Lecture 1
Population forecasting
Design Period
 Technical Lifetime : It represents the period during which it
operates satisfactorily in a technical sense.
 The
suggested periods for the main
components shown in the following table :
2
Component
Period ( Years )
Transmission main
30 - 50
Distribution main
30 - 100
Reservoirs
20 - 80
Pumping station - facilities
20 - 80
Pumping station - equipment
15 - 30
distribution
system
Technical Concerns
Economical Concerns
Estimation of design period
3
‫‪Estimation of population‬‬
‫النمو السكاين الطبيعي نتيجة للتناسل يرتاوح عادة مابني ‪ %1‬إىل ‪ %7‬حسب اجملتمعات وعاداهتا‬
‫وتقاليدها‪ ،‬وعلى سبيل املثال يف اجملتمع الفلسطيين ترتاوح معدل النمو الطبيعي من ‪ %3‬إىل ‪ ،%6‬ففي‬
‫مدينة كمدينة غزة يبلغ معدل النمو الطبيعي ‪ ، %4.5‬بينما تبلغ الزايدة يف معسكر جباليا حوايل‬
‫‪ .%6‬بينما تبلغ النسبة كمتوسط يف مجيع أحناء فلسطني ‪ ،%2.5‬ويوضح اجلدول التايل توقع مركز‬
‫اإلحصاء‬
‫الفلسطيين للسكان‪:‬‬
‫‪4‬‬
Population forecasting
 Arithmetic increase method ‫الطريقة الرايضية‬
This method assume that the population density increase uniformly
Population after time t
This equation can be
simplified to:
5
Present population
Pf  Po 1  kt 
population growth rate by each t unit
 Uniform percentages Method ‫طريقة ذات النسبة المنتظمة‬
(Geometric Increase)
dP
 kP
dt
population growth rate by each t unit
ln Pf  ln P0  k Δt
• Predicted curves Method ‫طريقة المنحنيات التوقعية‬
(curvilinear method)
Pf  Po 1  k 
t
6
population growth rate by each t unit
Example
 Estimate the population number after 30 year, if the initial
population number is 100,000. the population growth rate is 3%
Pf  Po 1  kt 
Pf  1000001  0.03(30)   190,000
Method 1
ln Pf  ln P0  k Δt
ln Pf  ln 100000  0.03 30
Pf  expln( 100000)  (0.03  30)  245,961 Method 2
Pf  Po 1  k 
t
7
Pf  Po 1  0.03  242,726
30
Method 3
Population Saturation and urban planning
 Master Plane
 Land-Use
 General laws
Although the population forecasting methods gives the population at
the end of the design period, the maximum possible population is also
estimated according to the number of apartment and stories per unit
area and the maximum family members.
The Saturation population can be determined using the Master plan for
the area and its general laws.
The minimum of the two values is taken in consideration when the
network is designed.
8
Example
A residential area 100,000 m² and population of 4200 capita has the
following Master Plan :
25 % of the total area is for Streets.
5 % of the total area is for green Areas.
70 % of the total area is for buildings.
60 % of the building area is for building.
If the family living there consists of members and the average
apartment area is 150 m², Calculate the Population needed for
designing a water distribution system to serve that area for 25
years if you are given that the growth rate of the population = 3 %.
9
Cont. example
Pt = P0 (1+k)n
Pt = 4200 (1+0.03)25
Pt = 8794 capita
10
Total Building Area (m²)
0.7 × 100,000 = 70,000
Building Area (m²)
0.6 × 70,000 = 42,000
5 Stories Building Area
42,000 × 5 = 210,000
No. of Apartments
210,000 / 150 =1400
Saturation Population
1,400 × 5 = 7000 capita
Area Study
Residential Area
Public Area
Commercial Area
Agricultural Area
Streets
Item
Area (m2)
%
Total Area
641 Duonm
100
Residential Area
519,917
81
Public Area
19,230
3
Commercial Area
25,640
4
Agricultural Area
12,820
2
Industrial Area
38,460
6
Streets
25,640
4
Population Density
Calculation of total number of residents:
1.
Normal Increase ( using equations)
Hypothetical
figures
Pf  Po 1  k 
t
Example:
Total Area of Southern Remal = 2,754 Dounm
- Total population in Southern Remal in 2002 = 17,500
- Total population in Southern Remal in 2008 = 29,300
k

P2008  P2002
6  P2002
29300  17500
 0.112
6 17500
Statistical
Data
P2038  P2008 (1  0.112)^30
 427,213 person
Continue..
 Population in Area under consideration:
Population
Area
427,213
2,754,000
?
641,000
 Population in T arg et Area 
427,213  641,000
 99,435 person
2,754,000
Population Density
2.
Saturation population
Total Residential Area = 519,917 m2
Area
Avg. no. of floors
Avg. area of flats (m2)
# of flats
Avg. no. of persons/flat
No. of residents
Regular Buildings
(5 floors or less)
Towers
(6 floors and more)
370,122
149,795
3
8
170
140
(370,122/170) x 3
(149,795/140) x 8
7
7
45,721
59,918
Total number of population under saturation condition = 45,721 + 59,918 = 105,639
Population Density
Therefore,
Population density= Population / Total Area
= 99,435 / 641,000 = 0.155 person/m2