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Calculations

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Calculations:
Part I: Variation of the Piezometric Head along the Venturi Meter.
Flow Rate Q:
𝑄=
𝐸𝑛𝑑 π‘‰π‘œπ‘™π‘’π‘šπ‘’ − π‘†π‘‘π‘Žπ‘Ÿπ‘‘ π‘‰π‘œπ‘™π‘’π‘šπ‘’
π‘‡π‘–π‘šπ‘’
𝑄=
10𝐿 − 0𝐿
= 0.16722 𝐿/𝑠
59.8𝑠
𝑄 = 0.16722 𝐿/𝑠 ×
1π‘š3
= 0.00016722π‘š3 /𝑠
1000𝐿
Converting Piezometric Head:
Piezometric Table No. 1
Piezometric Table No.2
1π‘š
1π‘š
β„Ž = 275 π‘šπ‘š × 1000π‘šπ‘š = 0.275π‘š
β„Ž = 215 π‘šπ‘š × 1000π‘šπ‘š = 0.215π‘š
Piezometric Table No. 3
Piezometric Table No.4
β„Ž = 142 π‘šπ‘š ×
1π‘š
1000π‘šπ‘š
= 0.142π‘š
β„Ž = 65 π‘šπ‘š ×
Piezometric Table No. 5
β„Ž = 5π‘šπ‘š ×
1π‘š
1000π‘šπ‘š
1π‘š
1000π‘šπ‘š
= 0.065π‘š
Piezometric Table No.6
= 0.005π‘š
β„Ž = 153π‘šπ‘š ×
1π‘š
1000π‘šπ‘š
= 0.153π‘š
Area of the Tube:
Piezometric Table No. 1
π‘Ž=
πœ‹(24.5π‘šπ‘š×
2
1π‘š
)
1000π‘šπ‘š
4
Piezometric Table No.2
= 0.0004714π‘š
2
Piezometric Table No. 3
π‘Ž=
πœ‹(11.6π‘šπ‘š×
2
1π‘š
)
1000π‘šπ‘š
4
Piezometric Table No. 1
π‘Ž=
πœ‹(13.6π‘šπ‘š×
2
1π‘š
)
1000π‘šπ‘š
4
= 0.0001452π‘š2
Piezometric Table No.4
= 0.0001057π‘š
2
π‘Ž=
πœ‹(10.5π‘šπ‘š×
2
1π‘š
)
1000π‘šπ‘š
4
Piezometric Table No.2
= 0.0000866π‘š2
π‘Ž=
πœ‹(9.8π‘šπ‘š×
2
1π‘š
)
1000π‘šπ‘š
4
= 0.0000754π‘š
2
π‘Ž=
πœ‹(24.5π‘šπ‘š×
2
1π‘š
)
1000π‘šπ‘š
4
= 0.0004714π‘š2
Velocity:
Piezometric Table No. 1
𝑉=
0.00016722π‘š3 /𝑠
0.0004714π‘š2
= 0.354713 π‘š/𝑠
Piezometric Table No. 3
𝑉=
0.00016722π‘š3 /𝑠
0.0001057π‘š2
= 1.58232 π‘š/𝑠
Piezometric Table No. 5
𝑉=
0.00016722π‘š3 /𝑠
0.0000754π‘š2
= 2.21696 π‘š/𝑠
Piezometric Table No.2
𝑉=
0.00016722π‘š3 /𝑠
0.0001452π‘š2
= 1.15115 π‘š/𝑠
Piezometric Table No.4
𝑉=
0.00016722π‘š3 /𝑠
0.0000866π‘š2
= 1.93121 π‘š/𝑠
Piezometric Table No.6
𝑉=
0.00016722π‘š3 /𝑠
0.0004714π‘š2
= 0.35471 π‘š/𝑠
Velocity Head:
Piezometric Table No. 1
Piezometric Table No.2
0.354713 π‘š/𝑠
1.15115 π‘š/𝑠
9.81π‘š
2( 2 )
𝑠
= 0.00685 π‘š
2(
9.81π‘š
)
𝑠2
= 0.07218 π‘š
Piezometric Table No. 3
Piezometric Table No.4
1.58232 π‘š/𝑠
1.93121 π‘š/𝑠
2(
9.81π‘š
)
𝑠2
= 0.13637 π‘š
2(
9.81π‘š
)
𝑠2
= 0.20314 π‘š
Piezometric Table No. 5
Piezometric Table No.6
2.21696 π‘š/𝑠
0.35471 π‘š/𝑠
9.81π‘š
2( 2 )
𝑠
= 0.2677 π‘š
Total Head:
2(
9.81π‘š
)
𝑠2
= 0.00685 π‘š
Piezometric Table No. 1
0.275π‘š +
0.354713 π‘š/𝑠
2(
9.81π‘š
)
𝑠2
Piezometric Table No.2
= 0.28185 π‘š
Piezometric Table No. 3
0.142π‘š +
1.58232 π‘š/𝑠
2(
9.81π‘š
)
𝑠2
2.21696 π‘š/𝑠
2(
9.81π‘š
)
𝑠2
1.15115 π‘š/𝑠
9.81π‘š
)
𝑠2
2(
= 0.28718 π‘š
Piezometric Table No.4
= 0.27837 π‘š
Piezometric Table No. 5
0.005π‘š +
0.215π‘š +
0.065π‘š +
1.93121 π‘š/𝑠
2(
9.81π‘š
)
𝑠2
= 0.26814 π‘š
Piezometric Table No.6
= 0.2727 π‘š
0.153π‘š +
0.35471 π‘š/𝑠
2(
9.81π‘š
)
𝑠2
= 0.15985 π‘š
Part 2: Calibration of the Venturi Meter:
Flow Rate at Each Trial:
Trial 1:
𝑄=
10𝐿−0𝐿
59.8𝑠
Trial 2:
1π‘š3
× 1000𝐿 = 0.000167 π‘š3 /𝑠
Trial 3:
𝑄=
10𝐿−0𝐿
74.99𝑠
10𝐿−0𝐿
66.57𝑠
1π‘š3
× 1000𝐿 = 0.00015 π‘š3 /𝑠
Trial 4:
1π‘š3
× 1000𝐿 = 0.000133 π‘š3 /𝑠
Trial 5:
10𝐿−0𝐿
𝑄=
𝑄=
10𝐿−0𝐿
79.28𝑠
1π‘š3
× 1000𝐿 = 0.000126 π‘š3 /𝑠
Trial 6:
1π‘š3
𝑄 = 119.03𝑠 × 1000𝐿 = 0.000084 π‘š3 /𝑠
𝑄=
10𝐿−0𝐿
253.1𝑠
1π‘š3
× 1000𝐿 = 0.000040 π‘š3 /𝑠
Height Difference:
Trial 1:
Trial 2:
1π‘š
1π‘š
βˆ†β„Ž = (275π‘šπ‘š − 5π‘šπ‘š) × 1000π‘š = 0.27π‘š
βˆ†h = (245π‘šπ‘š − 40π‘šπ‘š) × 1000π‘š = 0.205π‘š
Trial 3:
Trial 4:
βˆ†β„Ž = (220π‘šπ‘š − 50π‘šπ‘š) ×
1π‘š
1000π‘š
= 0.17π‘š
Trial 5:
βˆ†β„Ž = (180π‘šπ‘š − 113π‘šπ‘š) ×
1π‘š
1000π‘š
= 0.146π‘š
Trial 6:
1π‘š
βˆ†β„Ž = (180π‘šπ‘š − 113π‘šπ‘š) × 1000π‘š = 0.067π‘š
1π‘š
βˆ†β„Ž = (161π‘šπ‘š − 147π‘šπ‘š) × 1000π‘š = 0.014π‘š
Value of B:
Assuming trial 5 diameter as throat diameter and trial 1 diameter as pipe diameter
9.81π‘š
2( 2 )
𝑠
𝐡 = 0.000075π‘š √
2
0.000075π‘š2
1−(
)
0.000471π‘š2
2𝑔
𝐡 = π‘Ž5
√
π‘Ž 2
1 − (π‘Ž5 )
1
2
5
π‘š ⁄2
𝐡 = 0.0003385
𝑠
Discharge Coefficient:
Trial 1:
𝐢=
Trial 2:
0.000167 π‘š3 /𝑠
5⁄
π‘š 2
0.0003385
√0.27π‘š
𝑠
= 0.95081
= 0.98022
Trial 4:
0.000133 π‘š3 /𝑠
5⁄
π‘š 2
0.0003385
√0.17π‘š
𝑠
= 0.95554
0.000126 π‘š3 /𝑠
𝐢=
5⁄
π‘š 2
√0.146π‘š
𝑠
= 0.9753
0.0003385
Trial 5:
𝐢=
5⁄
π‘š 2
√0.205π‘š
𝑠
0.0003385
Trial 3:
𝐢=
0.00015 π‘š3 /𝑠
𝐢=
Trial 6:
π‘š3
𝑠
0.000084
5⁄
π‘š 2
0.0003385
√0.067π‘š
𝑠
= 0.95892
Average Discharge Coefficient:
𝐢=
0.000040 π‘š3 /𝑠
5⁄
π‘š 2
0.0003385
√0.014π‘š
𝑠
= 0.98655
πΆπ‘Žπ‘£π‘” =
0.95081+0.98022+0.95554+0.9753+0.95892+0.98655
6
=0.967890109
From the plot, Q vs. (h1-hT)1/2
y = 0.000326268x
Discharge Coefficient:
πΆπ‘π‘™π‘œπ‘‘ =
0.000326268
𝐡
=
0.000326268
0.0003385
= 0.9638641064
Average of Cavg and Cplot:
πΆπ‘“π‘–π‘›π‘Žπ‘™ =
0.967890109+0.9638641064
2
=0.96591681
Reynolds Number:
οƒ˜ Assuming throat diameter is 0.0098m, area is 0.00007543m2
οƒ˜ Temperature is at 25℃
οƒ˜ ρ of water is 997.08 kg/m3
οƒ˜ μ of water is 0.8937×10-3 Pa-s
Trial 1:
𝑁𝑅𝑒 =
Trial 2:
π‘š3 997.08π‘˜π‘”
0.0098π‘š(0.000167 )(
)
𝑠
π‘š3
0.00007543m2 (0.8937×10−3 π‘ƒπ‘Žβˆ™π‘ )
𝑁𝑅𝑒 =
π‘š3 997.08π‘˜π‘”
)(
)
𝑠
π‘š3
0.00007543m2 (0.8937×10−3 π‘ƒπ‘Žβˆ™π‘ )
0.0098π‘š(0.00015
𝑁𝑅𝑒 =24239.35761
𝑁𝑅𝑒 =21774.2765
Trial 3:
Trial 4:
𝑁𝑅𝑒 =
π‘š3 997.08π‘˜π‘”
0.0098π‘š(0.000133 )(
)
𝑠
π‘š3
0.00007543m2 (0.8937×10−3 π‘ƒπ‘Žβˆ™π‘ )
𝑁𝑅𝑒 =
π‘š3 997.08π‘˜π‘”
)(
)
𝑠
π‘š3
0.00007543m2 (0.8937×10−3 π‘ƒπ‘Žβˆ™π‘ )
0.0098π‘š(0.000126
𝑁𝑅𝑒 =19329.42506
𝑁𝑅𝑒 =18283.47105
Trial 5:
Trial 6:
𝑁𝑅𝑒 =
π‘š3 997.08π‘˜π‘”
0.0098π‘š(0.000084 )(
)
𝑠
π‘š3
0.00007543m2 (0.8937×10−3 π‘ƒπ‘Žβˆ™π‘ )
𝑁𝑅𝑒 =12177.71642
𝑁𝑅𝑒 =
π‘š3 997.08π‘˜π‘”
)(
)
𝑠
π‘š3
0.00007543m2 (0.8937×10−3 π‘ƒπ‘Žβˆ™π‘ )
0.0098π‘š(0.000040
𝑁𝑅𝑒 =5727.039056
Pressure Drop:
Trial 1:
βˆ†π‘ƒ = (997.08π‘˜π‘”/π‘š3 )(9.81π‘š/𝑠 2 )(0.27π‘š)
Trial 2:
βˆ†π‘ƒ = (997.08π‘˜π‘”/π‘š3 )(9.81π‘š/𝑠 2 )(0.205π‘š)
βˆ†π‘ƒ =2640.965796 Pa
βˆ†π‘ƒ =2005.177734 Pa
Trial 3:
βˆ†π‘ƒ = (997.08π‘˜π‘”/π‘š3 )(9.81π‘š/𝑠 2 )(0.17π‘š)
Trial 4:
βˆ†π‘ƒ = (997.08π‘˜π‘”/π‘š3 )(9.81π‘š/𝑠 2 )(0.146π‘š)
βˆ†π‘ƒ = 19329.42506 Pa
βˆ†π‘ƒ = 18283.47105 Pa
Trial 1:
βˆ†π‘ƒ = (997.08π‘˜π‘”/π‘š3 )(9.81π‘š/𝑠 2 )(0.067π‘š)
Trial 2:
βˆ†π‘ƒ = (997.08π‘˜π‘”/π‘š3 )(9.81π‘š/𝑠 2 )(0.014π‘š)
βˆ†π‘ƒ = 12177.71642 Pa
βˆ†π‘ƒ = 5727.039056 Pa
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