This document contains 5 questions regarding fluid mechanics. Question 1 involves calculating the torque and power required to overcome viscous resistance in a rotating shaft. Question 2 involves calculating pressure drop, head loss, and power required for a given water flow rate through a pipe and orifice system. Question 3 determines the necessary counterweight to balance a water gate. Question 4 calculates the water level in a tank given pump specifications and a triangular weir. Question 5 determines if a hydraulic machine is a pump or turbine and calculates its power output or input.

1. FINAL EXAM. 2017 - 2018
By: DR. EZZAT EL-SAYED G. SALEH

2. Question No. 1 “6 Marks”
The figure shows a bearing, in which a vertical shaft is rotating. An oil film
between the bottom surface of the shaft and a bearing is provided to
rotate. The viscous resistance is offered by the oil to the shaft. Let:
N = speed of the shaft, & R = radius of the shaft,
y = thickness of oil film & the area of the elementary ring = 2 r. dr
Estimate:
a) Total torque required to overcome
the viscous resistance,
a) The power “P” absorbed.

3. Solution: (6 Marks)
 Total torque required to overcome the viscouse resistance
𝑇 =
0
𝑇
𝑑𝑇 =
0
𝑅
𝜇 ×
𝑉
𝑦
. 𝑑𝐴 . 𝑟 =
0
𝑅
𝜇 ×
𝜔 . 𝑟
𝑦
. 2 𝜋 𝑟 𝑑𝑟 . 𝑟
=
𝜇 ∙ 𝜋2
∙ 𝑁
15 𝑦 0
𝑅
𝑟3 ∙ 𝑑𝑟 =
𝜇 ∙ 𝜋2
∙ 𝑁
15 𝑦
𝑟4
4 0
𝑅
=
𝜇 ∙ 𝜋2
∙ 𝑁
15 𝑦
×
𝑅4
4
=
𝜇 ∙ 𝜋2
∙ 𝑁
60 𝑦
∙ 𝑅4
 The power absorbed “P”= 𝐹 × 𝑉 = 𝑇 × 𝜔
=
𝜇∙𝜋2∙ 𝑁
60 𝑦
∙ 𝑅4 ×
2 𝜋 𝑁
60
𝑃 =
𝜇 ∙ 𝜋3 ∙ 𝑁2
1800 𝑦
∙ 𝑅4

4. Question No. 2 “16 Marks”
Water is pumped at a rate of 20 cfs through the system shown in the figure.
a) What differential pressure will occur across the orifice?
b) What power must the pump supply to the flow for the given conditions?
c) Also, draw the HGL and the EGL for the system. Assume f = 0.015 for the
pipe.

5. Solution:
𝑄 = 𝐶𝑑 ∙ 𝑎𝑛
2 𝑃𝑖 − 𝑃𝐿
𝜌𝐿 1 − 𝛽4 = 𝐶𝑑 ∙ 𝑎𝑛
2 ∆𝑃
𝜌𝐿. 1 − 𝛽4
where:
𝑑𝑛 = 1 𝑓𝑡 & 𝑎𝑛 = 𝜋 ∙ 12 4 = 3.14 4 = 0.785 𝑓𝑡2 and
𝛽 = 𝑑𝑛 𝐷𝑃 = 1 2 = 0.5
 Substituting for 𝑄 = 20 𝑐𝑓𝑠 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑤𝑒 ℎ𝑎𝑣𝑒
20 = 0.62 × 0.785
2 ∆𝑃
1.94 × 1 − 0.54
1688.64 =
2 ∆𝑃
1.94 × 1 − 0.54
Or the pressure drop across the orifice plate is:
∆𝑃 = 1535.61 𝐼𝑏 𝑓𝑡2 = 24.58 𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 (6 Marks)

6.  The average velocity of flow through the pipe is
𝑉 = 𝑄 𝐴 = 𝑄 𝐴 = 20 𝜋 ∙ 22
4 = 6.37 𝑓𝑡 𝑠𝑒𝑐
 The head loss through the pipe only
ℎ𝐿 =
8 𝑓∙𝐿
𝑔 ∙ 𝜋2∙ 𝐷𝑃
5 × 𝑄2 =
8 ×0.015×300
𝑔 ∙ 𝜋2∙ 25 × 202 = 3.54 𝑓𝑡
 Applying Bernoulli’s equation between points “1” and “2”
1
2

7. 𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 1
+ ℎ𝑝 = 𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 2
+ ℎ𝐿 1 →2
10 0 0 5 0 0
24.58 + 3.54
10 + ℎ𝑝 = 5 + 24.58 + 3.54
∴ ℎ𝑝= 23.12 𝑓𝑡 𝑎𝑛𝑑
 Power must the pump supply to the flow for the given conditions,
𝑃𝑜𝑤𝑒𝑟 = 𝜌 𝑔 × 𝑄 × 𝐻𝑝=
1.94×32.2×20×23.12
550
≅ 53 𝐻𝑝 (8 Marks)

8. H.G.L
E.G.L
HP
(2 Marks)

9. Question No. 3 “10 Marks”
The counterweight pivot gate shown in the figure, controls the flow from a
tank. The gate is rectangular and is 3m x 2 m.
Determine the value of the counterweight “W” such that the upstream
water can be 1.5 m.
1.5 m
w
Pivot

10.  The hydrostatic force on the gate
𝐹𝐺 = 𝜌 𝑔 ℎt∙ 𝐴 = 1000 × 9.81 × 1.50 2 × 3 × 2 =
44145 𝑁 =044145 𝑁 103 = 44.145 𝑘𝑁
 tIt acts at ℎ𝑐.𝑝
= ℎ +
𝐼𝑐.𝑔
𝐴 ∙ ℎ
=
2
3
× 3.0 = 2.0 𝑚 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 ℎ𝑖𝑛𝑔𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑛𝑐𝑙𝑖𝑛𝑒𝑑 𝑔𝑎𝑡𝑒
w
1.5 m
𝜃 = 𝑐𝑜𝑠−1
1.5
3
= 60𝑜
𝜃
hinge
W sin 60

11.  Taking moments about the hinge
𝐹𝐺 × 3 − ℎ𝑐.𝑝 = 𝑊 sin 60𝑜 × 0.60
∴ 𝑊 =
𝐹𝐺 × 3 − ℎ𝑐.𝑝
sin 60𝑜 × 0.60
=
44.145 × 3 − 2
sin 60𝑜 × 0.60
= 84.96 𝐾𝑁
 The value of the counterweight “W” such that the upstream water can
be 1.5 m is 84.96 KN. (10 Marks)

12. Question No. 4 “12 Marks”
A pump is used to deliver water from a well to a tank. The bottom of the tank is
2 m above the water surface in the well. The pipe is commercial steel 2.5 m
long with a diameter of 5 cm and f = 0.02. The pump develops a head of 20 m.
A triangular weir with an included angle of 60° “Cd = 0.58” is located in a wall
of the tank with the bottom of the weir 1 m above the tank floor. Find the level
of the water in the tank above the floor of the tank.

13. Solution:
 The discharge through a triangular
Weir is expressed as
 𝑄𝑤=
8
15
𝐶𝑑 2 𝑔 tan
𝜃
2
∙ 𝐻𝑤
5 2
∴ For 𝜃 = 60𝑜 & 𝐶𝑑 = 0.58
𝑄𝑤 =
8
15
× 0.58 × 2 × 9.81 × tan
60
2
∙ 𝐻𝑤
5 2
= 0.79 𝐻𝑤
5 2
 From the given data
𝑄𝑤 = 0.79 ∙ ℎ − 1 5 2
 To have a constant water level,
𝑄𝑝𝑢𝑚𝑝 = 𝑄𝑤𝑒𝑖𝑟
1
2

14.  Applying Bernoulli equation between points “1” and “2” gives,
𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 1
+ ℎ𝑝 = 𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 2
+ ℎ𝐿 1 →2
 Taking water surface at the lower tank as a reference datum
𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 1
+ ℎ𝑝 = 𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 2
+ ℎ𝐿 1 →2
 The head loss in the given system ℎ𝐿 1 →2 =
8 𝑓 𝐿
𝑔 𝜋2 𝐷5 × 𝑄2
=
8×0.02×2.5
9.81× 𝜋2 ×0.055 × 𝑄2
= 13220.3 𝑄2
0 0 0 2 + h 0 0
20 m

15. ∴ 20 = (2 + ℎ) + 13220.3 𝑄2
or
18 = ℎ + 13220.3 𝑄2
 But, 𝑄𝑝𝑢𝑚𝑝 = 0.79 ∙ ℎ − 1 5 2
∴ 18 = ℎ + 13220.3 × 0.79 ∙ ℎ − 1 5 2 2
𝑖. 𝑒. 18 = ℎ + 8250 ∙ ℎ − 1 5
∴ 𝐵𝑦 𝑡𝑟𝑖𝑎𝑙 𝑎𝑛𝑑 𝑒𝑟𝑟𝑜𝑟 "ℎ" = 1.28 𝑚
 The level of the water in the tank above the floor of the tank = 1.28 𝑚.
(12 Marks)

16. Question No. 5 “16 Marks”
i. For a hydraulic machine shown in the figure, the following data are available:
Flow : From “A” to “B”
Discharge: 200 L/s of water
Diameters: at “A” 20 cm, and at “B” 30 cm
Elevation (m): at “A” 105 m, and at “B” 100 m
Pressures : at “A” 100 kpa, and at “B” 200 kpa.
a) Is this machine a pump or a turbine?
b) Calculate the power input or output depending on whether it is pump or a
turbine.

17. For a hydraulic machine shown in the figure, the following data are available:
Flow : From “A” to “B”
Discharge : 200 L/s of water
Diameters : at “A” 20 cm, and at “B” 30 cm
Elevation (m) : at “A” 105 m, and at “B” 100 m
Pressures : at “A” 100 KPa, and at “B” 200 KPa.
a) Is this machine a pump or a turbine?
b) Calculate the power input or output depending on whether it is pump or a
turbine.
Solution:
Using the continuity equation
𝑉1 = 𝑄 𝐴1 =
(200 1000)
1000 × 𝜋 .0.20 2 4
= 6.37𝑚 𝑠
and 𝑉2 = 𝑄 𝐴2 =
(200 1000)
1000× 𝜋 .0.30 2 4
= 2.83 𝑚 𝑠

18. Applying Bernoulli’s equation to point “A”, gives the (TEL)A
𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 𝐴
= 105 +
100×103
103×9.81
+
6.372
2×9.81
= 117.26𝑚
and that at point “B”, (TEL)B
𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 𝐵
= 100 +
200 × 103
103 × 9.81
+
2.832
2 × 9.81
= 120.80𝑚
 Since the TEL at “B” is larger than that at “A” (the friction losses is
assumed to be zero), this machine a pump with a head = 120.8 −
117.26 = 3.54 𝑚
 To estimate the power
𝑃𝑜𝑤𝑒𝑟 = 𝜌 𝑔 × 𝑄 × ℎ𝑝=
103×9.81×(200 103)×3.54
103 = 6.95 𝑘𝑁. 𝑚 𝑠 = 6.95 𝑘𝑤
(8 Marks)

19. 19
ii- A vertical circular tank of 600mm
diameter and 2.5m height is full of
water. It contains two orifices each of
1300 mm2 area, one at the bottom of
the tank and the other at a height of
1.25m above the bottom as shown in
figure.
 Determine the time required to empty
the tank. Take coefficient of discharge
for both of the orifices as 0.62.

20. 20
Given:
• Diameter of the tank =600 = 0.60 m
• Height of the tank = 2.50 m
• Area of each orifice = 1300 mm2 = 1300x 10-6 m2
• C
d
= 0.60
 For the sake of simplicity, let use divide the tank into two parts, i.e.,
 first up to the center of the top orifice, and the to the bottom orifice.
 First of all, consider first part of the example. In this case, the water
is flowing through both orifices.
 Now consider an instant, when the height of water above center of
the top orifice be “h” meters and the height of the water above the
bottom orifice will be “h +1.25”.

21. 21
The surface area of the tank is constant, 𝐴 =
𝜋×0.62
4
= 0.283 𝑚2
 Now let us use the general equation for the time to empty the tank,
𝑑𝑡 = −
𝐴∙𝑑ℎ
𝐶𝑑∙ 𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 ℎ
→ (1)
or for this case (two identical orifices)
𝑑𝑡 = −
𝐴 ∙ 𝑑ℎ
𝐶𝑑 ∙ 𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 ℎ + 𝐶𝑑 ∙ 𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 ℎ + 1.25
=−
𝐴∙𝑑ℎ
2. 𝐶𝑑 × 𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 ℎ+ ℎ+1.25

22. 22
For the given data;
𝑇1= − 0
1.25 𝐴 ∙𝑑ℎ
2 . 𝐶𝑑 ×𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒∙ 2𝑔 ℎ+ ℎ+1.25
= −
𝐴
2 𝐶𝑑∙ 𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 0
1.25 𝑑ℎ
ℎ+ ℎ+1.25
Multiplying both numerator and denominator by:
ℎ − ℎ + 1.25
𝑇1 = −
𝐴
2 𝐶𝑑∙ 𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔 0
1.25
ℎ − ℎ + 1.25 . 𝑑ℎ
1.25
= −
0.283
2 × 0.60 × 0.0013 2 × 9.81 × 1.25 0
1.25
ℎ − ℎ + 1.25 . 𝑑ℎ
≅ 55.82 𝑠𝑒𝑐.

23. 23
Now consider the flow of water below the center of the top orifice. A
little consideration will show that now the water will be flowing through
the bottom orifice only. We know that the time required to empty the
tank,
𝑇2 =
2 𝐴 ℎ
𝐶𝑑∙ 𝑎𝑜𝑟𝑖𝑓𝑖𝑐𝑒 2𝑔
=
2 × 0.283 × 1.25
0.6 × 0.0013 × 2 × 9.81
= 183.16 sec
The total time,
𝑇1+ 𝑇2= 55.82 + 183.16 = 239 sec ≅ 4 min
(8 Marks)

24. 24
Question No. 6 “10 Marks”
A sluice gate is used to control the water flow rate over a dam. The
gate is 20 ft wide, and the depth of the water above the bottom of the
sluice gate is 16 ft. The depth of the water upstream of the gate is 20 ft,
and the depth downstream is 3 ft.
i. Estimate the flow rate under the gate and the force on the gate.
𝐹𝑥

25. Applying Bernoulli’s equation to point “1” and point “1”, gives,
(Neglecting minor losses and taking the bed of the channel as a
reference datum)
𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 1
= 𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 2
For unit discharge
𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 1
= 𝑍 +
𝑃
𝜌 𝑔
+
𝑉2
2 𝑔 2
=2
0
 0 𝑞2
2 𝑔 𝑦2
= 3  0 𝑞2
2 𝑔 𝑦2
20 + 0 +
𝑞2
2 𝑔 𝑦1
2 = = 3 + 0 +
𝑞2
2 𝑔 𝑦1
2

26. ∴ 20 − 3 ) × 2 × 32.2 = 𝑞2 ∙
1
32
−
1
202
 The flow rate under the gate "𝑞" ≅ 100 𝑓𝑡3 𝑠/𝑓𝑡 (5 Marks)
 Applying momentum equation between point “1” and point “1”, gives,
𝐹𝑥,1 − 𝐹𝑥,2 − 𝐹𝑥 = ρ 𝑄 𝑉2 − 𝑉1 ……… Eq. (1)
or
𝐹𝑥,1 − 𝐹𝑥,2 − 𝐹𝑥 = ρ 𝑞2
1
𝑦2
−
1
𝑦1
 To obtain 𝐹𝑥,1 − 𝐹𝑥,2 ,
𝐹𝑥,1 − 𝐹𝑥,2 = 0.5 𝜌 𝑔 𝑦1
2
− 𝑦2
2
= 0.50 × 1.94 × 32.2 × 202 − 32
= 12212.50 𝐼𝑏 𝑓𝑡 ⇉

27. 𝑎𝑛𝑑 ρ 𝑞2
1
𝑦2
−
1
𝑦1
= 1.94 × 1002
1
3
−
1
20
= 5496 𝐼𝑏 𝑓𝑡 ⇉
Eq. (1) gives,
12212.50 − 𝐹𝑥 = 5496
or
𝐹𝑥 = 12212.50 − 5496 = 6716.5 𝐼𝑏 𝑓𝑡 ⇇
 The total force on the gate.
 𝐹𝑥× 20 = 6716.5 × 20 = 134330 𝐼𝑏 ⇇ … (5 Marks)

29.  State the assumptions upon which simple tank drainage problems
are based.
 Water flows at a steady rate of 350 m3/ h into a vertical – sided
tank of area 10 m2. The water discharges continuously from a
sluice gate of area 0.0564 m2.
If the initial level in the tank is 2.50 m above the sluice, determine:
The final depth after 5 minutes assuming a discharge coefficient of
0.60 for the sluice.
( Ans. 0.60 m)

30. A long pipe is installed to carry water from one large reservoir to another (see the
attached figure). The total length of the pipe is of 10 km, its diameter is 0.5 m and
its roughness factor = 0.018. It must climb over a hill, so that the altitude changes
along with distance. The pump must be powerful to push 1 m3/s of water.
What is the pressure drop generated by the water flow?
What is the pumping power required to meet the design requirement,
What would be the power required for the same volume flow if the pipe is
doubled?
Layout of the water pipe (the vertical scale is greatly exaggerated). The
diameter of the pipe is also exaggerated.

31. 25 ft
L = 600 ft, D = ?? , f = 0.02
Main
A
B
Reference datum
The figure shows a pipe delivering water to the putting green on a golf
course. The pressure in the main is 80 p.s.i and it is necessary to
maintain a minimum of 60 p.s.i at point B to adequately supply a
sprinkler system. Specify the required size of steel pipe ( f = 0.02) to
supply 0.50 ft3/s.
Q = 0.5 ft3/s
Pipe Flow and Pipe Systems

32. Solution: Given:
Diameter of the circular disc 𝐷 = 200/1000 =0.20 m
∴ 𝑅 = D/2 =0.10 m
𝑦 = 0.0004 m
𝑁 = 1000 r.p.m
𝜇 = 1.05 poise = 1.05/10 = 0.105 N.s/m2
Then
The power required to rotate the disc is given as
𝑃 =
𝜇 𝜋2∙𝑁2𝑅2∙
1800 ×𝑦
=
0.105× 𝜋2×10002×0.12∙
18000×0.0004
= 452.1 𝑤
Find the power required to rotate a circular disc of diameter 200 mm at
1000 r.p.m. The circular disc has a clearness of 0.4 mm from the bottom
flat plate and the clearness contains oil of viscosity 1.05 poise.

33. For the system shown in figure, calculate the height “H” of oil
at which the rectangular hinged gate will just begin to rotate
counterclockwise.
Gate 1.5x 0.6 m
Oil
S.G.= 0.80
Air
H
1.5 m
30 KPa
Hinge
F1
F2
Oil
Pressure
Air Pressure
1.5 m
(a)
(b)

38. Deep sump catch basin 2
nonproprietary settling

43. ℎ𝐿 = 𝑘 .
𝑉2
2 𝑔
where: k =
𝐷𝑝𝑖𝑝𝑒
2
𝐶𝑑. 𝐷𝑜𝑟𝑖𝑓𝑖𝑐𝑒
2 − 1
2
∴ ℎ𝐿=
𝐷𝑝𝑖𝑝𝑒
2
𝐶𝑑. 𝐷𝑜𝑟𝑖𝑓𝑖𝑐𝑒
2 − 1
2
.×
8 𝑄2
𝑔 𝜋2 𝐷𝑝𝑖𝑝𝑒
4
ℎ𝐿 =
22
0.62 × 12
− 1
2
.
8 × 202
32.2 × 𝜋2 × 24
= 18.70 ft

44. 𝑸𝑽 =
𝝅𝟐
𝟒
× 𝒅𝟐 ×
𝟐 ∆𝑷
𝝆 𝟏 −
𝒅𝟒
𝑫𝟒

46. Flow
Internal Flow
Pipe Flow
Open Channel Flow
External Flow

48. Storm water flowing into pond from concrete sewage discharge pipe

49. The Bernoulli’s equation can be considered to be a statement of the
conservation of energy principle appropriate for flowing fluids.

57. A vertical circular tank of 600mm diameter and 2.5m height is full of
water. It contains two orifices each of 1300 mm2 area, one at the bottom
of the tank and the other at a height of 1.25m above the bottom as shown
in figure.
Determine the time required to empty the tank. Take coefficient of
discharge for both of the orifices as 0.62..
1.25 m
2.50 m
600 mm

58. Given,
 Diameter of the tank = 600 mm = 0.6 m & H = 2.5 m
Area of each orifice, a = 1300 mm2 & Cd = 0.62
For the sake of simplicity, let us divide the problem into two parts, i.e.,
first up to the center of the top orifice, and then up to the bottom orifice.
 First of all, consider the first part of the problem. In that case, the water
is flowing through both the orifices.
 Now consider an instant, when the height of water above the center of
the top orifice be h meters. At that instant, the height of water above
the bottom orifice will be (h + 1.25) meters.
 We know that the surface area of the tank,

59. ∴ 𝐴𝑇=
𝜋
4
× 𝐷𝑇
2
=
𝜋
4
× 0.602 = 0.283 𝑚2
 Now let us use the general equation for the time to empty a tank,
𝑑𝑡 =
− 𝐴 𝑑ℎ
𝐶𝑑 ∙ 𝑎 ∙ 2 𝑔 ℎ
=
− 𝐴 𝑑ℎ
𝐶𝑑 ∙ 𝑎 ∙ 2 𝑔 ℎ + 1.25 + 𝐶𝑑 ∙ 𝑎 ∙ 2 𝑔 ℎ
=
− 𝐴 𝑑ℎ
𝐶𝑑 ∙ 2 𝑎 ∙ 2 𝑔 ℎ + 1.25 + ℎ
 The time “T1” required to bring the water level up to the center of the
top orifice may be found by integrating the above equation between the
limits 1.25m and 0. Therefore

60. 𝑇1 =
1.25
0
− 𝐴 𝑑ℎ
𝐶𝑑 ∙ 2𝑎 ∙ 2 𝑔 ℎ + 1.25 + ℎ
or 𝑇1 =
− 𝐴
𝐶𝑑 ∙2𝑎 ∙ 2 𝑔 1.25
0 𝑑ℎ
ℎ+1.25 + ℎ
Multiplying the numerator and denominator by
ℎ + 1.25 − ℎ
𝑇1 =
0.283
0.62 × 2 × 0.0013 × 2 × 9.81 0
1.25
ℎ + 1.25 − ℎ
1.25
𝑇1 = 39.58 × 1.05𝟓 = 61.35 𝑆
 Now consider the flow of water below the center of the top orifice. A
little consideration will show that now the water will be flowing through
the bottom orifice only. We know that the time required to empty the
tank,

61. 𝑇2 =
2 𝐴 𝐻1
𝐶𝑑 ∙ 2𝑎 ∙ 2 𝑔
=
2 × 0.283 × 1.25
0.62 × 0.0013 × 2 × 9.81
= 177.25 𝑆
Total time,
𝑇 = 𝑇1 + 𝑇2 = 61.35 + 177.25 ≅ 239 𝑆

62. ‫االبعاد‬
:
120
*
45
*
40

63. Dimensions 122 x 67
5 shelf's

66. Name of Presentation
Company Name

67. Slide Master
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