Let's break down and solve the design problem step by step: --- ### **Given Data** - Discharge (\( Q \)) = **24 m³/s** - Depth of flow (\( h \)) = **1.8 m** - Bed width (\( b \)) = **20 m** - Upstream FSL = **118.20 m** - Downstream FSL = **117.00 m** - Drop = **1.20 m** - Upstream bed level = **116.40 m** - Downstream bed level = **115.20 m** - Bligh’s coefficient of creep (\(C\)) = **8** --- ## **(a) Design the dimensions of the crest and cistern downstream of the fall** ### **1. Crest Length and Height** The Sarda type fall is a simple weir-like structure. The crest should be designed to pass the entire discharge with a freeboard. #### **Crest Length (\( L \))** The length of crest is generally taken as the bed width of the channel: - **Crest length (\( L \)) = Bed width (\( b \)) = 20 m** #### **Crest Height** The crest is kept at the downstream bed level: - **Crest level = Downstream bed level = 115.20 m** #### **Crest Width** The width of the crest is generally taken as 0.6 to 1.0 times the depth of water over the crest. Let’s use 0.75 × depth of water over crest (\( h \)). First, compute the head over the crest (\( H \)): Upstream FSL = 118.20 m Crest level = 115.20 m Head over crest (\( H \)) = 118.20 – 115.20 = **3.00 m** But this is the total head. The actual depth over crest for discharge calculation is: - Upstream FSL - Crest level = 118.20 – 115.20 = **3.00 m** But this seems too high, let's check discharge formula for a broad-crested weir: \[ Q = CLH^{1.5} \] Where, - \( Q \) = Discharge (24 m³/s) - \( C \) = Coefficient (usually for broad crested weir, 1.7) - \( L \) = Crest length (20 m) - \( H \) = Head over crest (unknown) Plug in the values: \[ 24 = 1.7 \times 20 \times H^{1.5} \] \[ H^{1.5} = \frac{24}{34} = 0.7059 \] \[ H = (0.7059)^{2/3} \approx 0.795 \text{ m} \] So, the **head over crest (\(H\)) is about 0.80 m**. - **Crest width = 0.75 × H = 0.75 × 0.8 = 0.6 m (recommend: 0.6 to 1.0 m)** #### **Crest Level** - **Crest level = Downstream bed level = 115.20 m** #### **Cistern (Downstream Apron)** Length of cistern (\( L_c \)): \[ L_c = 4 \times H = 4 \times 0.8 = 3.2 \text{ m} \text{ (usually 3-4 times H)} \] Depth of cistern below downstream bed (\( d_c \)): \[ d_c = 0.3 \text{ to } 0.5 \text{ m} \] --- ### **Summary for (a):** - **Crest length:** 20 m - **Crest width:** 0.6 – 1.0 m (use 0.8 m for safety) - **Crest level:** 115.20 m - **Cistern length:** 3.2 m - **Cistern depth below D/S bed:** 0.3 – 0.5 m --- ## **(b) Design of the Floor Using Bligh's Creep Theory** Bligh’s theory states: \[ L = C \frac{H}{h} \] Where: - \( L \) = Required length of impervious floor - \( C \) = Bligh’s coefficient = 8 - \( H \) = Head causing seepage (difference between upstream and downstream water levels) - \( h \) = Difference in water levels = 1.2 m So, \[ L = C \times h = 8 \times 1.2 = 9.6 \text{ m} \] So, **provide a floor of length 9.6 m** (rounded to 10 m for safety) from the upstream end of the crest to the downstream end of the cistern. --- ## **Summary Table** | Item | Value | |---------------------|-------------| | Crest length | 20 m | | Crest width | 0.8 m | | Crest level | 115.20 m | | Cistern length | 3.2 m | | Cistern depth below D/S bed | 0.3–0.5 m | | Impervious floor length (Bligh's) | 10 m | --- Let me know if you need sketches or more details!