A rectangular channel width = 3 m, slope S = 0.001 S=0.001, and Manning’s roughness n = 0.015 n=0.015. Find the discharge (Q) when water depth is 1.2 m.

 A rectangular channel width = 3 m, slope 

$S = 0.001$, and Manning’s roughness $n = 0.015$. Find the discharge (Q) when water depth is 1.2 m.

Solution:

Step 1: Manning’s equation

$$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$

Where:

• $A = b \cdot h = 3 \cdot 1.2 = 3.6 m²$

• Wetted perimeter $P = b + 2h = 3 + 2(1.2) = 5.4 m$

• Hydraulic radius $R = A / P = 3.6 / 5.4 \approx 0.667 m$

Step 2: Substitute into Manning

$$Q = \frac{1}{0.015} \cdot 3.6 \cdot (0.667)^{2/3} \cdot (0.001)^{1/2}$$

Step 3: Compute

• $R^{2/3} = 0.667^{2/3} \approx 0.763$

• $S^{1/2} = \sqrt{0.001} \approx 0.03162$

$$Q = 66.67 \cdot 3.6 \cdot 0.763 \cdot 0.03162$$ $$Q \approx 5.8 \text{ m³/s}$$

✅ Answer: Discharge = 5.8 m³/s

A clay layer of thickness 4 m is subjected to a uniform load of 100 kPa. Coefficient of volume compressibility m v = 0.0004 m²/kN mv​=0.0004 m²/kN. Find the immediate consolidation settlement.

 A clay layer of thickness 4 m is subjected to a uniform load of 100 kPa. Coefficient of volume compressibility 

$m_v = 0.0004 \text{ m²/kN}$. Find the immediate consolidation settlement.

Solution:

Step 1: Settlement formula

$$S = m_v \cdot \sigma \cdot H$$

Where:

• $S$ = settlement

• $m_v$ = coefficient of volume compressibility

• $\sigma$ = applied pressure

• $H$ = thickness of layer

Step 2: Substitute values

$$S = 0.0004 \cdot 100 \cdot 4 = 0.16 \text{ m} = 160 \text{ mm}$$

✅ Answer: Settlement = 160 mm

A concrete sample has slump = 80 mm. Determine the workability grade and suitability for: Beams, columns, and slabs

 A concrete sample has slump = 80 mm. Determine the workability grade and suitability for:

• Beams, columns, and slabs

Solution:

Step 1: Identify slump grade (IS 456:2000)

Slump (mm)	Workability
0–25	Very low
25–50	Low
50–100	Medium
100–175	High

• Given slump = 80 mm → Medium workability

Step 2: Suitability

• Medium workability concrete is suitable for beams, columns, and slabs that are reinforced and have normal compaction.

✅ Answer: Medium workability; suitable for beams, columns, slabs.

A simply supported beam of length 8 m carries a central point load of 20 kN. Find: Reactions at supports Maximum bending moment

 A simply supported beam of length 8 m carries a central point load of 20 kN. Find:

1. Reactions at supports

2. Maximum bending moment

Solution:

Step 1: Determine reactions

• For a central point load $P$ on a simply supported beam of length $L$:

$$R_A = R_B = \frac{P}{2}$$ $$R_A = R_B = \frac{20}{2} = 10 \text{ kN}$$

Step 2: Maximum bending moment

• Maximum moment occurs at midspan (under the load):

$$M_{\text{max}} = \frac{P L}{4}$$ $$M_{\text{max}} = \frac{20 \times 8}{4} = 40 \text{ kNm}$$

✅ Answer:

• Reactions: RA = RB = 10 kN

• Max bending moment: 40 kNm

Water flows through a 100 m long, 200 mm diameter pipe with a head loss of 4 m. Find the flow rate (Q) using Darcy-Weisbach equation: h f = f L D v 2 2 g hf​=fDL​2gv2​ Assume friction factor f = 0.02 f=0.02, g = 9.81 m / s 2 g=9.81m/s2.

 Problem:

Water flows through a 100 m long, 200 mm diameter pipe with a head loss of 4 m. Find the flow rate (Q) using Darcy-Weisbach equation:

$$h_f = f \frac{L}{D} \frac{v^2}{2g}$$

Assume friction factor $f = 0.02$, $g = 9.81 m/s^2$.

Solution:

1. Velocity (v):

$$v = \sqrt{\frac{2 g h_f D}{f L}} = \sqrt{\frac{2 \cdot 9.81 \cdot 4 \cdot 0.2}{0.02 \cdot 100}}$$ $$v = \sqrt{\frac{15.696}{2}} = \sqrt{7.848} \approx 2.8 m/s$$

2. Flow rate (Q):

$$Q = A \cdot v = \pi \frac{D^2}{4} \cdot v = \pi \frac{0.2^2}{4} \cdot 2.8$$ $$Q = 0.0314 \cdot 2.8 \approx 0.088 m^3/s \approx 88 L/s$$

✅ Answer: Flow rate = 0.088 m³/s (88 L/s)

Label for blog: Fluid Mechanics / Pipe Flow

Calculate the Short Break Procedure from table.1 Table 1 Time period 3 mins VehicleCount 07:00 – 07:05 45 07:05 – 07:10 55 07:10 – 07:15 30 07:15 – 07:20 65 07:20 – 07:25 40 07:25 – 07:30 50

Q :                                                                                                                                     

(a)    Calculate the Short Break Procedure from table.1

Table 1

Time period	3 mins VehicleCount
07:00 – 07:05	45
07:05 – 07:10	55
07:10 – 07:15	30
07:15 – 07:20	65
07:20 – 07:25	40
07:25 – 07:30	50

                                                     

Answer:

 

Time period	3 mins VehicleCount (v)	Count adjustment (f)	Adjusted mins count (Va)
07:00 – 07:05	45	5/3	5/3 × 45 = 75
07:05 – 07:10	55	5/3	5/3 × 55 = 91.67
07:10 – 07:15	60	5/3	5/3 × 60 = 100
07:15 – 07:20	65	5/3	5/3 × 65 = 10.33
07:20 – 07:25	40	5/3	5/3 ×40  =  67
07:25 – 07:30	50	5/3	5/3 × 50 = 83.33