How to write a field report, calculate river discharge 158 width and 9ft depth, and velocity 0.02 m/s. Water used 21mm and infiltration rate 1 minutes 16 seconds

Geography

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Answer

2.6 m3/s\text{2.6 m}^3\text{/s}

0.28 mm/s\text{0.28 mm/s}

Step-by-step solution

5 steps

1. How to write a field report
A field report documents observations and analysis from a specific location or event. It typically includes an introduction (purpose, location, date), methodology (how data was collected), observations/results (detailed findings), discussion (analysis, interpretation, limitations), and a conclusion (summary of key findings).

2. Calculate river discharge
To calculate river discharge, we first need to find the cross-sectional area of the river and then multiply it by the velocity. Ensure all units are consistent.

  1. Convert width and depth from feet to meters.

    Given:
    Width = 158 ft
    Depth = 9 ft
    Velocity = 0.02 m/s
    Conversion factor: 1 ft=0.3048 m1 \text{ ft} = 0.3048 \text{ m}

    Width in meters=158 ft×0.3048 m/ft=48.1584 m\text{Width in meters} = 158 \text{ ft} \times 0.3048 \text{ m/ft} = 48.1584 \text{ m} Depth in meters=9 ft×0.3048 m/ft=2.7432 m\text{Depth in meters} = 9 \text{ ft} \times 0.3048 \text{ m/ft} = 2.7432 \text{ m}

  2. Calculate the cross-sectional area (AA).

    A=width×depthA = \text{width} \times \text{depth} A=48.1584 m×2.7432 m=132.1209 m2A = 48.1584 \text{ m} \times 2.7432 \text{ m} = 132.1209 \text{ m}^2

  3. Calculate the river discharge (QQ).

    The formula for discharge is Q=A×VQ = A \times V.

    Q=132.1209 m2×0.02 m/s=2.642418 m3/sQ = 132.1209 \text{ m}^2 \times 0.02 \text{ m/s} = 2.642418 \text{ m}^3\text{/s}

    Rounding to two significant figures based on the velocity (0.02 m/s):

    Q≈2.6 m3/sQ \approx \boxed{\text{2.6 m}^3\text{/s}}

    3. Calculate infiltration rate
    Assuming "Water used 21mm" is the depth of water that infiltrated and "1 minute 16 seconds" is the time taken for this infiltration.

  4. Convert the time to seconds.

    1 minute 16 seconds=(1×60 s)+16 s=60 s+16 s=76 s1 \text{ minute } 16 \text{ seconds} = (1 \times 60 \text{ s}) + 16 \text{ s} = 60 \text{ s} + 16 \text{ s} = 76 \text{ s}

  5. Calculate the infiltration rate (IRI_R).

    IR=Infiltrated depthTimeI_R = \frac{\text{Infiltrated depth}}{\text{Time}} IR=21 mm76 s≈0.2763 mm/sI_R = \frac{21 \text{ mm}}{76 \text{ s}} \approx 0.2763 \text{ mm/s}

    Rounding to two significant figures:

    IR≈0.28 mm/sI_R \approx \boxed{\text{0.28 mm/s}}
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