Type:
Solution Manual
Resource:
Water-Resources Engineering
Edition:
3rd Edition
Author(s):
David A. Chin
Chapter 1
Introduction 1.1. The mean annual rainfall in Boston is approximately 1050 mm , and the mean annual evapotranspiration is in the range of 380–630 mm (USGS). On the basis of rainfall, this indicates a subhumid climate. The mean annual rainfall in Santa Fe is approximately 360 mm and the mean annual evapotranspiration is < 380 mm . On the basis of rainfall, this indicates an arid climate.
1
Chapter 3
Design of Water-Distribution Systems 3.1. Use Drinking Cooking Bathing Washing of cloths Washing of utensils Washing and clearing of houses and residences Flushing of water, closets etc. Lawn watering and Gardening Total
Consumption in L/d/person 5 5 55 20 10 10 30 15 150
3.2. (a) By graphical extension P2030 = 100, 000 (b) For arithmetic growth: P = kt + P0 where k=
61000 − 52000 P1990 − P1980 = = 900 10 10
Therefore P2030 = 900t + P1990 = 900(40) + 61000 = 97, 000 (c) For geometric growth: P = P0 ekt Therefore P1990 = P1970 ek(20) 61000 = 40000e20k k = 0.021 50
51 and hence P2030 = P1990 ek(40) = 61000e(0.021)(40) = 141, 298 (d) For declining growth: P = Psat − Ce−kt
(1)
where Psat = 100,000 and 1970 : t = 0, P = 40000 1990 : t = 20, P = 61000 Substituting 1970 population into Equation 1 gives 40000 = 100000 − Ce−k(0) = 100000 − C which gives C = 60, 000 Substituting 1990 population into Equation 1 gives 61000 = 100000 − 60000e−k(20) which gives k = 0.0215 Equation 1 can now be used to predict the 2030 population (t = 60) as P2030 = 100000 − 60000e−0.0215(60) = 83, 483 (e) Equations 3.11 and 3.12 give the logistic-curve parameters a and b as Psat − P0 100 − 40 = 1.5 = P0 40 1 P0 (Psat − P1 ) 1 40(100 − 52) b= ln = ln = −0.0486 Δt P1 (Psat − P0 ) 10 52(100 − 40)
a=
The logistic curve for predicting the population is given by Equation 3.9 as P =
100, 000 Psat = bt 1 + ae 1 + 1.5e−0.0486 t
In 2030, t = 60 years and the population given by Equation 3.19 is P = 92, 487 people
(2)
52 3.3. According to Equation 3.10 – (a) 2P0 P1 P2 − P − 12 (P0 + P2 ) P0 P2 − P12 (2 × 25, 000 × 40, 000 × 70, 000) − (40, 000)2 × (25, 000 + 70, 000) 25, 000 × 70, 000 − (40, 000)2 = 80, 000
Psat =
According to Equation 3.11 a=
Psat − P0 80, 000 − 25, 000 = 2.2 = P0 25, 000
According to Equation 3.12 1 P0 (Psat − P2 ) ln t P1 (Psat − P0 ) 1 25, 000(80, 000 − 40, 000) = ln 10 40, 000(80, 000 − 25, 000) = −0.079
b=
(b) Using Equation 3.9 Psat 1 + aebt 80, 000 = 1 + 2.2 × e−0.079t
P( ) =
(c) When t = 30 yrs, 80, 000 1 + 2.2 × e−0.079×30 = 66, 251
P( ) =
Hence population after 10 years will be 66,251. 3.4. Diameter of single main pipe = 300mm = 0.3m Let the length of single pipe line be ‘L’ According to Equation 2.33– 2
Lv hf = f2gD
2
f Lv = 2g×0.3