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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
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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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
Transcribed Image Text: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
Transcribed Image Text: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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