Assignment 2 Solutions
.pdf
School
McGill University**We aren't endorsed by this school
Course
CIVE 290
Subject
Civil Engineering
Date
Dec 17, 2024
Pages
4
Uploaded by MajorInternet15662
1 Department of Civil Engineering & Applied Mechanics McGill University, Montreal, Quebec Canada CIVE 290 THERMODYNAMICS & HEAT TRANSFER Assignment #2 SOLUTIONS 1.2-lbm of water at 500 psia initially fill the 1.5-ft3left chamber of a partitioned system. The right chamber’s volume is also 1.5 ft3, and it is initially evacuated. The partition is now ruptured, and heat is transferred to the water until its temperature is 300 ºF. Determine the final pressure of water in psia, and the total internal energy, in Btu, at the final stage. Left chamber of a partitioned system contains water at a specified state while the right chamber is evacuated. The partition is now ruptured and heat is transferred to the water. The pressure and the total internal energy at the final state are to be determined. U2= 921 Btu (keep 3 significant digits)
2 2.One pound-mass of water fills a 2.649-ft3weighted piston-cylinder device at a temperature of 400 °F. The piston-cylinder device is now cooled in a constant-pressure process until its temperature is 100 °F. Determine the final pressure and volume of the water. Show this process on a P-υdiagram and on a T-υdiagram with respect to saturation lines. A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined.AnalysisThe initial specific volume is /lbmft649.2lbm1ft649.23311===mVvThis is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be 6E)-A(Table/lbmft649.2F40021311psia180===°=PPTvThe saturation temperature at 180 psia is 373.1°F. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation, 4E)-A(Table/lbmft01613.03F100@2==°fvvThe final volume is then 3ft0.0161===/lbm)ft01613.0)(lbm1(322vVm3.A 0.3-m3rigid vessel initially contains a saturated liquid-vapour mixture of water at 150°C. The water is now heated until it reaches the critical state. Determine the mass of the liquid water and the volume occupied by the liquid at the initial state. What can you conclude about the values of υfand υgat the critical state? Show this process on a T-υdiagram with respect to saturation lines. A rigid vessel that contains a saturated liquid-vapor mixture is heated until it reaches the critical state. The mass of the liquid water and the volume occupied by the liquid at the initial state are to be determined. AnalysisThis is a constant volume process (v= V /m= constant) to the critical state, and thus the initial P v2 1 H2O 400°F 1 lbm 2.649 ft3T v2 1
3 specific volume will be equal to the final specific volume, which is equal to the critical specific volume of water, /kgm0.003106321===crvvv(last row of Table A-4) The total mass is kg.6096/kgm0.003106m0.333===vVmAt 150°C, vf= 0.001091 m3/kg and vg= 0.39248 m3/kg (Table A-4). Then the quality of water at the initial state is 0.0051490.0010910.392480.0010910.00310611=−−=−=fgfxvvvThen the mass of the liquid phase and its volume at the initial state are determined from 3m0.105kg96.1====−=−=/kg)m91kg)(0.0010(96.1096.60)0.005149)((1)1(31ffftfmmxmvVAt the critical state, crgfvvv==4.One kilogram of R-134a fills a 0.14-m3weighted piston-cylinder device at a temperature of -26.4ºC. Determine the total enthalpy in this container. The container is now heated at constant pressure until the temperature is 100 °C. Determine the final volume and total enthalpy of the R-134a when the heating is completed. Show this process on a P-υdiagram with respect to saturation lines.A piston-cylinder device that is filled with R-134a is heated. The initial enthalpy, final volume, and final enthalpy are to be determined. AnalysisThe initial specific volume is /kgm14.0kg1m14.03311===mVvThis is a constant-pressure process. The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature 12)-A(TablekPa100C26.4-@sat 21===°PPPand kJ/kg94.174)16.217)(726.0(28.17726.0/kgm)0007259.019254.0(/kgm)0007259.014.0(113311=+=+==−−=−=fgffgfhxhhxvvvThe total enthalpy is then T vCP vcrH2O 150°C P v2 1 R-134a −26.4°C 1 kg 0.14 m3
4 175kJ===)kJ/kg94.174)(kg1(11mhHThe final state is superheated vapor and the specific volume and enthalpy are 13)-A(TablekJ/kg344.60h/kg,m30138.0C100kPa10023222==°==vTPThe final volume is then 3m0.301===/kg)m30138.0)(kg1(322vVmAnd the final enthalpy is kJ345===)kJ/kg60.344)(kg1(22mhH