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2. After measuring the distance across the river channel, one measures the water

ID: 118227 • Letter: 2

Question

2. After measuring the distance across the river channel, one measures the water depths at equally spaced intervals. The depths and distances are then used to calculate the area within each cross-section segment. Calculate the Total Discharge (Q) by adding up the discharge calculated in each segment (See table below). Average the depths of each polygon by adding up the two sides and dividing your answer by two. Discharge is calculated by multiplying the average depth times the velocity. Use appropriate units in your final answer Distance across stream (m) 0 0.50 1.0 1.5 2.0 2.5 2. Velocities (m/s) 0.10 | 0.23 | 0.41 | 0.35 | 0.21 10.05m/s Depth (m) at each 0.50m 0.18 0.24 0.27 0.42 0.90m :A«V Field data collected to calculate Discharge (Q) Cross- sectional area of m channel 2 city mls Discharge 26n' I 0.10 mis 2 M 2.0S s Total dis chovge

Explanation / Answer

Cross Sectional Area (m2)

for each subsection of 0.5m width

(0.5 x corresponding depth)

Velocity (m/s)

Discharge (Q) m3/sec

0.5 x 0.18 = 0.09

0.10

0.09 x 0.10 = 0.009

0.5 x 0.24 = 0.12

0.23

0.12 x 0.23 = 0.028

0.5 x 0.27 = 0.14

0.41

0.14 x 0.41 = 0.057

0.5 x 0.42 = 0.21

0.35

0.21 x 0.35 = 0.074

0.5 x 0.90 = 0.45

0.21

0.45 x 0.21 = 0.095

0.43 x 0.90 = 0.39

0.05

0.39 x 0.05 = 0.019

Total Discharge = 0.282

Cross Sectional Area (m2)

for each subsection of 0.5m width

(0.5 x corresponding depth)

Velocity (m/s)

Discharge (Q) m3/sec

0.5 x 0.18 = 0.09

0.10

0.09 x 0.10 = 0.009

0.5 x 0.24 = 0.12

0.23

0.12 x 0.23 = 0.028

0.5 x 0.27 = 0.14

0.41

0.14 x 0.41 = 0.057

0.5 x 0.42 = 0.21

0.35

0.21 x 0.35 = 0.074

0.5 x 0.90 = 0.45

0.21

0.45 x 0.21 = 0.095

0.43 x 0.90 = 0.39

0.05

0.39 x 0.05 = 0.019

Total Discharge = 0.282