Chapter 2 – Solutions to Odd Numbered Problems

 

 

Dr. Todd Rayne of Hamilton College provided these solutions.

 

 

  1. The standard U .S. Class A evaporation land pan has an inside diameter of 47.5 in. and a depth of 10.0 in.

 

Metric conversions

            Diameter = 47.5 in, therefore radius = r = 23.75 in

            r = 23.75 in x 2.54 cm/in = 60.33 cm or 0.6033 m

            Depth = d =10 in x 2.54 cm/in = 25.4 cm or 0.254 m

 

 

 

(A)  Calculate the surface area of water in the pan in square meters.

 

A = p r2

A = p x (0.6033 m)2 = 1.14 m2

 

 

(B)  Calculate the volume of the pan in cubic meters.

 

V = A x d

V  = 1.14 m2 x 0.254 m = 0.29 m3

 

                                   

(C)  If the initial volume of water in the pan is 11.5 U.S. gallons, what is the depth of the water in millimeters?

 

Convert gallons to m3

V = 11.5 gal x 3.78 x 10-3 m3/gal = 0.04347 m3

d =V/a = 0.04347 m3/1.14 m2 = 0.0381 m or 3.81 mm

 

 

(D)  If after a 24-h period with no precipitation the volume of water in the pan is measured and found to be 10.2 U.S. gallons, what is the evaporation rate in millimeters/ day?

 

Initial volume = 0.04347 m3

Final volume = 10.2 gal x 3.78 x 10-3 m3/gal = 0.03856 m3

Change = 0.04347 m3 – 0.03856 m3 = 0.0049 m3 / day

d=V/A = 0.0049 m3 / 1.14 m2  = 0.00430 m/day = 4.30 mm/day

 

 

(E)   What would be the depth of water in millimeters?

     

Initial depth = 38.1 mm

Change = - 4.3 mm

New depth = 33.8 mm

 

 

(F)   During the succeeding day there was a 3-h period of precipitation at a constant rate of 5 mm/h. Assuming that the 24-h evaporation rate calculated in step D also occurs during this 24-h period, what would be the depth of water in the pan?

 

Precipitation = 5 mm/hr x 3 hr = 15 mm

Evaporation = 4.3 mm/day

New depth = 33.8 mm + 15 mm – 4.3 mm = 44.5 mm

 

 

(G)  If there is no further rain, and no water is added to the pan, how long would it take for the water in the pan to totally evaporate, assuming the net 24-h evaporation rate of step D?

 

To evaporate all water

44.5 mm / (4.3 mm/day) = 10.3 days

 

3. Figure 2.26 is a map of a drainage basin and the rainfall amounts during a storm at a number of precipitation stations both within and outside the drainage basin. Make a Thiessen network drawing for the drainage basin. The exact station location is the decimal point in the rainfall amount. The relative size of the area associated with each Thiessen polygon can be measured with a planimeter or estimated by tracing the Thiessen network on cross-section paper and counting the number of squares in each polygon. Estimate the effective uniform depth of precipitation.

 

Precipitation at station (L)

Polygon area (L2)

Fraction of total area

Weighted Precipitation (L)

1.78

0.0562

0.089

0.159

1.55

0.0752

0.119

0.185

1.07

0.0540

0.086

0.092

1.35

0.0454

0.072

0.097

0.23

0.0472

0.075

0.017

1.63

0.0439

0.070

0.114

2.11

0.0354

0.056

0.119

0.92

0.0422

0.067

0.062

0.55

0.0390

0.062

0.034

0.72

0.0469

0.074

0.054

2.01

0.0257

0.041

0.082

1.12

0.0280

0.044

0.050

1.78

0.0268

0.045

0.081

2.08

0.0110

0.017

0.036

1.43

0.0091

0.014

0.021

0.06

0.0072

0.011

0.001

0.16

0.0079

0.013

0.002

1.46

0.0079

0.013

0.018

1.13

0.0061

0.010

0.011

0.42

0.0042

0.007

0.003

2.21

0.0036

0.006

0.013

2.37

0.0023

0.004

0.009

1.46

0.0029

0.005

0.007

0.83

0.0001

0.000

0.000

Totals

0.0630

 

1.2638

The net E.U.D. is therefore 1.26 (L) where L is the unit of length.

 

 

5. A pond has a surface area of 35 ac. If the mean daily air temperature is 66°F, the mean daily dew-point temperature is 55°F, the solar radiation is 480 langleys, and the daily wind movement is 115 mi, what is the daily lake evaporation in acre-feet?

 

 

                Using the nomograph in figure 2.1 the evaporation rate is 0.15 in/day

            Evaporation = 0.15 in/day x 0.083 ft/in x 35 ac = 0.44 ac-ft/day

 

 

7. Consider an air mass that has an absolute humidity of 10 g/m3 at a temperature of 22°C. Using the graph that you created for Problem 2.6, find (a) the dew point, and (b) the relative humidity.

 

 

        The dew point is the temperature corresponding to the intersection of the curve with the line extending parallel to the x=axis from the absolute humidity.

                  Dew point = 11o C

      The relative humidity is the ratio of the absolute humidity to the saturation humidity for a given temperature. At 22 oC the saturation humidity is 19.5 g/m3.

                  Relative humidity = (10 gm/m3) / (19.5 gm/m3) x 100

                  Relative humidity = 51 %

 

 

 

9. The flow of a river at the start of a base flow recession was 712 m3 / s; after 60 d the flow declined to 523 m3 / s. (A) What is the recession constant? (B) What would be the flow after 112 d?

 

(A)

 

            Given:

                        Qo = 712 m3/s,             t = 60 days,      Q = 523 m3/s

                       

                        Q = Qo e-at

 

                        Q/Qo = e-at

 

                        ln (Q/Qo) = -at

 

                        a = -(1/t) ln (Q/Qo)

 

                        a = - (1/60 day) ln (523/712)

 

                        a = 5.14 x 10-3 day-1

 

(B)

 

      Given:

                  Qo = 712 m3/s,             t = 112 days,                a = 5.14 x 10-3 day-1

     

                  Q = Qoe-at

 

                        Q = 712 e –at

 

                        Q = 712 x e – 5.14 x 10^-3 x 112

 

                  Q = 712 x 0.562

 

                  Q = 400 m3/s

 

11. A 90 o V-notch weir is placed in a road culvert to measure the flow of a stream passing through the culvert. The value of H is 2.72 ft. Compute the discharge of the stream.

 

                        Q = 2.5 H5/2

 

                    H  = 2.72 ft

 

                  Q = 2.5 x 2.725/2

 

                  Q = 30.5 ft3/s

 

 

 

13. An industrial park with flat-roofed buildings, large parking lots, and little open area has a drainage basin area of 398 ac. The 25-year rainfall event (the amount that would on an average occur once in 25 years) has a precipitation intensity of 2.382 in./h. If the C factor is 0.75, what is the maximum rate that overland flow will drain from the industrial park?

 

       Q = C I A

     

      A  = 398 ac x 43,560 ft2/ac = 17,336,880 ft2

 

      I = 2.382 in/hr x 0.083 ft/in x 0.01667 min/s = 5.5139 x 10-5 ft/s

 

      C = 0.75

 

      Q = 0.75 x 17,336,880 ft2 x 5.5139 x 10-5 ft/sec

 

      Q = 717 ft3/sec

 

 

15. Figure 2.28 is the hydrograph of a river with a long summer baseflow recession. Compute the volume of annual recharge that occurs between runoff year 1 and runoff year 2.

 

      Qo in year 1 is 280 ft3/s

 

0.1  Qo in  year 1 is reached in 5.5 months

 

t1 = 5.5 mo x 8.64 x 104 sec/day x 30 day/mo = 1.4256 x 107 s

 

Vtp = Qo t1 / 2.3026

 

Vtp = (280 x 1.4256 x 107) / 2.3026 = 1.73 x 109 ft3

 

Total recession in year 1, t, is 6.4 months.

 

Vt = Vtp / 10t(t/t1)

 

Vt = 1.73 x 109 ft3 / 10(6.4/5/5)

 

Vt = 1.19 x 108 ft3 Where Vt is the volume of water that could contribute to baseflow at the end of the first year’s recession.

 

Qo in year 2 is 320 ft3/s

 

0.1  Qo is reached in 5.9 mo

 

t2 is 5.9 mo or 1.53 x 107 s

 

Vtp2 = (320 ft3/s x 1.53 x 107s) / 2.3026 = 2.13 x 109 ft3

 

The amount of recharge between recessions is Vtp2 -Vt

 

Vtp2 -Vt = 2.13 x 109 ft3  - 1.19 x 108 ft3

 

 Vtp2 -Vt = 2.00 x 109 ft3                                                       

 

 

 

17. An aqueduct has smooth earthen sides and bottom. The slope of the water surface is 1.7 ft/mi. The channel is trapezoidal in shape with a 45° angle to the sides of the trapezoid and a bottom segment that is 8.5 ft wide. The water in the aqueduct is 3.6 ft deep in the center.

(A) What is the average velocity of water in the aqueduct?

(B) What is the volume of flow in the aqueduct?