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20.6. Relief is the highest elevation minus the lowest elevation. Step 3: Thickness of Sediment
5
V = 1.3 × 10 m 3 V _
summit = 5,930 m relief = summit − base 4 2 V = Az ∴ z =
A = 3.6 × 10 km A
base = 5,400 m = 5.,930 m − 5,400 m
z = ? Convert the area from km to m.
relief = ? = 530 m
2 )
(
6
2 1 × 10 m
4

3.6 × 10 km _ 2

20.7. Determine the stream gradient by using the slope equation. 1 km
10
3.6 × 10 m 2
Δy
Δx = 13.8 km _
gradient = __ 3
5
1.3 × 10 m
Δy = 5 m Δx z = 10 2

5 m
gradient = ? = _ 3.6 × 10 m

5
3
1.3 × 10 m _
13.8 km = _

m _ 3.6 × 10 10 m 2
= 0.36
km −6
= 3.6 × 10 m
20.8. This type of problem is best solved by subdividing it into steps 20.10. Step 1: Volume of Sediment

that first calculate the volume of sediment eroded each year. 5 _ 3 m _ m _

Mg

m = 4.5 × 10 (1 × 10 yr) ρ = ∴ V =

V
ρ

The mass of sediment being transported is then determined yr
8
8
4.5 × 10 Mg
from the volume. = 4.5 × 10 Mg __
V =
Mg
Mg _
_
Step 1: Volume Removed ρ = 1.2 1.2
m 3 m 3
z = 1 mm V = Az V = ? 8
4.5 × 10 Mg
2

A = 5.12 × 10 km 2 = _ _

Convert the thickness from mm to km. 1.2 Mg
V = ? _ 3

( _ m
6 )
1 km

1 mm

8
1 × 10 mm = 3.8 × 10 m 3
−6
1 × 10 km Step 2: Area of Delta
2
2
−6
V = 1 × 10 km(5.12 × 10 km ) L = 150 km Based on the delta being an isosceles triangle:
= 5 × 10 km 3 W = 240 km A = LW
−4
A = ? = 150 km(240 km)
Step 2: Mass Transported = 3.6 × 10 km 2
4
−4
V = 5 × 10 km 3 m _
ρ = ∴ m = Vρ
Mg
_ V Step 3: Thickness of Sediment
ρ = 1.35 8 3
Mg
Mg
m 3 _ _ V = 3.8 × 10 m V _

4
A
m = ? Convert m 3 to km 3 . A = 3.6 × 10 km 2 V = Az ∴ z =
3
2
z = ? Convert the area from km to m .
9
_ _ 3
Mg 1 × 10 m

1.35 3 ( 3 ) 4 2 1 × 10 m 2

6
(
_

m 1 km 3.6 × 10 km 1 km 2 )
Mg
9 _
10
1.35 × 10 3.6 × 10 m 2

km 3 __
8
3
3.8 × 10 m

(
9 _
10
−4
3
m = 5 × 10 km 1.35 × 10 Mg 3) z = 3.6 × 10 m 2

km _ 8 3
3.8 × 10 m _
= 10

(
Mg
3 _
−4
9
= 5 × 10 (1.35 × 10 ) km 3) 3.6 × 10 m 2

km = 1.1 × 10 m
−2
5
= 7 × 10 Mg
20.11. To determine the discharge, multiply the velocity of the stream

20.9. This type of problem is best solved by subdividing it into steps by its cross-sectional area.
that consider (1) the volume of sediment deposited in
m _
1,000 years, (2) the area of the delta, and (3) the thickness of v = 2.6 Q = vA m _ 2
s
sediment associated with this volume. A = 54.8 m 2 = 2.6 (54.8 m )
s
3
2 m _

Step 1: Volume of Sediment Q = ? = 1.42 × 10
s
m _
m _
Mg
2 _ 3 ρ = ∴ V =

m = 1.6 × 10 (1 × 10 yr) V ρ 20.12. To determine the velocity, divide the discharge by the

yr
5
= 1.6 × 10 Mg V = __ cross-sectional area.
1.6 × 10 Mg
5

Mg
Mg _ 3 m _ 3 Q
_
_

ρ = 1.2 1.2 3 Q = 1.13 × 10 Q = vA ∴ v =
s
m 3 m A
3
V = ? _ _ A = 3.68 × 10 m 2 3 m _ 3
1.6 × 10 Mg
5

1.13 × 10

= __
s
1.2 _ v = ? = 3 2
Mg

m 3 3.68 × 10 m
m _
5
= 1.3 × 10 m 3 _ s 3
3
1.13 × 10 _

=

Step 2: Area of Delta 3.68 × 10 m 2
3
L = 150 km Based on the delta being an isosceles triangle: −1 m _

s
W = 240 km A = LW = 3.07 × 10
A = ? = 150 km(240 km)
4
= 3.6 × 10 km 2
E-45 APPENDIX E Solutions for Group A Parallel Exercises 687