Week 2 Dimensional Analysis as applied to Fluid Machineries
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02 Week 2: Dimensional Analysis as applied to Fluid Machineries FME17 - Fluid Machineries Technological University of the Philippines Engr. Jay Mark P. Delos Reyes 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Similitude Geometric, kinematic, and dynamic qualities are defined in terms of the recognized definitions of similarity. Specifically, If all physical or bodily dimensions in all three axes (Cartesian) have the same linear ratio, a model and prototype are geometrically identical. For our objectives, all aspects are considered. It is necessary to "scale" flow directions, orientations, and even surface roughness qualities. Time similarity introduces components like velocity and acceleration, as in a fluid flow concern. Be aware that geometric similarity must exist for there to be similitude in a kinematic sense. Kinematic similarity leads to requirements in Reynolds number, Mach number, and perhaps Froude number in the context of fluid mechanics. The maintaining of Reynolds, Mach, and Froude numbers is often necessary. Furthermore, Weber and Cavitation numbers could be needed. When comparing the behavior of entities or machines of different "scales" (size, speed, fluid characteristics), it is often essential to maintain exact similarity. 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Turbo Machineries For Liquid • Pump Pump; adds energy to a fluid, resulting in an increase in pressure across the pump. 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Turbo Machineries For Liquid • Turbine Turbine; extracts energy from the fluid, resulting in a decrease in pressure across the turbine. 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Turbo Machineries For Gas • Fans Fans; Low pressure gradient, High volume flow rate. Examples include ceiling fans and propellers. 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Turbo Machineries For Gas • Blowers Blower; Medium pressure gradient, Low volume flow rate. Examples include centrifugal and squirrel-cage blowers found in Furnaces, leaf blowers and hair dryers. 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Turbo Machineries For Gas • Compressors Compressor; High pressure gradient, Low volume flow rate. Examples include air compressor for refrigerators and air conditioner 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Dimensionless Performance using Buckingham π Theorem Buckingham π Theorem – is a mathematical method that enables the development of a relationship between the model and the actual situation for a quantity of interest. If there are π dimensional variable in a dimensionally homogenous equation, described by π fundamentals dimensions, they may be grouped in π = π − π dimensionless group. π → π·πππππ ππππππ π πΊπππ’π 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Dimensionless Performance using Buckingham π Theorem The relationship among the dimensionless group will be written: π π1 , π2 , π3 , … . . ππ = 0 π1 = π π2 , π3 , … . . ππ π2 = π π1 , π3 , … . . ππ π3 = π π1 , π2 , … . . ππ } πππ‘βπππ ππ ππππππ‘πππ π£ππππππππ 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: Drag on Sphere 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 1: List the parameters in the problem Parameters Symbol Drag Force πΉ Sphere Diameter π· Fluid Viscosity π£ Fluid Density π Fluid Viscosity π 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 2: List the unit of parameters Parameters Symbol Unit Drag Force πΉ ππ π /π 2 Sphere Diameter π· π Fluid Viscosity π£ π/π Fluid Density π ππ/π3 Fluid Viscosity π ππ ππ 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 2: List the unit of parameters Parameters Symbol Unit Drag Force πΉ ππ π /π 2 Sphere Diameter π· π Fluid Viscosity π£ π/π Fluid Density π ππ/π3 Fluid Viscosity π ππ ππ πππ π π ≈ ππ ; πΏππππ‘β πΏ ; π‘πππ (π‘) 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 3: Calculate the Dimensionless Group Parameters Symbol Unit Repeating Variable (j) Drag Force πΉ ππ π /π 2 ππΏπ‘ −2 Sphere Diameter π· π πΏ Fluid Viscosity π£ π/π πΏπ‘ −1 Fluid Density π ππ/π3 ππΏ−3 Fluid Viscosity π ππ ππ ππΏ−1 π‘ −1 πππ π π ≈ ππ ; πΏππππ‘β πΏ ; } π‘πππ (π‘) π=3 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 3: Calculate the Dimensionless Group Parameters Symbol Unit Repeating Variable (j) Drag Force πΉ ππ π /π 2 ππΏπ‘ −2 Sphere Diameter π· π πΏ π‘βππ‘ πππππ π‘βπππ πππ 2 π·πππππ ππππππ π ππππ’π −1 Fluid Viscosity π£ π/π Fluid Density π ππ/π3 ππΏ−3 Fluid Viscosity π ππ ππ ππΏ−1 π‘ −1 πππ π π ≈ ππ ; πΏππππ‘β πΏ ; πΏπ‘ ∴π =π−π = 5−3 π=2 π π π‘πππ (π‘) πππ π π 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 4: Choose (π) repeating variables (mass, geometry, & kinematics) ∴π π· π£ Step 5: Construct the dimensionless group π1 = ππ1 π·π1 π£ π1 πΉ π2 = ππ2 π·π2 π£ π2 π For π1 π1 = ππ1 π·π1 π£ π1 πΉ π0 πΏ0 π‘ 0 = ππΏ−3 π1 πΏ π 0 = π π1 π 0 = π1 + 1 π1 = −1 π1 πΏπ‘ −1 π1 ππΏπ‘ −2 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 5: Construct the dimensionless group π1 = ππ1 π·π1 π£ π1 πΉ π2 = ππ2 π·π2 π£ π2 π For π1 π1 = ππ1 π·π1 π£ π1 πΉ π0 πΏ0 π‘ 0 = ππΏ−3 π1 πΏ π1 πΏπ‘ −1 π1 ππΏπ‘ −2 πΏ0 = πΏ−3π1 πΏπ1 πΏπ1 πΏ 0 = −3π1 + π1 + π1 + 1 → ππ. 1 π‘ 0 = π‘ −π1 π‘ −2 0 = −π1 − 2 π1 = −2 ππ. 1 0 = −3π1 + π1 + π1 + 1 0 = −3 −1 + π1 + −2 + 1 π1 = −2 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 5: Construct the dimensionless group π1 = ππ1 π·π1 π£ π1 πΉ π1 = −1 π1 = −2 π1 = ππ1 π·π1 π£ π1 πΉ π1 = π−1 π·−2 π£ −2 πΉ π1 = πΉ π π·2 π£ 2 π1 = −2 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 5: Construct the dimensionless group π1 = ππ1 π·π1 π£ π1 πΉ For π2 π2 = ππ2 π·π2 π£ π2 π π2 = ππ2 π·π2 π£ π2 π π0 πΏ0 π‘ 0 = ππΏ−3 π 0 = π π2 π 0 = π2 + 1 π2 = −1 π2 πΏ π2 πΏπ‘ −1 π2 ππΏ−1 π‘ −1 π‘ 0 = π‘ −π2 π‘ −1 πΏ0 = πΏ−3π2 πΏπ2 πΏπ2 πΏ−1 0 = −π2 − 1 0 = −3π2 + π2 + π2 − 1 π2 = −1 0 = −3 −1 + π2 − 1 − 1 π2 = −1 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 5: Construct the dimensionless group π1 = ππ1 π·π1 π£ π1 πΉ For π2 π2 = ππ2 π·π2 π£ π2 π π2 = ππ2 π·π2 π£ π2 π π2 = π−1 π·−1 π£ −1 π π2 = π ππ·π£ 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Step 5: Construct the dimensionless group π2 = π ; ππ·π£ ∴ π2 = π1 = π1 = π π2 1 π π π π = ; ππ·π£ π π π −1 πΉ π π·2 π£ 2 → πΉ = π π π π π·2 π£ 2 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Dimensionless Performance using Buckingham π Theorem There are several options for selecting the variables (for example, using the head H, pressure increase Ap, or usually pH to replace H) so that the analysis won't provide a single set of dimensionless variables. π2 ππ» − βπππ π£πππππππ, 2 π π3 π − π£πππ’ππ ππππ€ πππ‘π, π 2 ππ π π − πππ€ππ, πππ‘π‘π , π 3 πππ −1 π − π ππππ, ,π π π ππ», π, π, π, π·, π, π, π· − ππππππ‘ππ, π ππ π3 ππ π − πππ’ππ π£ππ πππ ππ‘π¦, ππ π − ππππππ ππππππ π ππππππ‘ππ πππ‘πππ π· π − πππ’ππ ππππ ππ‘π¦, π π· 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Dimensionless Performance using Buckingham π Theorem There are several options for selecting the variables (for example, using the head H, pressure increase Ap, or usually pH to replace H) so that the analysis won't provide a single set of dimensionless variables. π2 ππ» − βπππ π£πππππππ, 2 π π3 π − π£πππ’ππ ππππ€ πππ‘π, π 2 ππ π π − πππ€ππ, πππ‘π‘π , π 3 πππ −1 π − π ππππ, ,π π π· − ππππππ‘ππ, π ππ π3 ππ π − πππ’ππ π£ππ πππ ππ‘π¦, ππ π − ππππππ ππππππ π ππππππ‘ππ πππ‘πππ π· π − πππ’ππ ππππ ππ‘π¦, The symbol d/D serves as a reminder that geometric similitude is imposed to ignore shape- or proportion-related issues. There are three fundamental units: time, length, and mass (π , πΏ, πππ π‘ as before). The seven factors listed above can so be expected to be reduced to four. The following π − πππππ’ππ‘π are created by selecting π, π, πππ π· as primaries: 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries π1 = π2 = π3 = π4 = ππ1 ππ1 π·π1 π ππ2 ππ2 π·π2 π ππ3 ππ3 π·π3 ππ» ππ4 ππ4 π·π4 π For π1 π1 = ππ1 ππ1 π·π1 π π0 πΏ0 π‘ 0 = ππΏ−3 π1 π‘ −1 π1 πΏ π1 πΏ3 π‘ −1 π0 πΏ0 π‘ 0 = ππ1 πΏ−3π1 π‘ −1π1 πΏπ1 πΏ3 π‘ −1 π 0 = π π1 0 = π1 πΏ0 = πΏ−3π1 πΏπ1 πΏ3 0 = −3π1 + π1 + 3 0 = −3 0 + π1 + 3 π1 = −3 π‘ 0 = π‘ −1π1 π‘ −1 0 = −π1 − 1 π1 = −1 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries π1 = ππ1 ππ1 π·π1 π π1 = π0 π −1 π·−3 π π1 = For π2 π2 = ππ2 ππ2 π·π2 π π0 πΏ0 π‘ 0 = ππΏ−3 0 0 0 π πΏ π‘ π2 π‘ −1 π2 π2 πΏ π2 −3π2 −1π2 π2 =π πΏ π‘ −1 −1 πΏ ππΏ π‘ π0 = ππ2 +1 π2 = −1 ππΏ−1 π‘ −1 π π π·3 πΏ0 = πΏ−3π2 πΏπ2 πΏ−1 0 = −3π2 + π2 − 1 0 = −3 −1 + π2 − 1 π2 = −2 π‘ 0 = π‘ −1π2 π‘ −1 0 = −π2 − 1 π2 = −1 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries π2 = ππ2 π£ π2 π·π2 π π2 = π−1 π£ −1 π·−2 π π2 = For π3 π3 = ππ3 ππ3 π·π3 ππ» π0 πΏ0 π‘ 0 = ππΏ−3 0 0 0 π πΏπ‘ π3 π‘ −1 π3 πΏ π3 πΏ2 π‘ −2 π3 −3π3 −1π3 π3 2 −2 =π πΏ π‘ π0 = ππ3 π3 = 0 πΏ πΏπ‘ π π π π·2 πΏ0 = πΏ−3π3 πΏπ3 πΏ2 0 = −3π3 + π3 + 2 0 = −3 0 + π3 + 2 π3 = −2 π‘ 0 = π‘ −1π3 π‘ −2 0 = −π3 − 2 π3 = −2 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries π3 = ππ3 ππ3 π·π3 ππ» π3 = π0 π −2 π·−2 ππ» π3 = For π4 π4 = ππ4 ππ4 π·π4 π π0 πΏ0 π‘ 0 = ππΏ−3 0 0 0 π πΏπ‘ π4 π‘ −1 π4 πΏ π4 π4 −3π4 −1π4 π4 =π πΏ π‘ π0 = ππ4 +1 π4 = −1 ππΏ2 π‘ −3 2 −3 πΏ ππΏ π‘ ππ» π 2 π·2 πΏ0 = πΏ−3π4 πΏπ4 πΏ2 0 = −3π4 + π4 + 2 0 = −3 −1 + π4 + 2 π4 = −5 π‘ 0 = π‘ −1π4 π‘ −3 0 = −π4 − 3 π4 = −3 02 Week 1 : Dimensional Analysis as applied to Fluid Machineries π4 = ππ4 ππ4 π·π4 π π4 = π−1 π −3 π·−5 π π4 = π1 = π π π·3 π2 = π π π π·2 ππ» π 2 π·2 π π4 = π π 3π·5 π3 = π π π 3 π·5 π= π π π·3 π π π·2 π π = π ππ» π= 2 2 π π· π π= π π 3π·5 πΉπππ€ πΆππππππππππ‘ π ππ¦πππππ ππ’ππππ π»πππ πΆππππππππππ‘ πππ€ππ πΆππππππππππ‘ 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries As a description of turbomachine performance. This is really compact. For convenience. Recall the concept of the efficiency defined as. π= π π π·3 ππ» π 2 π·2 π π= π π 3 π·5 π= ππ = πππ‘ππ πππ€ππ πππ‘ππ πππ€ππ ππ = ππππ» ππ π πππ·3 ππ 2 π·2 ππ = ππ π 3 π·5 ππ = ππ π 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A test on a 1.2m fan diameter operating a speed of 1200 rpm and pressure π3 of 1kPa at a capacity of 70 . Determine the flow coefficient, head coefficient, power π coefficient and the efficiency of the machinery. The power input required is 90ππ. Estimate the power of if the fan were operated at 1000 rpm and the diameter of the fan is 35 inches. Assume normal operation at point of maximum efficiency in each case. 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A test on a 1.2m fan diameter operating a speed of 1200 rpm and pressure π3 of 1kPa at a capacity of 70 . Determine the flow coefficient, head coefficient, power π coefficient and the efficiency of the machinery. The power input required is 90ππ. Estimate the power of if the fan were operated at 1000 rpm and the diameter of the fan is 35 inches. Assume normal operation at point of maximum efficiency in each case. Given: π·1 = 1.2 π π1 = 70 π3 /π π1 = 1200 πππ π1 = 90 ππ π =? π =? π =? ππ = ? Solution: π= π π1 = π π·3 π1 π·1 3 70 π3 /π π= πππ£ 2π πππ 1 πππ (1200 π₯ π₯ ) 1.2π πππ 1 πππ£ 60 π ππ π = 0.322 3 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A test on a 1.2m fan diameter operating a speed of 1200 rpm and pressure π3 of 1kPa at a capacity of 70 . Determine the flow coefficient, head coefficient, power π coefficient and the efficiency of the machinery. The power input required is 90ππ. Estimate the power of if the fan were operated at 1000 rpm and the diameter of the fan is 35 inches. Assume normal operation at point of maximum efficiency in each case. Given: π·1 = 1.2 π 3 π1 = 70 π /π π1 = 1200 πππ π1 = 90 ππ π =? π =? π =? ππ = ? Solution: π= ππ» π 2 π·2 ; ππππ π π’ππ = 1πππ = 1000ππ ππππ π π’ππ πππ π = βπ = πΎπ» = πππ» ππ» = βπ π π 1 2 1 ππ π (1000 ππ π₯ π π₯ ) ππππ π π’ππ 1πππ 1 π π 2 π= = ππππ π1 2 π·1 2 (1.2 ππ ) (1200 πππ£ π₯ 2π πππ π₯ 1 πππ )2 1.2π πππ 1 πππ£ 60 π ππ π3 = 0.037 2 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A test on a 1.2m fan diameter operating a speed of 1200 rpm and pressure π3 of 1kPa at a capacity of 70 . Determine the flow coefficient, head coefficient, power π coefficient and the efficiency of the machinery. The power input required is 90ππ. Estimate the power of if the fan were operated at 1000 rpm and the diameter of the fan is 35 inches. Assume normal operation at point of maximum efficiency in each case. Given: π·1 = 1.2 π π1 = 70 π3 /π π1 = 1200 πππ Solution: π π1 π= = π π 3 π·5 ππππ π1 2 π·1 2 π =? π =? 1 ππ½ 1000 π½ 1 π π 1 ππ π π₯ π₯ π₯ 1π½ 1ππ π 1 ππ½ 1 π π 2 π= ππ πππ£ 2π πππ 1 πππ 3 (1.2 3 ) (1200 π₯ π₯ ) 1.2π πππ 1 πππ£ 60 π ππ π π =? ππ = ? π = 0.015 π1 = 90 ππ 90ππ π₯ 5 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A test on a 1.2m fan diameter operating a speed of 1200 rpm and pressure π3 of 1kPa at a capacity of 70 . Determine the flow coefficient, head coefficient, power π coefficient and the efficiency of the machinery. The power input required is 90ππ. Estimate the power of if the fan were operated at 1000 rpm and the diameter of the fan is 35 inches. Assume normal operation at point of maximum efficiency in each case. Given: π·1 = 1.2 π π1 = 70 π3 /π π1 = 1200 πππ π1 = 90 ππ π =? π =? π =? ππ = ? Solution: ππ π 0.322 0.037 ππ = 0.015 ππ = π π = 0.794 ≈ 0.8 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A test on a 1.2m fan diameter operating a speed of 1200 rpm and pressure of 1kPa at a π3 capacity of 70 π . Determine the flow coefficient, head coefficient, power coefficient and the efficiency of the machinery. The power input required is 90ππ. Estimate the power of if the fan were operated at 1000 rpm and the diameter of the fan is 35 inches. Assume normal operation at point of maximum efficiency in each case. Given: π·1 = 1.2 π π1 = 70 π3 /π π1 = 1200 πππ π1 = 90 ππ π = 0.322 π = 0.037 π = 0.015 π π = 0.8 Solution: π2 =? If, π2 = 1000 πππ ; π·2 = 35 πππβππ π = π1 = π2 = 0.015 π2 = π2 2 ππππ π2 π·2 2 π2 = π2 ππππ π2 2 π·2 2 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A test on a 1.2m fan diameter operating a speed of 1200 rpm and pressure of 1kPa at a π3 capacity of 70 π . Determine the flow coefficient, head coefficient, power coefficient and the efficiency of the machinery. The power input required is 90ππ. Estimate the power of if the fan were operated at 1000 rpm and the diameter of the fan is 35 inches. Assume normal operation at point of maximum efficiency in each case. Given: π·1 = 1.2 π π1 = 70 π3 /π π1 = 1200 πππ π1 = 90 ππ π = 0.322 π = 0.037 π = 0.015 π π = 0.8 Solution: π2 =? If, π2 = 1000 πππ ; π·2 = 35 πππβππ π = π1 = π2 = 0.015 π2 = π2 ππππ π2 2 π·2 2 ππ πππ£ 2π πππ 1 πππ 3 1 ππππ‘ 1π π2 = 0.015 (1.2 3 ) (1000 π₯ π₯ ) 35 ππ π₯ π₯ π πππ 1 πππ£ 60 π ππ 12 ππ 3.28 ππ‘ π½ π2 = 11,492.7 ≈ 11,492.7πππ‘π‘π π 5 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A centrifugal air compressor delivers 32,850 πππ of air with a pressure change of 4 ππ π. The compressor is driven by an 800 βπ motor at 3550 πππ. a.) What is the overall efficiency? b.) What will the flow and pressure rise be at 3000 πππ? c.) Estimate the impeller diameter 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A centrifugal air compressor delivers 32,850 πππ of air with a pressure change of 4 ππ π. The compressor is driven by an 800 βπ motor at 3550 πππ. a.) What is the overall efficiency? b.) What will the flow and pressure rise be at 3000 πππ? c.) Estimate the impeller diameter Given: π1 = 32850 ππ‘ 3 /π βπ = 4 ππ π π1 = 3550 πππ π1 = 800 βπ ππ ππππ = 1.2 3 π Solution: a.) π π =? ππ = πππ‘ππ πππ€ππ πππ‘ππ πππ€ππ ππ = π1 πΎπ» π1 ππ = ; βπ = πΎπ» π1 βπ = π1 ππ‘ 3 32,850 πππ ππ 144 ππ2 4 2π₯ ππ 1 ππ‘ 2 550 ππ ππ‘ π 800 βπ π₯ 1 βπ 1 πππ 60 π ππ = 0.717 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A centrifugal air compressor delivers 32,850 πππ of air with a pressure change of 4 ππ π. The compressor is driven by an 800 βπ motor at 3550 πππ. a.) What is the overall efficiency? b.) What will the flow and power be at 3000 πππ? c.) Estimate the impeller diameter Solution: Given: π1 = 32850 ππ‘ 3 /π βπ = 4 ππ π π1 = 3550 πππ π1 = 800 βπ ππππ = 1.2 ππ π3 b.) π2 πππ π2 ππ‘ 3000 πππ π2 =? π = π1 = π2 = π π1 π2 = = π π·3 π1 π·1 3 π2 π·2 3 Assuming same diameter; π·1 = π·2 π1 π2 = π1 π2 π1 π2 = π2 π1 32850 ππ‘ 3 /π = 3000 πππ 3550 πππ = 27,760.56 ππ‘ 3 /π 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A centrifugal air compressor delivers 32,850 πππ of air with a pressure change of 4 ππ π. The compressor is driven by an 800 βπ motor at 3550 πππ. a.) What is the overall efficiency? b.) What will the flow and power be at 3000 πππ? c.) Estimate the impeller diameter Solution: Given: π1 = 32850 ππ‘ 3 /π βπ = 4 ππ π π1 = 3550 πππ π1 = 800 βπ ππππ = 1.2 ππ π3 b.) π2 πππ π2 ππ‘ 3000 πππ π2 =? π = π1 = π2 = π1 ππππ π1 2 π·1 2 = π2 ππππ π2 2 π·2 2 Assuming same diameter; π·1 = π·2 π1 π2 = π1 2 π2 2 π2 = π2 2 π1 π1 2 = 3000 πππ 2 800 βπ 3550 πππ 2 = 571.32 βπ 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A centrifugal air compressor delivers 32,850 πππ of air with a pressure change of 4 ππ π. The compressor is driven by an 800 βπ motor at 3550 πππ. a.) What is the overall efficiency? b.) What will the flow and power be at 3000 πππ? c.) Estimate the impeller diameter Solution: Given: π1 = 32850 ππ‘ 3 /π βπ = 4 ππ π π1 = 3550 πππ π1 = 800 βπ ππππ = 1.2 ππ π3 c.) π· =? π = π΄π£ ;π΄ = π 2 π· ππ·π 4 π 2 π·3 π π= 4 π 2 π· 4 ; π£ = ππ·π π= πππ£ 2ππππ 1πππ. π 2 π·1 3 3550 π₯ π₯ ππ‘ 3 π 2 π·1 3 π1 πππ 1 πππ£ 60 π ππ = 32850 = π1 = π 4 4 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Example: A centrifugal air compressor delivers 32,850 πππ of air with a pressure change of 4 ππ π. The compressor is driven by an 800 βπ motor at 3550 πππ. a.) What is the overall efficiency? b.) What will the flow and power be at 3000 πππ? c.) Estimate the impeller diameter Solution: Given: π1 = 32850 ππ‘ 3 /π βπ = 4 ππ π π1 = 3550 πππ π1 = 800 βπ ππππ = 1.2 ππ π3 c.) π· =? π = π΄π£ ;π΄ = π 2 π· 4 ; π£ = ππ·π π 2 π· ππ·π 4 π 2 π·3 π π= 4 π= π·1 = 3.29 ππ‘ ≈ 40 πππβππ 02 Week 2 : Dimensional Analysis as applied to Fluid Machineries Seatwork: A commercially available fan has BEP (Best Efficiency Point) performance of π = 350 ππ‘ 3 /π ,βπ = 100 πππ /ππ‘ 2 .and π π = 0.86 .Density is ρ = 0.00233 π ππ’ππ /ππ‘ 3 ,and speed and size are given as π = 485 π −1, and π· = 2.25 ππ‘, respectively. a.) At what speed will the fan generate 120 πππ /ππ‘ 2 ? b.) What will the flow rate be at that speed? c.) Estimate the fan efficiency at that speed. d.) Calculate the required motor horsepower. Seatwork: Assume that the power required to drive a pump is a function of the fluid density (π), the rotational speed (π), the impeller diameter (π·), and the volume flow rate (π). Use dimensional analysis to show that the power coefficient is a function of the flow π coefficient π′ = π(Ο) and Ο = 3. ππ·
