ME449_OConnor_Presentation
- 1. Fan Speed Design Project Design Project for ME 449 By Dillon O’Connor 1
- 2. Duct System Technical Details The given dimensions are shown on in the Table 1. Other relevant information: 1) Duct recirculates air within the house 2) The ducts are made of galvanized iron sheeting 3) Outlet velocities may not be the same. 4) Outlet velocities should be close to 25 ft/s without exceeding that amount. Description Cross Section Length Duct 1 24 in. X 18 in. 25 ft 90o Turns (1 to 2; and 1 to 3) - - Duct 2 18 in. X 12 in. 12 ft Duct 3 12 in. X 12 in. 7 ft 90o Turns (2 to 4; and 2 to 5) - - Duct 4 w/ one 90o Turn 8 in. diameter (circular) 6 ft Duct 5 w/ one 90o Turn 8 in. diameter (circular) 6 ft 90o Turns (3 to 6; and 3 to 7) - - Duct 6 w/ one 90o Turn 8 in. diameter (circular) 6 ft Duct 7 w/ one 90o Turn 8 in. diameter (circular) 6 ft Return Duct w/ one 90o Turn 24 in. X 18 in. 22 ft Table 1: Given duct system dimensions 2
- 3. Duct Schematic • Figure 1 shows an isometric view of the system. • Figure 2 shows a dimensioned top down view (left) and a dimensioned side view (right) of the system. Figure 1. Isometric view of the duct system. Figure 2. Dimensioned Schematic of the duct system. 3
- 4. Fan Curve • The fan has already been selected. • However the required operating speed needs to be determined. • The fan curve for the given fan has been provided in Figure 3. • The curve fit equation to this fan curve will be used in later calculations. Figure 3. Fan curve for the given system’s fan. 4
- 5. Circuit Diagram • Figure 4 shows the equivalent circuit diagram for the system. • Major loss sections are shown as long rectangular boxes. • Minor loss sections are shown as short rectangular boxes. • Each will have a hf that contributes to the resulting pressure loss. Figure 4. Equivalent circuit diagram of the duct system. 5
- 6. Minor Loss Coefficients The minor loss coefficients used are shown on in Table 2. Tee and Elbow K values are taken from Table 6.5 of the Fluid Mechanics (5th edition) text by F. M. White. Flanged connections were assumed. K values for sudden expansion and sudden contraction come from equations 6.80 and 6.81 respectively from the same textbook. 𝐾 = 1 − 𝑑2 𝐷2 2 𝐾 = 0.42 1 − 𝑑2 𝐷2 Duct Minor Loss Coefficient Value Description K1 0.41 Tee (d = 20 in.) K2 0.7035 Sudden Contraction (d1 to d2) + Tee (d = 14.4 in.) K3 0.8004 Sudden Contraction (d1 to d3) + Tee (d = 12 in.) K4 1.5504 Sudden Contraction (d2 to d4) + Elbow (90 degree, regular, d = 8 in.) + Sudden Expansion (d4 to large d) K5 1.5504 Sudden Contraction (d2 to d5) + Elbow (90 degree, regular, d = 8 in.) + Sudden Expansion (d5 to large d) K6 1.4933 Sudden Contraction (d3 to d6) + Elbow (90 degree, regular, d = 8 in.) + Sudden Expansion (d6 to large d) K7 1.4933 Sudden Contraction (d3 to d7) + Elbow (90 degree, regular, d = 8 in.) + Sudden Expansion (d7 to large d) K8 0.63 Sudden Contraction (large d to d8) + Elbow (d = 20 in.) Table 2: Minor loss coefficients (6.80) (6.81) 6
- 7. Constants • The other constants used in this design are shown in Table 3. • Air properties, including density, were assumed constant throughout the duct system. • Air properties were evaluated at standard room temperature (70 ºF) and pressure (1 bar). • Acceleration due to gravity was taken at sea level. symbol Description Value g Acceleration due to gravity 32.2 ft/s² ε Duct roughness 0.0005 ft ν Kinematic viscosity of air 1.64e-4 ft²/s γ Specific weight of air 7.492e-2 lb/ft³ Table 3: Essential constants. 7
- 8. Equations (1st Part) 𝑓 = 1 −1.8𝑙𝑜𝑔 6.9 𝑅𝑒 𝐷 + 𝜀/𝐷ℎ 3.7 1.11 2 ℎ 𝑓,𝑚𝑎𝑗𝑜𝑟 = 𝑓 𝐿 𝐷ℎ 1 2𝑔𝐴2 𝑄2 ℎ 𝑓,𝑚𝑖𝑛𝑜𝑟 = 𝐾 1 2𝑔𝐴2 𝑄2 ℎ 𝑓 = ℎ 𝑓,𝑚𝑎𝑗𝑜𝑟 + ℎ 𝑓,𝑚𝑖𝑛𝑜𝑟 ∆𝑃 = 𝛾ℎ 𝑓 𝑇𝑃𝑡𝑜𝑝 𝑙𝑒𝑔 = ∆𝑃1 + ∆𝑃2 + ∆𝑃4 + ∆𝑃8 𝑇𝑃𝑏𝑜𝑡𝑡𝑜𝑚 𝑙𝑒𝑔 = ∆𝑃1 + ∆𝑃3 + ∆𝑃7 + ∆𝑃8 𝐷ℎ = 4 ∙ 𝐴𝑟𝑒𝑎 𝑓𝑙𝑜𝑤 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑒𝑡𝑡𝑒𝑑 𝑅𝑒 𝐷 = 𝑄𝐷ℎ 𝐴𝜈 8
- 9. Method • Guess a Q1 value. • Calculate total pressure lost through the top line for Q2 = 0 to 100% of Q1. • Calculate total pressure lost through the bottom line for Q3 = 0 to 100% of Q1. • Plot ΔPtop leg vs. increasing% of Q1 and ΔPbottom leg vs. decreasing % of Q1. • Independent axis value at intercept = % of Q1 to Q2. • Q2 = %*Q1 and Q3 = (1oo-%)*Q1. • Q4 = Q5 = 0.5*Q2 and Q6 = Q7 = 0.5*Q3 • Intercept can be better tuned by comparing resulting ΔPtop leg and ΔPbottom leg values until they are equivalent. • Find exit velocities using v = Q/A • Change Q1 until max v (of v4-v7) is close to 25 ft/s 9
- 10. Q1 And Where It Goes • A Q1 guess of 34 ft³/s yielded the plot shown in Figure 5. • The intercept in this case can be seen at about 51% of Q1 to Q2 and the remaining 49% of Q1 to Q3. • Fine tuning the % by ΔP values yields a 51.26% of Q1 to Q2. The remaining 48.74% of Q1 goes to Q3. Figure 5. Plot of losses in each line as a function of decimal fraction of Q1 that travels into Q2 for a guessed Q1 of 34 ft³/s. 10
- 11. Verifying Accuracy Of Guess • A guess of Q1 = 34 ft³/s yields a maximum exit velocity of 24.96 ft/s in ducts 4 and 5. • The lesser exit velocity is 23.74 ft/s in ducts 6 and 7. • This satisfies the given criteria of an exit velocity close to 25 ft/s without exceeding it. • The resulting TP = ΔP = 0.2761 in.H2O 11
- 12. Equations (2nd Part) 𝑄1 = 𝛼 𝑇𝑃2 1 𝑄2 2 2 + 𝛽 𝑇𝑃2 𝑁2 = 𝑄2 𝑄1 𝑁1 For these equations: • Q2 is our Q1 from the previous slide converted to ft³/min. • TP2 is our TP from the previous slide. • N2 is the operating speed to achieve it. • Q1 is the Q at N=2000 RPM as shown on the Fan Curve. • N1 = 2000 RPM • α = 5.95 in.H2O • β = 8.5e-7 (in.H2O/CFM²) 12
- 13. Final Results • The required Q1 = 2040 cfm • The required N = 1607 RPM • The resulting TP = ΔP = 0.2761 in.H2O • The maximum exit velocity is 24.96 ft/s in ducts 4 and 5. • The lesser exit velocity is 23.74 ft/s in ducts 6 and 7. 13
- 14. REFERENCES White, F.M., Fluid Mechanics, 5th Edition, McGraw-Hill, New York, 2003 Department of Mechanical, Aerospace, and Biomedical Engineering, ME 449 A Preliminary Design Problem, University of Tennessee, Knoxville, 2015 14