i need help with the calculations. i am messing up something in the math Orifice MeterFrom the theory, the equation governing the flow rate through a orifice meter is shown as follows:The data obtained for orifice meter from the experiment is summarized as followsΔΡNo.unitsThen,12345Flow rate (gpm)Gallons Per Minute545432Sample CalculationsInner diameter of the tube = 0.590/n = 0.014986mQ₂Orifice Bore = 0.311in = 0.078994m.A₂ =inV₂ =7032MD₂²Q₂ = A₂V₂ CA₂C-Fin200 160150 125 25127 100 27= A₂V₂ = A₂60P1-P223in1092(P₁-P₂)√(¹-())40 1.016 9963.912????2(P₁-P₂)√(1-3))2(P₁-P₂)√(1-3))AP=P1-P2T(0.0078994m)²4=m????(2)-(-AP=P1-P2Pa????????velocity0.078994m0.014986mm/s5.25312??????= 49.00925 x 10-6m²??flow rateQorim³/s0.0002574512(9963.912)kgm/s²1000kg/m³(1-(0.277854)²)????????= 0.277854Actualflow rateGotm³/s0.0003330.00030.000250.0001830.000117= 5.25312m/s(5.25312m (49.00925 x 10-6m²) = 0.000257451m²/sThe flow rate in this case is lower than the actual flow rate. This is due to the loss in the flow rate caused by the turbulence in the orifice.

The inner diameter is The outer diameter is The difference in pressure is

From the theory, the equation governing the flow rate through a orifice meter is shown as follows: 2(P₁-P₂) Q₂-A₂V₂ CA₂ *√(¹-(;;)) The data obtained for orifice meter from the experiment is summarized as follows No. units Then, 1 2 3 4 5 Flow rate (gpm) Gallons Per Minute 5 45 4 3 2 Sample Calculations Inner diameter of the tube =0.590 in = 0.014986m Orifice Bore = 0.311in = 0.078994m. = om in 32 200 160 150 127 70 60 TD₂² C-F P1- P2 in Q₂ = A₂V₂ = A₂ 125 100 AP 2 in 25 27 10 9 P2 2(P₁-P₂) √(¹-(3)) 40 1.016 9963.912 5.25312 2 (P₁-P₂) √(¹-(A)) 1- E T(0.0078994m)² ?? ?? ?? ?? (-) - AP= P1-P2 Pa ?? ?? ?? ?? velocity m/s 0.078994m 0.014986m 5.25312m ?? = 49.00925 x 10-6m² ?? ?? ?? flow rate Qori m³/s 0.000257451 2(9963.912)kgm/s² 1000kg/m³(1-(0.277854)²) 3333 ?? = 0.277854 Actual flow rate Qact m³/s 0.000333 0.0003 0.00025 0.000183 0.000117 = 5.25312m/s (49.00925 x 10-6m²) = 0.000257451m³/s The flow rate in this case is lower than the actual flow rate. This is due to the loss in the flow rate caused by the turbulence in the orifice.

From the theory, the equation governing the flow rate through a orifice meter is shown as follows: 2(P₁-P₂) Q₂-A₂V₂ CA₂ *√(¹-(;;)) The data obtained for orifice meter from the experiment is summarized as follows No. units Then, 1 2 3 4 5 Flow rate (gpm) Gallons Per Minute 5 45 4 3 2 Sample Calculations Inner diameter of the tube =0.590 in = 0.014986m Orifice Bore = 0.311in = 0.078994m. = om in 32 200 160 150 127 70 60 TD₂² C-F P1- P2 in Q₂ = A₂V₂ = A₂ 125 100 AP 2 in 25 27 10 9 P2 2(P₁-P₂) √(¹-(3)) 40 1.016 9963.912 5.25312 2 (P₁-P₂) √(¹-(A)) 1- E T(0.0078994m)² ?? ?? ?? ?? (-) - AP= P1-P2 Pa ?? ?? ?? ?? velocity m/s 0.078994m 0.014986m 5.25312m ?? = 49.00925 x 10-6m² ?? ?? ?? flow rate Qori m³/s 0.000257451 2(9963.912)kgm/s² 1000kg/m³(1-(0.277854)²) 3333 ?? = 0.277854 Actual flow rate Qact m³/s 0.000333 0.0003 0.00025 0.000183 0.000117 = 5.25312m/s (49.00925 x 10-6m²) = 0.000257451m³/s The flow rate in this case is lower than the actual flow rate. This is due to the loss in the flow rate caused by the turbulence in the orifice.

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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