heat loss t
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
Related questions
Question
![Solving for A's,
A1 - Aį
A, - A1
A = 2nr,L;
Am,tm
Ains,Im
A, = 2nr,L ;
A = 2nr,L
%3D
In ()
In )
(0.412
A = 2nrL
(1 ft) = 0.2152 ft²
12
A:=
Ao =
Am,im =.
Ains,Im
Solving for R's,
Rm
ro - ri
kinsAins,tm
Rins
R.
h,A.
T – To
267°F – 80°F
q =
_Btu/hr
Rị + Rm + Rins + R,
b. Using Ui and A; in equation (3.9)
q = UĻA{(T{ – To)
1
Ut
A ER
_Btu/hr ft²°F
Then,
q = ULA{(T{ – To) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F426042b7-854b-4316-8bab-10794fa80326%2Fa9572b6b-a86d-4e7a-a2bf-78a9a58221be%2Fqnc9mv_processed.png&w=3840&q=75)
Transcribed Image Text:Solving for A's,
A1 - Aį
A, - A1
A = 2nr,L;
Am,tm
Ains,Im
A, = 2nr,L ;
A = 2nr,L
%3D
In ()
In )
(0.412
A = 2nrL
(1 ft) = 0.2152 ft²
12
A:=
Ao =
Am,im =.
Ains,Im
Solving for R's,
Rm
ro - ri
kinsAins,tm
Rins
R.
h,A.
T – To
267°F – 80°F
q =
_Btu/hr
Rị + Rm + Rins + R,
b. Using Ui and A; in equation (3.9)
q = UĻA{(T{ – To)
1
Ut
A ER
_Btu/hr ft²°F
Then,
q = ULA{(T{ – To) =
![PDE 2. CONDUCTION HT.pdf
PDE 3. CONVECTION HEAT TRANSFEF X
O File | C:/Users/Kayleinzer/Desktop/3rd%20year%201st%20sem/CHE%20154/3.%20CONVECTION%20HEAT%20TRANSFER.pdf
(D Page view A Read aloud
V Draw
E Highlight
4
of 13
Erase
Thus,
T1- T4
q = UțA;(T1 – T4) = U,A,(T1– T4) =
ΣR
(3.9)
U can be based on inside area or outside area of the tube
Where:
1
Uj =
(3.10)
(ro - ri)A¡
+
hị
A;
1
U, =
(r. – ri)A.
hịA
1
+
(3.11)
ho
Example:
Saturated steam at 267°F is flowing inside a %-in steel pipe having an ID of 0.824 in. and an OD of
1.050 in. The pipe is insulated with 1.5 in of insulation on the outside. The convective coefficient for the
inside steam surface of the pipe is estimated as h; = 1000 Btu/hr-ft2 °F, and the convective coefficient on
the outside of the lagging is estimated as ho = 2 Btu/hr-ft2 °F. The mean thermal conductivity of the metal
is 26 Btu/hr-ft °F and 0.037 Btu/hr-ft °F for the insulation
a. Calculate for the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F
b. Repeat using the overall U¡ based on the area A
Given:
To= 80°F
Ti = 267°F
Dipipe = 0.824 in →r; = 0.412 in
Do,pipe = 1.050 in→rį = 0.525 in
ins= 2.025"
insulation
r,-0.525"
rins = 1.5 in + 0.525 in = 2.025 = ro
-1.5"-
h; = 1000 Btu/hr-ft2 °F
ho = 2 Btu/hr-ft² °F
km = 26 Btu/hr-ft °F
Kins = 0.037 Btu/hr-ft °F
I = 267 F
n= 0.412"
Peguired. a for - 1ft at I
200E:aucina baces on A
11:05 AM
0 Type here to search
Po
W
24°C Rain showers
9/23/2021
21](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F426042b7-854b-4316-8bab-10794fa80326%2Fa9572b6b-a86d-4e7a-a2bf-78a9a58221be%2Fe1e3ec_processed.png&w=3840&q=75)
Transcribed Image Text:PDE 2. CONDUCTION HT.pdf
PDE 3. CONVECTION HEAT TRANSFEF X
O File | C:/Users/Kayleinzer/Desktop/3rd%20year%201st%20sem/CHE%20154/3.%20CONVECTION%20HEAT%20TRANSFER.pdf
(D Page view A Read aloud
V Draw
E Highlight
4
of 13
Erase
Thus,
T1- T4
q = UțA;(T1 – T4) = U,A,(T1– T4) =
ΣR
(3.9)
U can be based on inside area or outside area of the tube
Where:
1
Uj =
(3.10)
(ro - ri)A¡
+
hị
A;
1
U, =
(r. – ri)A.
hịA
1
+
(3.11)
ho
Example:
Saturated steam at 267°F is flowing inside a %-in steel pipe having an ID of 0.824 in. and an OD of
1.050 in. The pipe is insulated with 1.5 in of insulation on the outside. The convective coefficient for the
inside steam surface of the pipe is estimated as h; = 1000 Btu/hr-ft2 °F, and the convective coefficient on
the outside of the lagging is estimated as ho = 2 Btu/hr-ft2 °F. The mean thermal conductivity of the metal
is 26 Btu/hr-ft °F and 0.037 Btu/hr-ft °F for the insulation
a. Calculate for the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F
b. Repeat using the overall U¡ based on the area A
Given:
To= 80°F
Ti = 267°F
Dipipe = 0.824 in →r; = 0.412 in
Do,pipe = 1.050 in→rį = 0.525 in
ins= 2.025"
insulation
r,-0.525"
rins = 1.5 in + 0.525 in = 2.025 = ro
-1.5"-
h; = 1000 Btu/hr-ft2 °F
ho = 2 Btu/hr-ft² °F
km = 26 Btu/hr-ft °F
Kins = 0.037 Btu/hr-ft °F
I = 267 F
n= 0.412"
Peguired. a for - 1ft at I
200E:aucina baces on A
11:05 AM
0 Type here to search
Po
W
24°C Rain showers
9/23/2021
21
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