REVIEW IN FLUID MECHANICS
CE board May 1994
Assuming specific weight of air to be at constant at 12 N/m 3. What is the approximate height of
Mount Banahaw if a mercury barometer at the base of the mountain reads 654mm and at the
same instant, another barometer at the top of the mountain reads 480mm?
Solution:
Pressure at the top of the mountain
Ptop = whg x hbarometer rdg at the top of the mountain
= (9810N/m3 x 13.6)(0.48m)
= 64039.68 Pa
Pressure at the base of the mountain
Pbase = whg x hbarometer rdg at the base
= (9810N/m3 x 13.6)(0.654m)
= 87254.06 Pa
Pbase - Ptop = (wair) (hmountain)
87254.06 Pa - 64039.68 Pa = 12N/m3(hmountain)
hm = 1934.53m
CE Board November 2000
For the tank shown in the figure, h1=3m and h3=4m. Determine the value of h2.
Solution:
Note that the pressure at the liquid surface expose to atmosphere is zero. By summing up the pressure
heads, we have
𝑃2
𝑃1
+ 4 − 3 − ℎ2 (0.80) =
𝑤
𝑤
P1 and P2 are zero, because they are both exposed to atmosphere
0 + 4 − 3 − ℎ2 (0.80) = 0
𝒉𝟐 = 𝟏. 𝟐𝟓𝒎
CE BOARD MAY 1993
In the figure shown, when the funnel is empty, the water surface is at level A and the mercury
(s=13.6) shows a deflection of 15cm. Determine the new deflection of mercury when the funnel
is filled with water up to level B.
Solution:
When the water is at level A (Figure A), adding the pressure heads
0 + y – 0.15(13.6) = 0
y=2.04m
When the water increases up to level B, the water will push the mercury downward by distance
x, as a result the mercury will rise by distance x on the other part of the tube, therefore the new
deflection of mercury will be 0.15+2x (refer on Figure B). By summing up the pressure heads,
this will lead to
0 + 0.80 + 2.04 + x – x(13.6) – 0.15(13.6) – x(13.6) = 0
x = 0.03m
Therefore new deflection = 0.21m or 21cm
SAMPLE PROBLEMS
1. A vertical rectangular gate 2m wide and 3m
high is submerged in water with its top edge
2.4m below the water surface. Find the total
pressure acting on one side of the gate and
its location from the water surface.
Solution:
𝐹 = 𝑤ℎ̅𝐴 = 9.81(3.9)(2𝑥3) = 𝟐𝟐𝟗. 𝟓𝟓𝑲𝑵 (answer)
𝑒=
𝐼𝑔
𝐴𝑦̅
ℎ̅ = 𝑦̅ = 3.9𝑚
𝑏ℎ3 2(3)3
=
= 4.5𝑚4
12
12
4.5
𝑒=
= 0.19𝑚
6(3.9)
𝐼𝑔 =
ℎ𝑝 = 3.9 + 0.19 = 𝟒. 𝟎𝟗𝒎 (answer)
2. A rectangular gate 2.4m wide and 3.6mhigh is inclined at 600 with respect to horizontal. It is
attached to a hinged support on top of the gate. The water stands 2m above the top of the
gate. What force P is required acting horizontally at the bottom of the gate to maintain
equilibrium.
Solution:
b=2.4m
h=3.6m
A = 2.4(3.6) = 8.64m2
Using the triangle above:
𝑥
sin 600 = 1.8
𝑥 =1.56m
ℎ̅ = 1.56𝑚 + 2𝑚 = 3.56𝑚
3
𝐼𝑔 =
3
𝑏ℎ
2.4(3.6)
=
= 9.33𝑚4
12
12
= 0.26𝑚
̅
ℎ
sin 600 = 𝑦̅
𝑒=
𝐹 = 𝑤ℎ̅𝐴 = 9.81(3.56)(8.64) = 301.74𝐾𝑁
𝑦̅ =4.11m
𝐼𝑔
9.33
=
̅ 8.64(4.11)
𝐴𝑦
By taking summation of moments at hinged:
F (0.26+1.8) = P(2)(1.56)
301.74 2.06) = P(3.12)
P=199.23KN (answer)
3. A vertical gate 2.6m wide and 4.8m high is subjected to hydrostatic forces on both sides.
The gate is supported by hinged on the top of the gate. Water stands 2.2m above the
hinged on the right side and 1.2m below the hinged on the left side. A wooden stopper is
placed at the bottom of the gate to prevent it from opening. Calculate the stress acting on
the wooden stopper if it has a sectional area of 0.01m 2.
Solution:
A1 = 2.6(4.8) = 12.48m2
ℎ̅1 = 2.2 + 2.4 = 4.6𝑚
A2 = 2.6(3.6) = 9.36m2
ℎ̅2 = 1.8𝑚
F1 = (9.81)(4.6)(12.48)
F1 = 563.17kN
𝑏ℎ3 2.6(4.8)3
=
= 23.96𝑚4
12
12
𝐼𝑔2 =
𝑏ℎ3 2.6(3.6)3
=
= 10.11𝑚4
12
12
𝐼𝑔1
23.96
=
= 0.42𝑚
̅ 1 12.48(4.6)
𝐴1 𝑦
𝑒2 =
𝐼𝑔2
10.11
=
= 0.60𝑚
̅ 2 9.36(1.8)
𝐴2 𝑦
𝐼𝑔1 =
𝑒1 =
F2 = (9.81)(1.8)(9.36) = 165.28 kN
By taking summation of moments at hinged:
F1(1.2+1.8+0.60) + Fstopper(4.8)=F2(2.4+0.42)
165.28(3.29) + Fstopper(4.8)=563.17(2.82)
Fstopper=206.90 KN
Stress on the stopper P = Fstopper / Area of stopper
Stress on the stopper P = 206.90KN / 0.01m2 = 20690 KPa
4. The submerged curve gate BC as shown in the figure is a quarter of a circular gate of
radius 3.0m. The length of the gate perpendicular to the drawing is 4m. Find the
magnitude horizontal and vertical components of the total force acting on the gate and
their location from the hinged.
Solution:
Fh = (9.81KN/m3)( 2m +1.5m)(12m2) = 412.02KN
Fv = wV = 9.81KN/m3 [(2m x 3m) + (1/4)(π)(3)2] 4m = 512.81KN
For their location from hinged: (distance y and x respectively)
y= 1.5 + e 𝐼𝑔 =
𝑏ℎ3
12
=
4(3)3
12
= 9𝑚4
𝐼𝑔
9
𝑒 = 𝐴𝑦̅ = 12(3.5) = 0.21𝑚
y = 1.5 + 0.21 = 1.71m
By taking summation of moment at the hinged:
412.02 (1.71) = 512.81 (x)
x = 1.37m
5. A crest gate BC as shown below is a part of a circle of radius 3.6m and subtended 60 º
from the hinged at point D. If the width of the gate is 2.4m, determine the following .
a) The magnitude of the horizontal hydrostatic force FH in kN
b) The magnitude of the vertical hydrostatic force Fv in kN
c) The location of Fh from the hinged in meters
d) The location of Fv from the hinged in meters.
Solution:
Sin 600 = h / 3.6
h = 3.12m
Cos 600 = DE / 3.6
DE = 1.8m
BE = 3.6 – 1.8 = 1.8m
Fh = (9.81KN/m3) (1.56m)(3.12m x 2.4m) = 114.59KN
ABCE = ASECTOR BDC - A▲CED
ABCE =
600
3600
(𝜋)(3.6)2 −
1
2
(3.12)(1.8) = 3.98𝑚2
AACB = AABEC - ABCE
AACB = 3.12m (1.8m) – 3.98m2 = 1.64m2
Fv = wV = w (AACB) 2.4m
Fv = 9.81KN/m3 (1.64m2) 2.4m
Fv = 38.61KN
𝐼𝑔 =
𝑏ℎ3 2.4(3.12)3
=
= 6.07𝑚4
12
12
𝑒=
𝐼𝑔
6.07
=
= 0.52𝑚
̅ (3.12𝑥2.4)(1.56)
𝐴𝑦
y = 1.56m – 0.52m
y = 1.04m
By taking summation of moment at the hinged:
Fh (y) = Fv (x)
114.59KN (1.04m) = 38.61KN (x)
x = 3.09m