# Step-by-Step Solution: Pipes in Parallel Flow Rate ## Given Data | Pipe | Length \( L \) (m) | Diameter \( d \) (cm) | Roughness \( \epsilon \) (mm) | \( \epsilon/d \) | |------|-------------------|-----------------------|-------------------------------|-----------------| | 1 | 100 | 8 | 0.24 | 0.003 | | 2 | 150 | 6 | 0.12 | 0.002 | | 3 | 80 | 4 | 0.20 | 0.005 | - **Fluid**: Water - **Density** \( \rho \) = 1000 kg/m³ - **Kinematic viscosity** \( \nu \) = \( 1.02 \times 10^{-6} \) m²/s - **Total head loss** \( h_L \) = 20.3 m --- ## Step 1: **Head Loss Formula for Each Pipe** The Darcy-Weisbach equation for head loss: \[ h_L = \frac{V^2}{2g} \cdot f \frac{L}{d} \] Given all pipes have the same head loss (\( h_L \)), for each pipe: \[ 20.3 = \frac{V_i^2}{2g} \cdot f_i \frac{L_i}{d_i} \] --- ## Step 2: **Convert Diameters to Meters** - Pipe 1: \( d_1 = 0.08 \) m - Pipe 2: \( d_2 = 0.06 \) m - Pipe 3: \( d_3 = 0.04 \) m --- ## Step 3: **Solve for Velocity \( V_i \) for Each Pipe** Rearrange the equation: \[ V_i = \sqrt{ \frac{2g \cdot h_L \cdot d_i}{f_i L_i} } \] --- ## Step 4: **Guess and Check for f (Moody Chart or Formula)** ### Pipe 1: - Initial guess: \( f_1 = 0.0262 \) - Compute velocity: \[ V_1 = \sqrt{ \frac{2 \cdot 9.81 \cdot 20.3 \cdot 0.08}{0.0262 \cdot 100} } \] \[ V_1 = \sqrt{ \frac{31.8368}{2.62} } = \sqrt{12.1579} \approx 3.49 \, \text{m/s} \] - Reynolds number: \[ Re_1 = \frac{V_1 \cdot d_1}{\nu} = \frac{3.49 \cdot 0.08}{1.02 \times 10^{-6}} \approx 273,000 \] - Re-check \( f_1 \) from the Moody chart: \( f_1 = 0.0267 \) - Re-calculate \( V_1 \): \( V_1 = 3.46 \) m/s ### Pipe 2: - Initial guess: \( f_2 = 0.0234 \) - \[ V_2 = \sqrt{ \frac{2 \cdot 9.81 \cdot 20.3 \cdot 0.06}{0.0234 \cdot 150} } \] \[ V_2 = \sqrt{ \frac{23.8776}{3.51} } = \sqrt{6.8056} \approx 2.61 \, \text{m/s} \] - \( Re_2 = \frac{2.61 \cdot 0.06}{1.02 \times 10^{-6}} \approx 153,000 \) - Re-check \( f_2 = 0.0246 \) - \( V_2 = 2.55 \) m/s ### Pipe 3: - Initial guess: \( f_3 = 0.0304 \) - \[ V_3 = \sqrt{ \frac{2 \cdot 9.81 \cdot 20.3 \cdot 0.04}{0.0304 \cdot 80} } \] \[ V_3 = \sqrt{ \frac{15.9184}{2.432} } = \sqrt{6.547} \approx 2.56 \, \text{m/s} \] - \( Re_3 = \frac{2.56 \cdot 0.04}{1.02 \times 10^{-6}} \approx 100,000 \) - Re-check \( f_3 = 0.0313 \) - \( V_3 = 2.52 \) m/s --- ## Step 5: **Calculate Flow Rate \( Q_i \) for Each Pipe** \[ Q_i = A_i V_i = \left( \frac{\pi}{4} d_i^2 \right) V_i \] - **Pipe 1**: \[ Q_1 = \frac{\pi}{4} (0.08)^2 \cdot 3.46 = 0.01744 \cdot 3.46 = 0.0604 \, \text{m}^3/\text{s} \] \[ Q_1 = 0.0604 \times 3600 = 62.5 \, \text{m}^3/\text{h} \] - **Pipe 2**: \[ Q_2 = \frac{\pi}{4} (0.06)^2 \cdot 2.55 = 0.00707 \cdot 2.55 = 0.01802 \, \text{m}^3/\text{s} \] \[ Q_2 = 0.01802 \times 3600 = 25.9 \, \text{m}^3/\text{h} \] - **Pipe 3**: \[ Q_3 = \frac{\pi}{4} (0.04)^2 \cdot 2.52 = 0.001257 \cdot 2.52 = 0.00317 \, \text{m}^3/\text{s} \] \[ Q_3 = 0.00317 \times 3600 = 11.4 \, \text{m}^3/\text{h} \] --- ## Step 6: **Total Flow Rate** \[ Q = Q_1 + Q_2 + Q_3 = 62.5 + 25.9 + 11.4 = 99.8 \, \text{m}^3/\text{h} \] --- ## Final Answer \[ \boxed{Q = 99.8 \ \text{m}^3/\text{h}} \] --- ### **Key Formulas Used** - Darcy-Weisbach: \( h_L = \frac{V^2}{2g} f \frac{L}{d} \) - Velocity: \( V = \sqrt{ \frac{2g h_L d}{f L} } \) - Reynolds number: \( Re = \frac{Vd}{\nu} \) - Flow rate: \( Q = \frac{\pi}{4} d^2 V \) - Sum for parallel pipes: \( Q_{total} = Q_1 + Q_2 + Q_3 \) --- **Note:** Iterative adjustment of the friction factor \( f \) may be necessary using the Moody chart or Colebrook equation. The above steps use the values as per the provided solution for convergence.

# Shear Strength and Mohr's Circle Analysis ## Given Data - Cohesion \( c' = 5 \, \text{kPa} \) - Angle of internal friction \( \phi = 32^\circ \) - Effective stress \( \sigma' = 125 \, \text{kPa} \) --- ## Part A: Compute Shear Strength ### Shear Strength Formula The shear strength \( \tau \) of the soil can be calculated using the Mohr-Coulomb failure criterion: \[ \tau = c' + \sigma' \tan(\phi) \] ### Calculation 1. **Convert angle to radians for calculation:** \[ \phi \text{ (in radians)} = \frac{32 \times \pi}{180} \approx 0.5585 \, \text{radians} \] 2. **Calculate \( \tan(\phi) \):** \[ \tan(\phi) = \tan(32^\circ) \approx 0.6249 \] 3. **Substitute values into shear strength formula:** \[ \tau = 5 + 125 \times 0.6249 \] \[ \tau = 5 + 78.1125 = 83.1125 \, \text{kPa} \] ### Result \[ \boxed{\tau = 83.11 \, \text{kPa}} \] --- ## Part B: Mohr's Circle Analysis ### Definitions - **Effective Stress (\( \sigma' \))**: \( 125 \, \text{kPa} \) - **Shear Stress (\( \tau \))**: \( 83.11 \, \text{kPa} \) ### Step 1: Principal Stresses Calculation 1. **Center of Mohr's Circle:** \[ \sigma_{avg} = \sigma' = 125 \, \text{kPa} \] 2. **Radius of Mohr's Circle:** \[ R = \tau = 83.11 \, \text{kPa} \] 3. **Principal Stresses:** \[ \sigma_1 = \sigma_{avg} + R = 125 + 83.11 = 208.11 \, \text{kPa} \] \[ \sigma_2 = \sigma_{avg} - R = 125 - 83.11 = 41.89 \, \text{kPa} \] ### Step 2: Maximum Shear Stress The maximum shear stress \( \tau_{max} \) is equal to the radius \( R \): \[ \tau_{max} = R = 83.11 \, \text{kPa} \] ### Result - **Principal Stresses:** \[ \sigma_1 = 208.11 \, \text{kPa} \quad \text{and} \quad \sigma_2 = 41.89 \, \text{kPa} \] - **Maximum Shear Stress:** \[ \tau_{max} = 83.11 \, \text{kPa} \] ### Summary \[ \boxed{\sigma_1 = 208.11 \, \text{kPa}, \quad \sigma_2 = 41.89 \, \text{kPa}, \quad \tau_{max} = 83.11 \, \text{kPa}} \]

# Suitable Delivery Method for the Owner ## Considerations: - Minimizing overall project cost - Reducing management risk by limiting the number of contracts - Ability to conduct design work in-house ## Analysis of Options: ### A. Design-Build (D-B) - Single contract for design and construction - **Advantages:** Reduced interface issues, faster delivery - **Disadvantages:** Less owner control over design details, reliance on a single entity ### B. Design-Bid-Build (D-B-B) - Separate contracts for design and construction - **Advantages:** Greater owner control, competitive bidding - **Disadvantages:** Multiple contracts increase management complexity and risk ### C. Construction Manager At Risk (CMAR) - Construction manager acts as consultant during design and then as general contractor during construction - **Advantages:** Owner retains more control, early cost estimates - **Disadvantages:** Potentially higher management effort, more complex process ### D. Construction Manager As Agency - Construction manager acts solely as an agent, overseeing and coordinating work - **Advantages:** High owner control, flexible - **Disadvantages:** No direct contract with trade contractors, may increase owner management burden ## Recommended Choice: **Design-Build (D-B)** - Aligns with owner's goal to minimize management complexity (single contract) - Facilitates in-house design work integration - Potentially reduces overall project cost and delivery time --- # **Final Answer:** **A. Design-Build (D-B)**