---
title: "Slope-area flood discharge, Marin River at Kusuntar"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-marin-kusuntar"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-marin-kusuntar
site: BCT Engineering
---

# Slope-area flood discharge, Marin River at Kusuntar

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2080 Baisakh, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-marin-kusuntar
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

Calculate the flood discharge of Marin River at Kusuntar by slope area method using the following data:

| | Upstream | Downstream |
|---|---|---|
| Flow area | 175 m$^2$ | 224 m$^2$ |
| Wetted perimeter | 75 m | 166 m |

Reach length = 120 m
Fall in water surface elevation = 1.46 m
Manning's coefficient = 0.1

## Answer

**Given:** $A_1=175$ m$^2$, $P_1=75$ m, $A_2=224$ m$^2$, $P_2=166$ m, $L=120$ m, fall $=1.46$ m, $n=0.1$. No velocity-head or eddy-loss coefficient is given, so $\alpha=1$ and $h_e=0$ are assumed.

**Method:** Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

#### Step 1: Conveyance
$$
R_1=\frac{175}{75}=2.3333\ \text{m},\qquad R_2=\frac{224}{166}=1.3494\ \text{m}
$$
$$
K_1=\frac{1}{0.1}\times175\times2.3333^{2/3}=3078.6,\qquad K_2=\frac{1}{0.1}\times224\times1.3494^{2/3}=2735.3
$$
$$
K_m=\sqrt{K_1K_2}=2901.9
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{1.46}{120},\qquad Q_1=K_m\sqrt{S_f}=320.087\ \text{m}^3/\text{s}
$$

#### Step 3: Correct for velocity head and eddy loss
$$
h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|
$$
With $h_e=0$: $h_f = 1.46 + (h_{v1}-h_{v2})$.

| Trial | $Q$ assumed | $v_1$ | $v_2$ | $\alpha_1 v_1^2/2g$ | $\alpha_2 v_2^2/2g$ | $h_e$ | $h_f$ | $Q=K_m\sqrt{h_f/L}$ |
|---|---|---|---|---|---|---|---|---|
| 1 | 320.087 | 1.8291 | 1.4290 | 0.17051 | 0.10407 | 0.00000 | 1.52644 | 327.289 |
| 2 | 327.289 | 1.8702 | 1.4611 | 0.17827 | 0.10881 | 0.00000 | 1.52946 | 327.613 |
| 5 | 327.628 | 1.8722 | 1.4626 | 0.17864 | 0.10904 | 0.00000 | 1.52961 | 327.628 |

The trials converge to $h_f$ and $Q$ shown in the last row.

**Answer: Discharge $Q \approx 327.6\ \text{m}^3/\text{s}$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is assumed since no coefficients are given?
   A. alpha = 1.5 and he = 0.3
   B. alpha = 1 and he = 0
   C. n = 0.025
   D. Fall = 0
   Answer: B. The answer takes no velocity-head coefficient or eddy loss. (Key point: Assumptions)

2. What is the first trial discharge?
   A. Q1 = Km x 1.46 = 4237 m^3/s
   B. Q1 = 2901.9 m^3/s
   C. Q1 = Km sqrt(1.46/120) = 320.1 m^3/s
   D. Q1 = 175 x 224
   Answer: C. Km = sqrt(K1 K2) = 2901.9 and Sf = fall/L. (Key point: Conveyance)

3. Why does the discharge increase to 327.6?
   A. Eddy loss reduces it
   B. n changes
   C. The reach length changes
   D. Velocity head drops downstream (hv1 > hv2), raising hf above the fall
   Answer: D. hf = 1.46 + (0.1705 - 0.1041) = 1.526 m, so Q increases. (Key point: Convergence)

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