3. The inner pipe of a double-pipe heat exchanger has an OD of 1.9 in. and contains 36 rectangular fins of the type shown in the sketch below. The fins are made of steel (k = 34.9 Btu/h ft.°F) and are 0.5 in. high and 0.035 in. thick. The pipe wall temperature is 250°F and the fluid in the annulus surrounding the fins is at 150°F with a heat-transfer coefficient of 30 Btu/h ft2 °F. Calculate: (a) The fin efficiency. (b) The rate of heat transfer from one fin per foot of pipe length. (c) The prime surface area per foot of pipe length. (d) The rate of heat transfer from the prime surface per foot of pipe length. (e) The weighted efficiency of the finned surface.
3. The inner pipe of a double-pipe heat exchanger has an OD of 1.9 in. and contains 36 rectangular fins of the type shown in the sketch below. The fins are made of steel (k = 34.9 Btu/h ft.°F) and are 0.5 in. high and 0.035 in. thick. The pipe wall temperature is 250°F and the fluid in the annulus surrounding the fins is at 150°F with a heat-transfer coefficient of 30 Btu/h ft2 °F. Calculate: (a) The fin efficiency. (b) The rate of heat transfer from one fin per foot of pipe length. (c) The prime surface area per foot of pipe length. (d) The rate of heat transfer from the prime surface per foot of pipe length. (e) The weighted efficiency of the finned surface.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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
Transcribed Image Text:3. The inner pipe of a double-pipe heat exchanger has an OD of 1.9 in. and contains 36
rectangular fins of the type shown in the sketch below. The fins are made of steel (k = 34.9
Btu/h ft °F) and are 0.5 in. high and 0.035 in. thick. The pipe wall temperature is 250°F and the
fluid in the annulus surrounding the fins is at 150°F with a heat-transfer coefficient of 30
Btu/h ft2 °F. Calculate:
(a) The fin efficiency.
(b) The rate of heat transfer from one fin per foot of pipe length.
(c) The prime surface area per foot of pipe length.
(d) The rate of heat transfer from the prime surface per foot of pipe length.
(e) The weighted efficiency of the finned surface.
(f) The total rate of heat transfer (from fins and prime surface) per foot of pipe length.
(g) The thermal duty for the exchanger is 390,000 Btu/h. What length of pipe is required to
satisfy this duty?
![1
Fin efficiency
m=
m =
A
b= 0.035 inf
.: Find M = |hp
кас
we
to = 250¹ F
Qin
30x 2.0058
34.9 x 2.9 166x10-3
24-31
Efficiency of Actual heat
=
fin
transfered by fin
Maximum heat
transfer by fin
if entire fin area
is at fin base
tip.
Can
fin
Omax
Here, fin formula used is
Beef in
our
case
af
20.
Per foot
pipe
=
=
=
$
Qmax per
foot A pipe
using
use insulated tip formulaes.
fin efficiency =
D = 1.9 in
Fins
36
K = 34-g Btu/h ft F
h heat transfer coefficient
= 30 Btu/h. ft2. of
P = perimeter of fin
= 2x (0.035 + L)
= 2x 2x (0.035 +12) in = 2006 75
= 258-74
ShpkAc ton h (m2) (to-ta)
ha eto-ta)
h A x 0.
=
= 30x 0.0862x
3
k Thermal conductivity of
fin
A₁ = Area of cross section
corrective length ie. 1= LA
2.472 x 0.7799 x 100
= g. 192-8 Btu/n
to = 250°F
tambent = 150°F
h = 30 Btu/h ft² po
=
for insulated fin tip.
Fin is of finite length a convecting heat
Fin
Q max
Shpk Ac ton h (mLc) (tota)
30 x2-0058 x 349x2.g1xi6³
[MA = 74.5 11
fin
ar bout
(250-150)
txL= 0.035 x 12 in 2
= 0.42 in 20
= 2-9166 x 103 f1²
=
5
0.42
= 0.5+
24-07
= 0.517 in = 0·040
Pt.
tanh (24-31x0.043)
* (250-150)
0o = to-ta
A total surface area
of fin
Px Le (Perimeter x height]
2.0058x 0.043
A fin
= 0·0862 f42
192.8
258-74
~ 75 1.
transfer from one fin per foot A Pipe
= 0.745](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd47fc2c3-8f4b-4085-b234-bc909be774e3%2F46afac52-c089-44f3-95c1-c40cb6836309%2F22ybidx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1
Fin efficiency
m=
m =
A
b= 0.035 inf
.: Find M = |hp
кас
we
to = 250¹ F
Qin
30x 2.0058
34.9 x 2.9 166x10-3
24-31
Efficiency of Actual heat
=
fin
transfered by fin
Maximum heat
transfer by fin
if entire fin area
is at fin base
tip.
Can
fin
Omax
Here, fin formula used is
Beef in
our
case
af
20.
Per foot
pipe
=
=
=
$
Qmax per
foot A pipe
using
use insulated tip formulaes.
fin efficiency =
D = 1.9 in
Fins
36
K = 34-g Btu/h ft F
h heat transfer coefficient
= 30 Btu/h. ft2. of
P = perimeter of fin
= 2x (0.035 + L)
= 2x 2x (0.035 +12) in = 2006 75
= 258-74
ShpkAc ton h (m2) (to-ta)
ha eto-ta)
h A x 0.
=
= 30x 0.0862x
3
k Thermal conductivity of
fin
A₁ = Area of cross section
corrective length ie. 1= LA
2.472 x 0.7799 x 100
= g. 192-8 Btu/n
to = 250°F
tambent = 150°F
h = 30 Btu/h ft² po
=
for insulated fin tip.
Fin is of finite length a convecting heat
Fin
Q max
Shpk Ac ton h (mLc) (tota)
30 x2-0058 x 349x2.g1xi6³
[MA = 74.5 11
fin
ar bout
(250-150)
txL= 0.035 x 12 in 2
= 0.42 in 20
= 2-9166 x 103 f1²
=
5
0.42
= 0.5+
24-07
= 0.517 in = 0·040
Pt.
tanh (24-31x0.043)
* (250-150)
0o = to-ta
A total surface area
of fin
Px Le (Perimeter x height]
2.0058x 0.043
A fin
= 0·0862 f42
192.8
258-74
~ 75 1.
transfer from one fin per foot A Pipe
= 0.745
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