2 PdV P1V1ln W12 1 vG v1 7 384 19 52 4 8 kPa 30 V2 mv2 0 1 19 52 1 952 m3 v T C 3 30 m3 kg vG 0 1 V2 V1 4 8 3 0 ln 1 952 6 19 kJ 3 v3 0 001008 0 5 19 519 9 7605 V3 mv3 0 976 m3 3 P C P g W23 PdV Pg V3 V2 7 384 0 976 1

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