2081 Baisakh · 8 marks

Discharge of Jhiku khola from stream gauging data

Short answer

Mid-section method: mean velocity v_m = (v_0.2 + v_0.8)/2 for each vertical, then q = v_m × depth × segment width. Summed across all strips, discharge Q ≈ 8.93 m³/s.

Following data is the discharge measurement of Jhiku khola using stream gauging, compute the discharge of Jhiku khola.

Distance from left bank (m)	Depth (m)	Velocity at 0.2d (m/s)	Velocity at 0.8d (m/s)
0	-	-	-
1.5	1.1	0.7	0.42
3.0	2.7	0.85	0.63
5.0	1.9	0.75	0.54
7.5	0.9	0.65	0.46
10.0	0.5	0.45	0.35
12.5	-	-	-

Mid-section method with the mean velocity of each vertical $v_m=(v_{0.2}+v_{0.8})/2$ :

q_i=v_m\,d\,\frac{b_{i+1}-b_{i-1}}{2}

The two banks (0 m and 12.5 m) have zero depth.

x (m)	d (m)	$v_{0.2}$	$v_{0.8}$	$v_m$ (m/s)	w (m)	Area (m $^2$ )	q (m $^3$ /s)
1.5	1.1	0.7	0.42	0.560	1.5	1.650	0.924
3	2.7	0.85	0.63	0.740	1.75	4.725	3.497
5	1.9	0.75	0.54	0.645	2.25	4.275	2.757
7.5	0.9	0.65	0.46	0.555	2.5	2.250	1.249
10	0.5	0.45	0.35	0.400	2.5	1.250	0.500
Total						14.150	8.927

Mean velocity $= 0.631$ m/s.

Answer: Discharge of Jhiku Khola $Q \approx 8.93\ \text{m}^3/\text{s}$ .

✓ Answer checked by Claude Sonnet 5.5

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