pl help :( 2. 18,000 lb/h of a petroleum fraction are to be cooled from 250°F to 150°F using cooling waterwith a range of 85-120°F. Properties of the petroleum fraction may be assumed constant at thefollowing values:PropertyCp (Btu/lbm. °F)k (Btu/h ft. °F)& (cp)p (lbm/ft³)A shell-and-tube heat exchanger with the following configuration is available:Type: AESTubes: 3/4 in. OD, 16 BWG, 20 ftlongNumber of tubes: 128Number of tube passes: 4Tube pitch: 1.0 in. (square)Tube material: Admirality brass (k =64 Btu/h ft °F)Step 1Shell ID: 15.25 in.Baffle type: segmentalBaffle cut: 20%Number of baffles: 50Shell material: carbon steelFouling factors of 0.001 and 0.002 h ft² °F/Btu are required for the cooling water and petroleumfraction, respectively. Is the heat exchanger thermally suitable for this service?Q = mcpATm =Value0.520.0742.7551.218,00036005lb Step 2EquationsThe inside heat transfer coefficient (hi) can be estimated using the Dittus-Boelter correlation:Step 3where it is the number of tubes, Do is the outside diameter of the tubes in inches, and Np is the tube pitch in inches.The viscosity of water at the average temperature of 102.5°F can be found from standard tables to be approximately:The Prandtl number can be assumed to be approximately 4 for both the petroleum fraction and the cooling water.Substituting these values into the Dittus-Boelter correlation, we get:ਘਟU₂ =1/7 = 1/1 + R²₁m =h₂D₁AT = 250-150 = 100°Flb 0.52BtuQ = 5 +100°F =*S lbm³FQ = mCpAT1618,000 5lbS3600BtuCppetro = 0.52 Ibm°F·+R+R+SDoin ( + 1 + RaDa +D₁2k1=R₂ = 5-R₂ =N₁ = 0.023 Rep0.4& R₁₁1 =₁₁A₁mDμD₁ (4128-260Btu11;hath Ry=h₂Aofeh. AufeD₁, is hydraulic diameter of the tubes. D₁ =Mwater = 2.6.lb 0.625ft2.6 1050.62512(+1)2+Roo10-5lbmft.s4.Nt ((π + N₂)=Tom 1.2 10%Ibmft.sN₁ = 0.023+ (1.2 в 106)ª s 4⁰4 = 2,923.457N₁ = 0.023 Repa

2. 18,000 lb/h of a petroleum fraction are to be cooled from 250°F to 150°F using cooling water with a range of 85-120°F. Properties of the petroleum fraction may be assumed constant at the following values: Property Cp (Btu/lbm - °F) k (Btu/h - ft. °F) & (cp) p (lbm/ft³) Step 1 A shell-and-tube heat exchanger with the following configuration is available: Type: AES Tubes: 3/4 in. OD, 16 BWG, 20 ft long Number of tubes: 128 Shell ID: 15.25 in. Baffle type: segmental Baffle cut: 20% Number of baffles: 50 Shell material: carbon steel Fouling factors of 0.001 and 0.002 h-ft²-F/Btu are required for the cooling water and petroleum fraction, respectively. Is the heat exchanger thermally suitable for this service? Value 0.52 0.074 2.75 51.2 Number of tube passes: 4 Tube pitch: 1.0 in. (square) Tube material: Admirality brass (k = 64 Btu/h. ft. °F) Q = mcpAT lb 18,000 75lb 3600 m = S

2. 18,000 lb/h of a petroleum fraction are to be cooled from 250°F to 150°F using cooling water with a range of 85-120°F. Properties of the petroleum fraction may be assumed constant at the following values: Property Cp (Btu/lbm - °F) k (Btu/h - ft. °F) & (cp) p (lbm/ft³) Step 1 A shell-and-tube heat exchanger with the following configuration is available: Type: AES Tubes: 3/4 in. OD, 16 BWG, 20 ft long Number of tubes: 128 Shell ID: 15.25 in. Baffle type: segmental Baffle cut: 20% Number of baffles: 50 Shell material: carbon steel Fouling factors of 0.001 and 0.002 h-ft²-F/Btu are required for the cooling water and petroleum fraction, respectively. Is the heat exchanger thermally suitable for this service? Value 0.52 0.074 2.75 51.2 Number of tube passes: 4 Tube pitch: 1.0 in. (square) Tube material: Admirality brass (k = 64 Btu/h. ft. °F) Q = mcpAT lb 18,000 75lb 3600 m = S

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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Concept explainers

Heat Exchangers

Heat exchangers are the types of equipment that are primarily employed to transfer the thermal energy from one fluid to another, provided that one of the fluids should be at a higher thermal energy content than the other fluid.

Heat Exchanger

The heat exchanger is a combination of two words ''Heat'' and ''Exchanger''. It is a mechanical device that is used to exchange heat energy between two fluids.

Question

pl help :(

2. 18,000 lb/h of a petroleum fraction are to be cooled from 250°F to 150°F using cooling water
with a range of 85-120°F. Properties of the petroleum fraction may be assumed constant at the
following values:
Property
Cp (Btu/lbm. °F)
k (Btu/h ft. °F)
& (cp)
p (lbm/ft³)
A shell-and-tube heat exchanger with the following configuration is available:
Type: AES
Tubes: 3/4 in. OD, 16 BWG, 20 ft
long
Number of tubes: 128
Number of tube passes: 4
Tube pitch: 1.0 in. (square)
Tube material: Admirality brass (k =
64 Btu/h ft °F)
Step 1
Shell ID: 15.25 in.
Baffle type: segmental
Baffle cut: 20%
Number of baffles: 50
Shell material: carbon steel
Fouling factors of 0.001 and 0.002 h ft² °F/Btu are required for the cooling water and petroleum
fraction, respectively. Is the heat exchanger thermally suitable for this service?
Q = mcpAT
m =
Value
0.52
0.074
2.75
51.2
18,000
3600
5lb

Step 2
Equations
The inside heat transfer coefficient (hi) can be estimated using the Dittus-Boelter correlation:
Step 3
where it is the number of tubes, Do is the outside diameter of the tubes in inches, and Np is the tube pitch in inches.
The viscosity of water at the average temperature of 102.5°F can be found from standard tables to be approximately:
The Prandtl number can be assumed to be approximately 4 for both the petroleum fraction and the cooling water.
Substituting these values into the Dittus-Boelter correlation, we get:
ਘਟ
U₂ =
1/7 = 1/1 + R²₁
m =
h₂D₁
AT = 250-150 = 100°F
lb 0.52Btu
Q = 5 +100°F =
*
S lbm³F
Q = mCpAT
16
18,000 5lb
S
3600
Btu
Cppetro = 0.52 Ibm°F
·+R+R+
S
Doin ( + 1 + RaDa +
D₁
2k
1
=
R₂ = 5-
R₂ =
N₁ = 0.023 Rep0.4
& R₁₁1 =₁₁A₁
mD
μ
D₁ (4128-
260Btu
1
1
;
hath Ry=h₂Aofe
h. Aufe
D₁, is hydraulic diameter of the tubes. D₁ =
Mwater = 2.6.
lb 0.625ft
2.6 105
0.625
12
(+1)
2
+Roo
10-5lbm
ft.s
4.Nt (
(π + N₂)
=
Tom 1.2 10%
Ibm
ft.s
N₁ = 0.023+ (1.2 в 106)ª s 4⁰4 = 2,923.457
N₁ = 0.023 Repa

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