MARKING GUIDELINE
NATIONAL CERTIFICATE
FLUID MECHANICS N6
15 April 2021
This marking guideline consists of 6 pages.
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MARKING GUIDELINE
-2FLUID MECHANICS N6
QUESTION 1
1.1
1.1.1
=
+
+
=
,
12,2 =
+
,
,

,
.z
a
Q = 0,0887 m3/s
= 88,738 l/s
1.1.2
(3)
=
=
,
=

,
,
.c
o
=

,
1.1.3
=
+
=
,
+
=
,
,
+
,
+
+
,
1.1.4
,
,
m1 =
,

tp
=
+
,
,

(3)
m2 =
=

= 0,0255 m 
i1 =
i2 =
i1 =
,
=

= 0,0788 
1.2
,
= 0,0455 
tv
e
1.1.5
(4)

ap
= 9,16 m
,
=
er
s
= 3,411 m/s
10,86 m/s
,
(4)

= 1,426 
(4)
=
=
=
=

= 15,588
alternatively
=
+
1,5 = 15,588
+

3
= 0,0904 m /s 
= 1,41 m3/s 
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= 0,0642
=
+
1,5 = 0,0642 + 
= 1,41 m3/s 
= 0,0904 m3/s
(4)
[22]
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QUESTION 2
2.1
2.1.1
=
,
=
,

,
Reaction of the jet = ρQ
,
= 103 x
 x 5,375 
= 11,647 N 
2.1.2
=
(5)
0,035 
=
,
.c
o
=
= 0,000962 m2

,
.z
a
= 5,375 m/s 
= 0,000403 m2 
er
s
=
=
,

,
= 0,419
x=
,
= 3,981 m 
ap
2.2
,
tp
=
tv
e
× 2
4
3,981
2 
= 15,963 m2
= 1,236 m
Q = AC√
1,236
= 4,456 m 
Alternatively
Area =
,
= 15,963 x 55
,
Area = ½ 4
4 7,963
= 15,963 m2 
Perimeter = 2(4,456) + 4
= 12,912 m 
m=
r=
(6)

= 25,207 m3/s 
It will not cope since the designed channel can handle less discharge than the
expected.
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(9)
[20]
QUESTION 3
Hat =
=
Hsep =
,
=

,
= 13,371 m 
,
,

,
= 0,185 m 
Ha = Hat – Hs – Hsep
xω R
Ha =
,
8,186 =
= 13,371 – 5 – 0,185 
= 8,186 m 
,
,
×ω ×
ω = 9,749 rad/s 
,

= 93,096 rev/s
3.2.1
Q = ALSE
=
0,13

(10)
er
s
3.2
,
.c
o
ω=
N =
.z
a
3.1
× 0,28 × 1 × 1 ×

= 0,00242 m3/s
3.2.2
P = ρgASHN
0,13
× 0,28 x 25 ×
ap
=103 × 9,81 ×

= 592,457 W
Alternatively
tp
P = ρgQH
= 103 × 9,81 × 0,00242 × 25
tv
e
= 592,457 W 
3.3
Q=
= 0,05 m3/s 
=
Tanᶲ =
x=
,
(2 × 2)
(4)
–x
= 9 – 1,919
= 7,08 m/s 

= 1,919 m/s
T = ρQ
= 103 × 0,05 × 7,08 ×
= 212,403 Nm
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,

(6)
3.4
0,793
a =
S=
= 0,494 m2
,
v=
= 18,83 m/s
,
× 0,793 × L = 2,491L
Pr =
198,3 =
,
,
,

,
L = 49,279 m
.z
a
(5)
[25]
QUESTION 4
4.1.1
0,17 2
= 0,17√2
9,81 14 
= 2,817 m/s 
E=
E=
,
=
= √11,34

= 11,34 m
14,956º
4.1.3
9,81
=
,
,

=
(6)
Internal diameter = ½ External diameter
ap
= ½(10,547) 
tp
= 5,274 m/s 
=
tan
=
=
,
,

= 28,114º
(4)
=
,
tv
e
D=
d = ½(682,842)

= 682,842 mm
4.1.5
(2)
=
= 10,547 m/s
=½
4.1.4
tan
er
s
4.1.2
.c
o
4.1
Q=
A=
= 341,421 mm
(3)
xA
,
,

= 0,142 m2 
But A =
DW
Wi =
,
,
,

= 73,533 mm 
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Wo =
,
,
,

= 147,067 mm 
(6)
4.2
4.2.1
U=
V = 0,98 2
H=
,
=
0,98 2
=

9,81
,

=
71,356 
= 14,923 m/s 
= 71,356 m 
= 36,668 m/s 
Q=VxA

.z
a
0, ,075
= 36,668 ×
= 0,162 m3/s 
P = ρQU (V – U)(1 + nCosy)
.c
o
= 103 × 0,162 × 14,923(36,668 - 14,923)(1 + 0,86Cos(1800 -1600)
=95,05 kW
(V – U)(1 + nCosy) × 100%
η=
=
,
,
,
(36,668 - 14,923)(1 + 0,86Cos(180º - 160º) × 100%
tv
e
tp
ap
=83,820%
er
s
4.2.2
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(10)
(2)
[33]
TOTAL:
100