Tweet
Login
Mathematics Crystal
You may switch between
tex
and
pdf
by changing the end of the URL.
Home
About Us
Materials
Site Map
Questions and Answers
Skills
Topic Notes
HSC
Integration
Others
Tangent
UBC
UNSW
Calculus Advanced
Challenges
Complex Numbers
Conics
Differentiation
Integration
Linear Algebra
Mathematical Induction
Motion
Others
Polynomial Functions
Probability
Sequences and Series
Trigonometry
/
QnA /
UBC /
UBC MECH327 A4 Q5.tex
--Quick Links--
The Number Empire
Wolfram Mathematica online integrator
FooPlot
Calc Matthen
Walter Zorn
Quick Math
Lists of integrals
List of integrals of trigonometric functions
PDF
\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \usepackage{cancel} \begin{document} {\large UBC MECH327 Assignment 4 Q5} \begin{align*} &\text{Air enters an insulated compressor at 1.05 bar and 23$^\circ$C at rate of 1.8 kg/s and exits at 2.9 bar. Assuming that the}\\ &\text{compressor is operating at steady state, and that the changes in kinetic and potential energy can be neglected:}\\ \\ &\text{(a) What is the minimum theoretical power input required for compression and what is the associated outlet temperature?}\\ \\ &P_1=1.05\text{ bar},\quad T_1=23^\circ C=296\text{ K},\quad P_2=2.9\text{ bar}.\\ \\ &\text{With reference to Example 7-13 p.367 and the Isentropic formula 7-57a p.365, with $k=1.4$ and $(k-1)/k=0.2857$:}\\ &w_{comp,in}=\frac{kRT_1}{k-1}\left[\left(\frac{P_2}{P_1}\right)^{(k-1)/k}-1\right] =\frac{0.287\times 296}{0.2857}\left[\left(\frac{2.9}{1.05}\right)^{0.2857}-1\right] =100.134\text{ kJ/kg}.\\ \\ &\text{Also, }w_{comp,in}=\frac{kR(T_2-T_1)}{k-1},\quad T_2=\frac{w_{comp,in}(k-1)}{kR}+T_1=\frac{100.134\times 0.2857}{0.287}+296=395.68\text{ K}=123^\circ\text{C}.\\ \\ %&\text{First, find $T_2$ using the adiabatic relations 7-46 p.358:}\quad %T_1P_1^{(1-k)/k}=T_2P_2^{(1-k)/k},\quad %\text{where $k=1.4$ and $(1-k)/k=-0.2857$.}\\ %&T_2=T_1\cdot\left(\frac{P_1}{P_2}\right)^{(1-k)/k}=221.43\text{ K}=-51.569^\circ\text{C}.\\ \\ &\text{(b) If the isentropic compressor efficiency is $\eta_c$=80\%, what is the required power input and what is the outlet temperature?}\\ \\ &\eta_C=\frac{w_{comp,in}}{w_{actual,in}},\quad w_{actual,in}=\frac{w_{comp,in}}{\eta_C}=\frac{100.134}{0.80}=125.168\text{ kJ/kg}.\\ \\ &T_2=\frac{125.168\times 0.2857}{0.287}+296=420.60\text{ K}=148^\circ\text{C}.\\ \end{align*} \end{document}