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In a pilot unit, illustrated in Figure Q3, an air blower supplies 172 g mol of air/min to a reactor. The blower uses 1.2 kW. The gas enters at a temperature of 307 K and a pressure of pl. Determine the final temperature of the gas assuming the process is adiabatic, and that the blower is 100% efficient. Given R = 8.314 Pa. m/ (mol. K), and Cp = 3.5x R

In a pilot unit, illustrated in Figure Q3, an air blower supplies 172 g mol of air/min to a reactor. The blower uses 1.2 kW. The gas enters at a temperature of 307 K and a pressure of pl. Determine the final temperature of the gas assuming the process is adiabatic, and that the blower is 100% efficient. Given R = 8.314 Pa. m/ (mol. K), and Cp = 3.5x R

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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In a pilot unit, illustrated in Figure Q3, an air blower supplies 172 g mol of air/min to a reactor. The blower uses 1.2 kW. The gas enters at a temperature of 307 K and a pressure of pl. Determine the final temperature of the gas assuming the process is adiabatic, and that the blower is 100% efficient. Given R = 8.314 Pa. m/ (mol. K), and Cp= 3.5× R
Transcribed Image Text:In a pilot unit, illustrated in Figure Q3, an air blower supplies 172 g mol of air/min to a reactor. The blower uses 1.2 kW. The gas enters at a temperature of 307 K and a pressure of pl. Determine the final temperature of the gas assuming the process is adiabatic, and that the blower is 100% efficient. Given R = 8.314 Pa. m/ (mol. K), and Cp= 3.5× R
Air 172 g mol/min T= Tf p = P1 Air AH= 988J/g mol 172 g mol/min = 307K p = P1 Power = 1.2 kW (100% efficient) Figure Q3
Transcribed Image Text:Air 172 g mol/min T= Tf p = P1 Air AH= 988J/g mol 172 g mol/min = 307K p = P1 Power = 1.2 kW (100% efficient) Figure Q3
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