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Weirs
- 1. By: Dr. Ezzat El-Sayed G. SALEH Civil Eng. Dept. Minia University
- 3. هللا رمحه الوراق باسلعا بوأ العالمة قال : .1 فاتأ امرالك : خمالطة ئامللا . .2 سةلجما ذوي اخلري و الفضل : زايدة يف ادلين و الع قل . .3 من بديأ كل مودته : فقد جزلأ كل تهيعط . .4 نفوسلا حتب من سنحأ ،لهياا و القلوب ملتأت مم ن ربج جَت علهيا . نظرأ ( : ص ،الوراق ابن حمك من اقرتالاف يفلتأ : 37 )
- 6. A weir is a small overflow-type dam commonly used to raise the level of a river or stream. Weirs have been used to create mill ponds in such places. Water flows over the top of a weir.
- 7. Some weirs have sluice gates which release water at a level below the top of the weir. The crest of an overflow spillway on a large dam is often called a weir.
- 9. Sharp-crested weirs are used in the measurement of irrigation water. The sharp edge in the crest causes the water to spring clear of the crest, and thus accurate measurements can be made.
- 10. Broad-crested weirs: are commonly incorporated in hydraulic structures of various types and, although sometimes used to measure water flow, this is usually a secondary function.
- 11. In simples state weir consists of: A bulkhead of timber, Metal or concrete with an opening of fixed dimensions cut in its top edge (weir notch) Its bottom edge is the weir crest. The depth of flow over the crest is called the head (H). The overflowing sheet of water is known as the nappe.
- 12. Sharp- crested Weirs Broad Crested Weirs
- 14. Open-channel discharge measurement structures (a) Side view of a sharp-crested weir, (b) Front view of a rectangular weir and (c) Front view of a triangular (V-notch) weir. Broad-crested weir
- 15. Broad-crested Weir Crump Weir Solid Weir
- 18. An Example of a Dam: The dam was constructed of stone blocks with a wooden spillway
- 19. A Weir Example: (Rectangular)
- 25. Trapezoidal Weir
- 27. When the sides of the flow channel act as the ends of a rectangular weir, no side contraction exists, and the nappe does not contract from the width of the channel A Suppressed Weir is:
- 28. Partially contracted rectangular weirs (including sharp-crested), Partially contracted 90o V-notch weirs. Common Partially Contracted Weirs are:
- 29. Cipolletti Weir Operating with Full Contractions at the End and on the Crest Cipolletti Weir
- 30. For a rectangular sharp-crested weir that takes end contractions into account and neglecting the velocity of approach, the discharge is determined by: Where: L is the crest length (l), and n is the number of end contractions
- 32. A Gated Weir Radial Gate D.S.W. L U.S.W. L
- 34. In some cases, the weir is made significantly longer than the width of the river by forming it in a 'U' shape or running it diagonally, instead of the short perpendicular path. Weirs are used in conjunction with locks, to render a river navigable and to provide even flow for navigation. Function
- 35. Weirs are used in conjunction with locks A long weir is made to allow a lot of water with a small increase in over flow depth.
- 36. Broad crested weirs are robust structures ( نشأم هيلكه قوي ) that are generally constructed from reinforced concrete and which usually span the full width of the channel. They are used to measure the discharge of rivers, and are much more suited for this purpose than the relatively flimsy sharp crested weirs. Additionally, by virtue of being a critical depth meter, the broad crested weir has the advantage that it operates effectively with higher downstream water levels than a sharp crested weir. Board Crested Weir
- 37. Board Crested Weir Broad-crested rectangular weir.
- 38. Broad-crested rectangular weir. Board Crested Weir
- 41. A duck bill weir is a type of long-crested weir that is designed to control water levels. As shown in the next slide, the design of a duck bill weir involves a staggered weir crest that has the appearance of teeth or duck bills. This staggering effect increases the weir crest length while minimizing the footprint of the weir. The result is a smaller weir that is effective at the upstream water depth. What is a Duck-bill Weir?
- 42. Duck- Bill
- 44. Napp e
- 45. Even though the water around weirs can often appear relatively calm, they are dangerous places to boat, swim or wade; the circulation patterns on the downstream side can submerge a person indefinitely.
- 46. Weirs allow hydrologists and engineers a simple method of measuring the rate of fluid flow in small to medium-sized streams, or in industrial discharge locations. Since the geometry of the top of the weir is known, and all water flows over the weir, the depth of water behind the weir can be converted to a rate of flow.
- 47. The calculation relies on the fact that fluid will pass through the critical depth of the flow regime in the vicinity of the crest of the weir. A weir will artificially reduce the upstream water velocity, which can lead to an increase in siltation (see next slide).
- 49. The following general rules should be observed in the construction and installation of weirs. A weir should be set at right angles to the direction of flow in a channel that is straight for a distance upstream from the weir at least ten times the length of the weir crest. Construction and Placement
- 50. The crest and sides of the weir should be straight and sharp-edged. The crest of the rectangular and Cipolletti weirs should be level and the sides should be constructed at exactly the proper angle with the crest. Each side of the V-notch weir should make a 45° angle with a vertical line through the vertex of the notch.
- 51. Avoid restrictions in the channel below the weir that would cause submergence. The crest must be placed higher than the maximum downstream water surface to allow air to enter below the nappe. The channel upstream should be large enough to allow the water to approach the weir in a smooth stream, free from eddies, and with a mean velocity not exceeding 0.3 foot per second.
- 52. How to Measure the Discharge by Weirs???
- 54. 54 V - notch
- 57. Venturi- Flume
- 58. Sharp-Crested Rectangular Weir EGL HGL H Vo 2/ 2g B.L. Pw Vo B H Nappe Pw Definition Sketch of Flow Over a Rectangular, Sharp-Crested Weir
- 59. Qact Actual discharge, B Width of the weir, H Head on the weir, Cd Coefficient of discharge. Rectangular Weir Equation
- 60. A V-shaped notch is a vertical thin plate which is placed perpendicular to the sides and bottom of a straight channel is defined as a V-notch sharp-crested weir. The line which bisects ( ينصف ) the angle of the notch should be vertical and at the same distance from both sides of the channel . The V-notch sharp-crested weir is one of the most precise ( دقة ) discharge measuring devices suitable for a wide range of flow. In international literature, the V-notch sharp-crested-weir is frequently referred to as the ‘Thomson weir’. Triangular or V-Notch Weir
- 61. V- Notch
- 62. H dh h b h b/2 /2 Free Water Surface Crest Level Sharp-Crested Triangular Weir H
- 63. Triangular Weir Equation Qact Actual discharge, B Width of the weir, H Head on the weir, Cd Coefficient of discharge.
- 66. Dam Spillway “Morning Glory Spillway” Tunnel spillway with a morning glory entrance.
- 67. water works to divert water
- 68. sharp crested weir fayoum weir
- 69. Side Weir
- 70. A side-flow weir is a structure installed along the side of a main channel or pipe. They are used to divert flow during high flow conditions. These types of weirs are commonly seen in irrigation and sewer systems.
- 73. A V-notch weir is to be designed to measure an irrigation channel flow. For case in reading the upstream water-level gage, a reading is desired for the design flow rate of 150 m3/h. . What is the appropriate angle for the V notch? Take Cd = 0.62. Worked Example
- 74. The designed flow rate The actual flow rate for a sharp-crested triangular ( V notch angle) is H
- 75. or Substituting the given data into Eq. (1) gives, Then and the appropriate angle …… (1)
- 76. The rectangular sharp-crested weir shown in the below figure is used to maintain a relatively constant depth in the channel upstream of the weir. I. How much deeper will the water be upstream of the weir during a flood when the flowrate is 45 ft3/s compared to normal conditions when the flowrate is 30 ft3/s? Assume the weir coefficient remains constant at Cd = 0.62. II.Repeat the calculations if the weir of part (a) is replaced by a rectangular sharp-crested “duck bill” weir that is oriented at an angle of 30o relative to the channel centerline as shown in Figure. The weir coefficient remains the same. Worked Example
- 77. For case (a): Q = 30 ft3/s when B = 20 ft, Cwr = 0.62 ……………(1) B Worked Example
- 78. If the discharge is increased to Q =45ft3/s If the discharge is increased to Q =45ft3/s with keeping “B”, Cwr …………………………….(2) …………………………….(3) Worked Example
- 79. For case (b): ……………(1) 10 ft 30o Worked Example
- 80. If the discharge is increased to Q =45ft3/s If the discharge is increased to Q =45ft3/s with keeping “B”, Cwr …………………………….(2) …………………………….(3) Worked Example
- 86. y1 y2 V2 V1 Using a Sluice Gate E1 E2 yG
- 88. Critical-depth or Venturi flume (Parshall type)
- 89. Using a Parchall Flume
- 90. Fiberglass 12-inch Parshall Flume suitable for Water Rights use
- 95. 95
- 97. 97 Specific Energy “Specific energy: is the total mechanical energy with respect to the local invert elevation of the channel”. We can write 𝐸𝑆 = 𝑦 + 𝛼 𝑉2 2 𝑔 = 𝑦 + 𝛼 𝑄 𝐴 2 2 𝑔 ≅ 𝑦 + 𝑄 𝐴 2 2 𝑔 (since 𝑉 = 𝑄 𝐴 )
- 98. For a given cross section, the flow area, A is a function of y, therefore the specific energy is a function of Q and y, We can thus study the variation of 98 Specific Energy 𝐸𝑆 ≅ 𝑦 + 𝑄 𝐴 2 2 𝑔 𝑦 𝑎𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸𝑆 (Specific energy curve) 𝑦 𝑎𝑠 𝑎𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑄 (Specific Discharge curve) 𝐸𝑆 ≅ 𝑦 + 𝑄 𝐴 2 2 𝑔 For Q = const For y = const.
- 99. 99 Let us consider a specific energy for a steady flow 𝒁 + 𝑷 𝝆 𝒈 + 𝑽𝟐 𝟐 𝒈 𝟏 = 𝒁 + 𝑷 𝝆 𝒈 + 𝑽𝟐 𝟐 𝒈 𝟐 + 𝒉𝑳 𝟏 →𝟐 Note that since: 𝑄 = 𝐴. 𝑉 , we can write 𝐸𝑆 = 𝑦 + 𝛼 ∙ 𝑉2 2 𝑔 = 𝑦 + 𝛼 ∙ 𝑄2 2 𝑔∙ 𝐴2 ≅ 𝑦 + 𝑄2 2 𝑔∙ 𝐴2 Concept of Specific Energy 𝑃 𝜌 𝑔 + 𝑉2 2 𝑔 Specific energy Specific energy: Is the total mechanic energy with the local invert elevation of the channel.
- 100. 100 For a given cross sectional, the flow area ”A” is a function of ”y”. Therefore, the specific energy is a function of “A” & “y” 𝑬𝑺 ≅ 𝒚 + 𝑸𝟐 𝟐 𝒈 ∙ 𝑨𝟐 We can thus study the function of: 𝒚 as a function of 𝑬𝑺 {Specific energy curve} 𝒚 as a function of 𝑸 {Specific discharge curve} Concept of Specific Energy for Q constant for Es constant
- 101. 101 We wish to plot the specific energy curve ( i.e. ”y” as a function of “Es“ for constant “Q”) . 𝑬𝑺 ≅ 𝒚 + 𝑸𝟐 𝟐 𝒈∙ 𝑨𝟐 One immediately sees that the curve has two asymptotes For 𝑦 → 0, we have 𝐴 → 0, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸𝑆 → ∞ 𝐹𝑜𝑟 𝑦 → ∞, we have 𝐴 → ∞, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸𝑆 → 𝑦 Specific Energy Curve
- 102. 102 In addition, for a given discharge “Q”, the curve has a minimum value 𝑦𝑐𝑟. We will see about this minimum later in details. Some observations imposed: For a given 𝐸𝑆, there are always ( except when 𝑦𝑐𝑟 = 𝑦) two depths 𝑦1 & 𝑦2. They are called alternate depths. The depth corresponding to minimum specific energy 𝐸𝑆 𝑚𝑖𝑛. is called critical depth 𝑦𝑐𝑟. Minimum specific energy 𝐸𝑆 𝑚𝑖𝑛. increases with increase discharge Q. There are three possible flow regimes: 1. Sub-critical (𝑦 > 𝑦𝑐𝑟), 2. Critical (𝑦 ≥ 𝑦𝑐𝑟), and 3. Super-critical (𝑦 < 𝑦𝑐𝑟).
- 103. 103
- 104. 104
- 105. 105
- 106. 106
- 107. 107
- 108. 108
- 109. 109 Critical Depth and its Importance Critical depth in a channel “𝑦𝑐𝑟” is the flow depth at which: The specific energy at which “𝐸𝑆 𝑚𝑖𝑛." for a given discharge “Q”. The discharge is maximum “𝑄𝑚" for a give specific energy “𝐸𝑆 ". 𝐸𝑆 − ℎ𝑐𝑟 = 𝑦ℎ 2 Recall that: 𝑄𝑚 = 2 𝑔 ∙ 𝐸𝑆 − ℎ𝑐𝑟 & 𝑄𝑚 = 2 𝑔 ∙ 𝑦ℎ The average velocity corresponding to the critical depth 𝑉 𝑐𝑟 = 2 𝑔 ∙ 𝑦ℎ & 𝑉𝑐𝑟 2 2 𝑔 = 𝑦𝑐𝑟 2
- 110. 110
- 111. 111
- 112. 112 Chezy and Manning Coefficients Attention Dimensional Coefficients Chezy Equation Manning-Strickler Equation 𝐶 𝐿1 2 ∙ 𝑆−1 𝑛 𝐿−1 3 ∙ 𝑆−1 Tables are available for various surfaces Tables are available for various surfaces
- 113. 113 Depth of flow for a given discharge, where the specific energy is at a minimum, Occurs when 𝑑𝐸𝑆 𝑑𝑦 = 0 and 𝐹𝑟= 1, It is important to calculate 𝑦𝑐𝑟 in order to determine if the flow in the channel will be sub-critical or super-critical, Can be found through specific energy diagram Critical Depth
- 114. 114
- 115. 115
- 116. 116 Best Hydraulic Section (Most Efficient Cross-section) What is best hydraulic section? A section that gives the largest flow area for the smallest perimeter. For a rectangular channel, 𝐴 = 𝐵 𝑦 & 𝑃 = 𝐵 + 2 𝑦 ∴ 𝑃 = 𝐴 𝑦 + 2 𝑦 Let us keep “A” as constant “P” is only a function of “y”. Let us vary “y” to minimize the perimeter: 𝑑𝑃 𝑑𝑦 = − 𝐴 𝑦2 + 2 = 0 → 𝐴 𝑦2 = 2 → 𝐵 𝑦 𝑦2 = 2 → ∴ 𝐵 = 2𝑦 To obtain best hydraulic section the width must equal twice the depth.
- 117. 117 y B = 2 y T = 2 r r Rectangular Channel Semi-circle Channel
- 118. 118 The triangular drainage ditch shown in the figure, has a side slope of Z = 2, Find: The critical depth, ycr, for a discharge of Q = 0.35 m3/s and the corresponding minimum specific energy. Calculate the discharge if the flow depth is y = 0.60 m (The channel has a Manning’s coefficient of n = 0.025 m-1/3/ S and a bed slope of So = 0.001. y Z 1 T
- 119. Solution Froude number for a triangular channel is given by : 𝐹𝑟 = 𝑉 𝑔 ∙ 𝑦ℎ When the flow is critical, we have: 𝐹𝑟 2 = 1.0 = 𝑄2∙ 𝑇 𝑔 ∙ 𝐴3 𝑦= 𝑦𝑐𝑟 ∴ 𝑄2 𝑔 = 𝐴3 𝑇 𝑦= 𝑦𝑐𝑟 For a triangular channel, we have 𝐴 = 𝑍 ∙ 𝑦2 & 𝑦ℎ = 𝐴 𝑇 = 𝑍 𝑦2 2 𝑍 𝑦 = 𝑦 2 Substituting into Eq. (1) gives: 𝑄2 𝑔 = 𝐴3 𝑇 𝑦= 𝑦𝑐𝑟 = 𝑍 𝑦2 3 2 𝑍 𝑦 𝑦= 𝑦𝑐𝑟 = 𝑦𝑐𝑟 5 2 ∴ 𝑦𝑐𝑟 = 5 2 𝑄2 𝑔 = 5 2 ×0.352 9.81 = 0.362 𝑚 119
- 120. 120 Corresponding minimum specific energy 𝐸𝑆 𝑚𝑖𝑛 = 𝑦𝑐𝑟 + 𝑉 𝑐𝑟 2 2 𝑔 = 𝑦𝑐𝑟 + 1 2 𝑔 𝑄 𝐴 𝑦= 𝑦𝑐𝑟 2 = 0.365 + 1 2×9.81 × 0.35 2× 0.3652 2 = 𝟎. 𝟒𝟓𝟑 𝑚 Uniform flow discharge is calculated using Manning– Strickler equation as For y = 0.60 m 𝐴𝑟𝑒𝑎 𝐴 = 𝑍 𝑦2 = 2 × 0.602 = 0.72 𝑚2 ∴ 𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑅𝑎𝑑𝑢𝑖𝑠 𝑅ℎ = 𝐴 𝑃 = 0.72 2.68 = 0.268 𝑚𝑊𝑒𝑡𝑡𝑒𝑑 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑃 = 2 𝑦 𝑍2 + 1 = 2 × 0.60 × 22 + 1 = 2.68 𝑚 and the discharge 𝑄 = 𝑘 𝑛 × 𝑅ℎ 2 3 × 𝑆𝑜 1 2 × 𝐴 = 1,0 0.025 ∙ 0.268 2 3 ∙ 0.001 1 2 ∙ 0.72 = 0.38 𝑚3/s
- 121. Solution The geometric relationships for a trapezoidal channel can be calculated follows: 𝑇 = 𝐵 + 2 𝑍 𝑦 = 2.0 + 2 × 1.5 × 0.56 = 3.68𝑚 𝐴 = 𝑦 𝐵 + 𝑍 𝑦 = 0.56 (2.0 + 1.5 × 0.56) = 1. 59 𝑚2 𝑃 = 𝐵 + 2 𝑦 𝑍2 + 1 = 2.0 + 2 × 0.56 1.5 2 + 1 = 4.02 𝑚 𝑅ℎ = 𝐴 𝑃 = 1.59 4.02 = 2.0 + 2 × 1.5 × 0.56 = 0.396 𝑚 𝑦ℎ = 𝐴 𝑇 = 𝐴 𝑃 = 1.59 3.65 = 0.43 𝑚 121 A trapezoidal channel having a bottom width of B = 2.0 m and side slope of Z = 1.5 carries a uniform flow with a depth of y = 0.56 m. The channel has a bed slope of So = 0.005 and the coefficient of Manning is 0.03. What is the discharge of a uniform flow? What is the regime of flow?
- 122. 1.5 1.0 y = 0.56 m T = 0.56 m B = 2.0 m The discharge using Manning equation 𝑄 = 𝑘 𝑛 ∙ 𝑅ℎ 2 3 ∙ 𝑆𝑜 1 2 ∙ 𝐴 or 𝑄 = 𝑘 𝑛 ∙ 𝑅ℎ 2 3 ∙ 𝑆𝑜 1 2 ∙ 𝐴 = 1 0.03 × 0.3962 3 × 0.0051 2 × 1.59 = 2.02 𝑚3 /𝑆 To determine the regime of the flow, Froude number for a trapezoidal channel is given by: 𝐹𝑟 = 𝑉 𝑔×𝑦=𝑦ℎ = 𝑄 𝐴. 𝑔× 0.435 = 2.02 1.59 × 9.81×0.43 = 0.62 Since 𝐹𝑟 = 0.62 < 1.0, the uniform flow is sub-critical
- 123. 123 Depth of flow for a given discharge, where the specific energy is at a minimum, Occurs when 𝑑𝐸𝑆 𝑑𝑦 = 0 and 𝐹𝑟= 1, It is important to calculate 𝑦𝑐𝑟 in order to determine if the flow in the channel will be sub-critical or super-critical, Can be found through specific energy diagram Computation of Critical Depth
- 124. A rectangular laboratory channel has a width of B=2.0 m wide and a Manning coefficient n = 0.020 m-1/3/ s. What should be the bed slope to achieve a critical uniform flow in this channel for a discharge of Q = 3.0 m3/s? (Hint: critical uniform flow is achieved when uniform flow depth for a given discharge is equal to critical depth for the same discharge Froude number for a rectangular channel is given by: 𝐹𝑟 = 𝑉 𝑔 ∙ 𝑦ℎ=𝑦 or 𝐹𝑟 2 = 1.0 = 𝑄2∙ 𝑇 𝑔 ∙ 𝐴3 𝑦= 𝑦𝑐𝑟 = 𝑄2∙𝐵 𝑔 ∙ 𝐵∙𝑦 3 𝑦= 𝑦𝑐𝑟 𝑦𝑐𝑟 = 3 𝑄2 𝑔 ∙ 𝐵2 𝑦𝑐𝑟 = 3 32 9.81×22 =𝟎. 𝟔𝟏𝟐 𝑚 124 Solution
- 125. 125 Uniform flow discharge is calculated using Manning– Strickler equation as For y = ycr =0.612m 𝐴𝑟𝑒𝑎 𝐴 = 𝐵. 𝑦 = 2 × 0.6122 = 1.224 𝑚2 ∴ 𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑅𝑎𝑑𝑢𝑖𝑠 𝑅ℎ = 𝐴 𝑃 = 1.224 2.68 = 0.457 𝑚 𝑊𝑒𝑡𝑡𝑒𝑑 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑃 = 𝐵 + 2 𝑦 = 2 + 2 × 0.612 = 2.68 𝑚 and the discharge 𝑄 = 𝑘 𝑛 ∙ 𝑅ℎ 2 3 ∙ 𝑆𝑜 1 2 ∙ 𝐴 𝑡ℎ𝑒𝑛 𝑏𝑒𝑑 𝑠𝑙𝑜𝑝𝑒 𝑆𝑜 = 𝑄 ∙ 𝑛 𝑘 ∙ 1 𝑅ℎ 2 3 . 𝐴 2 = 3 × 0.020 1.0 × 1 0.457 2 3 × 1.224 2 = 𝟎. 𝟎𝟎𝟖𝟕𝟒 (−− −)
- 126. Characteristics: Unstable surface Series of standing waves Occurrence Broad crested weir (and other weirs), Channel controls (rapid changes in cross- section), Over falls, Change in channel sloe from mild to steep slope, Used for flow measurements. 126 Critical Depth 𝑑𝐸𝑠 𝑑𝑦 = 0
- 127. Sharp-crestedWeir V – NotchWeir Broad- crestedWeir Sluice Gate
- 128. Specific Energy : Sluice Gate 𝐸𝑆,1 = 𝐸𝑆,2 𝐸𝑆 Given: downstream depth and discharge Find : Upstream depth ? 𝑦1& 𝑦2 are alternate depths (same specific energy) Why not use momentum conservation to find 𝑦1????
- 129. The Hazen–Williams equation is used in the design of water pipe systems such as fire sprinkler systems, water supply networks, and irrigation systems. It is named after Allen Hazen and Gardner Stewart Williams. The Hazen-Williams equation has the advantage that: the coefficient C is not a function of the Reynolds number, but it has the disadvantage that it is only valid for water.
- 130. Also, it does not account for the temperature or viscosity of the water. The Hazen-Williams equation is limited to the flow of water in pipe larger than 2.0 in. and smaller than 6.0 ft in diameter. Velocity of flow should not exceed 10.0 ft/s
- 132. Typical C factors used in design, which take into account some increase in roughness as pipe ages are as follows: Material C factor range Asbestos-cement Cast iron 10 years Cast iron 20 years Cast iron 30 years Cast iron 40 years Cast iron new Cement-Mortar Lined Ductile Iron Pipe Concrete Copper Fiber-reinforced plastic Galvanized iron Polyethylene Polyvinyl chloride (PVC) Steel 140 107 - 113 89 - 100 75 - 90 64 - 83 130 140 100 - 140 130 - 140 150 120 140 150 90 - 110
- 133. The general form can be specialized for full pipe flows. Taking the general form, Exponentiating each side by 1/0.54 gives (rounding exponents to 3-4 decimals) Rearranging gives: and the discharge Q = A , V
- 134. Sharp-crested Weir
- 141. ةةةكرةسلاب ةةةوةيربال ةةةلةمع ةةةةةيقرط ا طلل حمبب وهو غذائية ميةق وهل بيةرالع البالد ىف ذكل بعدو نسارف ىف صناعته هترتش ىف فال فطاراال العشاء وأ ،وخفيفةة يطةيسة كوجبةة العمةل ىف الكبةارو الفيدة يعةرسال الوجبات من كنوع النادى ىف وأ و عةىل نةهأ مةن الةر وهو معروف غري امسه نأ الا شالك معروف " يوربال " القادير : 500 امرج دقيق 200 & دافىئ حليب مل 70 & امرج سكر & ملح رشة & يلياناف & بيض صفار حليب ملعقة مع خمفوق 20 & فورية مخرية امرج 100 & ذبدة امرج 2 & بيضة يقةرالط : متزج لدة يرتكو ادلائف احلليب ىف السكر مع امخلرية 10 دقائق & متزج امخلو احلليب مع بيضلاو بدةزال رية الزجي هل يضافو وعاء ىف اللح ورشة يلياناالفو ادلقيق يوضع السابق & اخلطى ج الكوانت عةىل حتصةىل حةحت يةدا يتضةاعف حةحت كهيارات وغطهيا وعاء ىف ينةلعجا ضعى ينةل ينةجع جحمهةا & قسةمى لة ينةةلعجا ة 10 اتركة يةوتسةام . شلكى من ختمر ححت كهيارات نيةلصا ىف وضعهيا سهانف حول لفهياو حبل ىلا كرة لك جديد « خسن درجةة عةىل الفرن خمفوق بيض بصفار يوربال وجه ادهىن متوسطة ارةرح نرثىاو عليه عىل السكر من القليل نيةلصةا ادخةىل الوجةه لدة الفرن 30 ححت دقيقة ينضج »
- 147. 147``
- 148. A tank of 5 m2 plan area is fitted with a sharp-edged orifice of 5cm diameter in its base. The coefficient of discharge of the orifice is 0.62. Calculate: The time taken for the level in the tank to fall from 2 meters depth to 0.5 meters depth. If water is now admitted to the tank at a rate of 0.01 m3/s: calculate the rate at which the surface will be rising when the depth in the tank is 1 meter and the depth in the tank when the level becomes steady.
- 149. An oil (S. G. = 0.92) flows through a vertical tube of 8 in diameter. The flow is measured by a Venturi tube of 4 in. diameter throat with a U-tube manometer, containing mercury, as shown in the figure. If Cd = 0.98, what is:o (a) the flow, for a manometer reading of 9 in ? (b) the manometer reading, for a flow of 2 ft3/sec ?
- 150. A flow nozzle of 6 in diameter is placed in an 18 in diameter pipe. During calibration the pressure differential was measured to be 10 psi for a flow of 7.2 cfs and 17 psi for a flow of 9.3 cfs. Determine the discharge coefficient of the measuring nozzle (see Figure)
- 151. A 4-in. by 1-in. nozzle, shown in the figure, is attached to the end of a 4-in hose line. The velocity of the water leaving the nozzles is 96 fps, the coefficient of velocity, Cv, is 0.96 and the coefficient of contraction, Cc, is 0.80. Determine the necessary pressure at the base of the nozzle.
- 152. An orifice of area A0 and velocity coefficient Cv = 0.8 is installed in a pipe of area Ap = 2A0. The pipe is attached to a dam as shown and the water level in the dam is 100 ft. higher than the outlet from the reservoir. The contraction coefficient of the orifice is unity. There are no other losses in the pipe nor between the end of the orifice and the pipe. The velocity leaving the pipe Vp is to be determined.
- 153. A. A water jet of velocity Vj impinges normal to a flat plate which moves to the right at velocity Vc, as shown in Fig. 1a. Find: the force required to keep the plate moving at constant velocity if the jet density is 1000 kg/m3, the jet area is 3 cm2, and Vj and Vc are 20 and 15 m/s, respectively. “Neglect the weight of the jet and plate and assume steady flow with respect to the moving plate with the jet splitting into an equal upward and downward half-jet. B. Repeat part (A) with the plate and its cart unrestrained horizontally and thus allowed to accelerate to the right. Derive (a) the equation of motion for cart velocity Vc(t) and (b) the time required for the cart to accelerate from rest to 95 percent of the jet velocity and (c) compute actual numerical values from (b) for the conditions of part (A) and a cart mass of 3 kg. “Neglect cartwheel friction”.
- 155. The water level is 4 ft from the top of the cylindrical gate shown in Fig. (1). Calculate: the resultant water pressure force acting on the gate per unit length. (See Fig. 2)
- 156. A rectangular gate 4 by 4 ft in size is set on a 45° plane with respect to the water surface; the centroid of the gate is 10 ft below the water surface (see Figure). Compute the required location of the horizontal axis at which the gate is to be hinged if the hydrostatic pressure acting on the gate is to be balanced around the hinge so that there will be no moment causing rotation of the gate under the loading specified.