DBS721S - Drilling and Blasting 324 - 1st Opp - Nov 2022


DBS721S - Drilling and Blasting 324 - 1st Opp - Nov 2022



1 Pages 1-10

▲back to top


1.1 Page 1

▲back to top


n Am I BI A u n IVE RS ITV
OF SCIEnCE Ano TECHnOLOGY
FACULTY OF ENGINEERING AND THE BUILT ENVIRONMENT
DEPARTMENT OF MINING AND PROCESSENGINEERING
QUALIFICATION : BACHELOR OF ENGINEERING IN MINING ENGINEERING
QUALIFICATION CODE: 08BMIN
LEVEL: 7
COURSE CODE: DBS721S
COURSE NAME: DRILLING AND BLASTING 323
SESSION: NOV 2022
DURATION: 3 HOURS
PAPER: THEORY
MARKS: 100
EXAMINER(S)
MODERATOR:
FIRST OPPORTUNITY EXAM PAPER
Mr L. Madziwa
Mr P. Shava
INSTRUCTIONS
1. Answer all questions.
2. Read all the questions carefully before answering.
3. Marks for each question are indicated at the end of each question.
4. Please ensure that your writing is legible, neat and presentable.
PERMISSIBLEMATERIALS
1. Examination paper
2. Scientific calculator
THIS QUESTION PAPER CONSISTS OF 6 PAGES (Including this front page)
1

1.2 Page 2

▲back to top


1.3 Page 3

▲back to top


Question 1 [20 marks]
a) What are the effects of the delay timing and the initiation sequence of a blasting are
not adequate?
[4]
b) Blast designs are undertaken using controllable and uncontrollable factors. Describe
how five key controllable factors and five uncontrollable factors affect blast designs.
[10]
c) Explain how drillhole diameter, burden and powder factors variables affect
fragmentation.
[6]
Question 2 [12 marks]
A blast round was designed in a way that the diameters of the drilled holes were 50 mm,
75mm and 115mm with spacing to burden ratio 1:2. The bench height was determined by the
length of the drill rods coupled together which are 3 meters each. 0.45 m deviation was
recorded after using four of those drill rods. If rock constant is 10, and the density of the ANFO
is 900kg/m 3 •
a) Determine the loaded length, quantity of explosives and powder factor
required for the three-hole diameters, taken the mean fragment size to be 30
cm, Lis the loaded length above the toe, J is the hole stemming, H is the
bench height, D is the diameter of drilled hole and SANPCJ is relative specific
weight of ammonium fuel oil. Calculate the loaded hole length of explosive,
the quantity of explosive as well as powder factor? Each parameter carries
one mark.
Hole diameter,
D(mm)
so
75
115
Charge Length, L Amount
Powder factor, K (Kg/m 3)
(m)
explosives,
Oe
(Kg/hole)
................
..................
.....................
................
................
··················
..................
.....................
.....................
b) Draw your conclusion from the answers in (a)
[3]
2

1.4 Page 4

▲back to top


1.5 Page 5

▲back to top


Question 3 [14 marks]
The open pit mine has massive fine-grained lava with a skarn gold mineralization and there is
little jointing closer than anticipated drilling pattern. The mine management has requested
you as a mining engineer to determine the rock factor. The massive grained lava rock has the
following data:
UCS = SOOMPA,
Young's Modulus= 80 GPa
Density= 2.8 t/m ,
RDM =50,
J=O
Determine the following;
a) Rock density influence (RDI)
[2]
b) Hardness factor (HF),
[2]
c) Rock Factor (A)
[3]
Question 4 [20 marks]
In blasting operations, drilling, and how explosives are charged, and the quantity of explosives
used are major contributors to the output of a blast.
a) Outline the effects of fragmentation on drilling, blasting, loading and hauling
cost?
[4]
b) Explain what you understand by the term initiation system in blasting. [2]
c) Explain what you understand by the term initiation sequence in blasting. [2]
d) Why is hole orientation important in drilling?
[2]
e) Define cushion blasting using open pit mining and underground mining
example.
[4]
f) What is smooth wall blasting?
[2]
g) What is pre-splitting?
[2]
h) What is post-splitting?
[2]
Question 5 [14 marks]
3

1.6 Page 6

▲back to top


1.7 Page 7

▲back to top


If a mine drift is to be excavated in rock (c =0.4) by means of blasts with parallel blast holes
and four sections cut, knowing that the geometric dimensions and drilling data are:
• Tunnel width AT =4.5 m
• Abutment height= 4.0m
• Height of arch =0.5 m
• Relief hole diameter 0 2 = 102mm
• Drilling diameter 0 1 =45mm
The explosive to be used has a Relative Weight Strength with respect to ANFO of 1.09 (109%)
and the available cartridges have diameter of 25, 32 and 38mm, which give lineal charge
concentrations
for
density
of
1.2
g/cm
3
,
of
0,59, 0.97,
and
1.36
kg/m
respectively.
a) You require to determine the advance per round blasted, the development tunnel is
about 60m, how many advances do you need to complete this tunnel?
[6]
b) Calculate your first, second, third and fourth section quadrangle cut?
[8]
Question 6 [20 marks]
B&E Mining and Development contract has been awarded a contract for underground tunnel
development. The contractor has requested you as mining engineer to determine certain
parameters for their underground tunnel development. Some of the known parameters are
as follows:
• Tunnel height= 6.5 m
• Targeted powder factor= 1.7 kg/m 3
• 28 perimeter holes loaded with 0.5 kg load.
• The chosen burn cut requires 7 loaded holes with 2.8 kg load.
• Blasthole load weight =5.5 kg/hole
• The round length = 3.4m
Calculate the following;
a) Round volume (V),
[4]
b) Total Explosive weight (TW),
[4]
c) Blasthole weight (BW),
[4]
d) The cut weight (CT), and
[4]
4

1.8 Page 8

▲back to top


1.9 Page 9

▲back to top


e) The number of blastholes (N}.
[4]
End of Examination
5

1.10 Page 10

▲back to top


2 Pages 11-20

▲back to top


2.1 Page 11

▲back to top


Important formulae for DLB721S - 2022
1.
u K act
-
(L + )* M C
B *S *H
2.
3.
.\\{ =->C!' . _p_*d 1
- 100 1273
4.
B=
5.
=2
V rrD xRD
4
6.
Q
=
rr d
41000P
7.
P = 5.7 x 10-5 x (RF-28 x log (0.145 x Sc})W/</Jx RPM
P=(6l-28log 1
'f' ,•.- ) :
• 1310_'),·Q1_·_1
7.a.
Where: d = diameterof the chip in inches
p = density of chip in Ib/ft3
8.
9.
°· VM
* = 2
VP
0 65
·
= 2 * 50
= 65
10. ROI= 25*RD-50=25*2.8-50 = 20
11. HF= UCS/5 =500/5=100
12. Rock Factor (A)= 0.06(RMD+JF+RDl+HF) =0.06(50+ 20+100) =10.2
13.
p = 2.5 X pv2 X 10-6
6

2.2 Page 12

▲back to top