This preview shows pages 1–6. Sign up to view the full content.
UNIVERSITY
OF
FLORIDA
1
Thermodynamics
∆
E = Q – W =
∆
U +
∆
KE +
∆
PE
300 gal
20psig
60degF
air
1 Hp motor
for 15 min.
500 btu/hr
What is the
temperature after 15
minutes?
Problem 1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document UNIVERSITY
OF
FLORIDA
2
Thermodynamics
Q – Wother Wmb =
∆
U +
∆
KE +
∆
PE
300 gal
20psig
60degF
air
1 Hp motor
for 15 min.
500 btu/hr
Assumptions
Ideal gas
KE is zero
PE is zero
Q= 500*15/60=
125 btu
Wmb = zero
Wother= 1Hp*2544btu/
hrHp*15/60hr=
636.1btu
UNIVERSITY
OF
FLORIDA
3
Thermodynamics
125 – (636.1)0=
∆
U +
0
+
0 =
m*Cv(T2T1)
300 gal
20psig
60degF
air
1 Hp motor
for 15 min.
500 btu/hr
P1V1=mRT
m= P1V1/RT
P1=
20+14.7=34.7psia
V1=300gal*1
ft3/7.48gal= 40,1ft3
T1= 60+460=520R
R = 53.34ftlbf/lbm
R
(pg 840)
m= 7.22lbm
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document UNIVERSITY
OF
FLORIDA
4
Thermodynamics
125 – (636.1)0=
∆
U +
0
+
0 =
7.22*0.171(T2520)
Therefore T2= 933R or 474F
300 gal
20psig
60degF
air
1 Hp motor
for 15 min.
500 btu/hr
Cv = 0.171btu/lbm R
Page 840
UNIVERSITY
OF
FLORIDA
5
Thermodynamics
Q – Wother Wmb =
∆
U +
∆
KE +
∆
PE
110V
0.3A for
10min
Piston cylinder contains
0.2kg steam at 200kPa and
400deg C
What is the initial volume
and final temp. after 10min.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 08/03/2011 for the course EML 3100 taught by Professor Sherif during the Summer '08 term at University of Florida.
 Summer '08
 Sherif

Click to edit the document details