Problem 3-1

From Table 2-4, for T_a = 20°C, α = 2.19.
For Table A-1 (Appendix A), for T_a = 20°C, the heat of vaporization, H_v = 586 cal/g, and the density, ρ = 0.99821 g/cm³ The net radiation in evaporation units (solving from Eq. 2-24) is:

The saturation vapor pressure at the air temperature (Table A-1) is: e_o = 23.37 mb. Using the Dunne formula (Eq. 2-39), the mass-transfer rate of evaporation is:
E_a = [ 0.013 + (0.00016 • 180) ] (23.37) [ (100 - 30)/100 ] = 0.684 cm/d. Using Eq. 2-38: E = [ (2.19 • 0.898) + 0.684 ] / (2.19 + 1) = 0.831 cm/d.  ANSWER.

Problem 3-2

Using Eq. 2-44, the monthly heat indexes are:

Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
2.86	3.76	4.75	5.82	6.95	8.16	8.78	7.55	6.38	5.28	4.25	2.86

The temperature efficiency index, J, is the sum of the monthly heat indexes I: J = 67.39. Using Eq. 2-46: c = 1.556. Using Eq. 2-45 for the month of July: PET(0) = 9.385 cm/mo. Using Table A-4 (Appendix A): PET = 1.23 X 9.385 = 11.54 cm during the month of July.  ANSWER.

Problem 3-3

At x = 0 km, elevation y = 100 m. At x = L = 250,000 m, elevation y = 28.65 m. From Fig. 2-19, the area comprised between elevation 28.65 and the longitudinal profile is:   A_p = YL/2.

Therefore, the S₂ slope is: S₂ = Y/L = 2A_p/L².

The total area, A_t, below the longitudinal profile is obtained by integration:

A_t = ∫₀^250,000 y dx = ∫₀^250,000100e^-0.000005xdx = 14,269,904

The area comprised between elevation 0 and elevation 28.65 m is:

A_b = 28.65 X 250,000 = 7,162,500 m². Therefore, the area A_p is:

A_p = A_t - A_b = 7,107,404 m². And: S₂ = 2A_p/L² = 0.000227. ANSWER.

Problem 3-4

(a) Using Eq. 2-73, with C = 30:

q_p = 147.8 ft³/s/mi², and Q_p = 96,070 ft³/s. ANSWER.

(b) Using Eq. 2-73, with C = 100:

q_p = 492.7 ft³/s/mi², and Q_p = 320,300 ft³/s. ANSWER.