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Please help me understand how V2  was determined. The problem's context and solution given by another expert is given below

Please help me understand how V2  was determined. The problem's context and solution given by another expert is given below

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Please help me understand how V2  was determined. The problem's context and solution given by another expert is given below,

With a 50 m^3 capacity, a tank contains steam at 400 degrees Celsius and 4500 kPa. From the tank, steam is vented through a relief valve to the atmosphere until the pressure in the tank falls to 3500 kPa. Assuming the venting process is adiabatic, estimate the steam's final temperature in the tank and the mass of steam vented. Griven that: Volume = V = 5om³ P, = 4500kfa T, = 400°c Solution: From steem table at 4500 KPe and 400°c s, = 6-7093 KJ/19.K and v, = 0-06475 m³/kg. proces is adiobatic them s, = S2 tham. S2 = 6- 7093 KJ lky.k B = 35 00 kla 6. 6626 350 |from tade find out → final Temperature mess of steam vented. we ne G- 7563 375 by interpoletion : (6.70 3 - 6. 6626) (6-7563- 6.6626) (245-310) = 3624°c T = 362-4°c Ans T = 350 + How was it From table at 3500 kle and 362-4°C determined? If a = 0.078 73 m/ky. mess of steam at -1 = known value, what so table from what m, O-064 75 reference (book)? m, = 772 kg mara of steam at so 0.07873 - 2 m2 = 635 kg. mess of steem vented = m, -m2 m = 772 - 635 m = 137 ky As
Transcribed Image Text:With a 50 m^3 capacity, a tank contains steam at 400 degrees Celsius and 4500 kPa. From the tank, steam is vented through a relief valve to the atmosphere until the pressure in the tank falls to 3500 kPa. Assuming the venting process is adiabatic, estimate the steam's final temperature in the tank and the mass of steam vented. Griven that: Volume = V = 5om³ P, = 4500kfa T, = 400°c Solution: From steem table at 4500 KPe and 400°c s, = 6-7093 KJ/19.K and v, = 0-06475 m³/kg. proces is adiobatic them s, = S2 tham. S2 = 6- 7093 KJ lky.k B = 35 00 kla 6. 6626 350 |from tade find out → final Temperature mess of steam vented. we ne G- 7563 375 by interpoletion : (6.70 3 - 6. 6626) (6-7563- 6.6626) (245-310) = 3624°c T = 362-4°c Ans T = 350 + How was it From table at 3500 kle and 362-4°C determined? If a = 0.078 73 m/ky. mess of steam at -1 = known value, what so table from what m, O-064 75 reference (book)? m, = 772 kg mara of steam at so 0.07873 - 2 m2 = 635 kg. mess of steem vented = m, -m2 m = 772 - 635 m = 137 ky As
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