`
Question 1
From the notes, the expression of the horizontal component of the force is:
𝐹! = 𝜌𝑔 %
ℎ" + 𝑅
* [𝑊(ℎ" + 𝑅)]
2
From the notes, the expression of the vertical component of the force is:
𝐹# = 𝜌𝑔𝑊 0ℎ" 𝑅 +
𝜋𝑅$
3
4
By replacing the numbers in the expression of the horizontal force, we obtain:
ℎ" + 𝑅
𝐹! = 𝜌𝑔 %
* [𝑊(ℎ" + 𝑅)] = 6.13 × 10% 𝑁
2
The value of the vertical force is:
𝐹# = 𝜌𝑔𝑊 0ℎ" 𝑅 +
𝜋𝑅$
3 = 4.48 × 10% 𝑁
4
The magnitude of the resultant is:
|𝐹& | = >𝐹!$ + 𝐹#$ = 7.59 × 10% 𝑁
The angle of the force with the vertical axis is expressed as:
𝐹(
𝜃 = 𝑡𝑔'" % * = 36.16°
𝐹!
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Question 2
The energy balance is:
−𝐻 − ℎ) + ℎ*+,* = 0.
This equation can be rearranged in terms of average velocity as:
-
𝑉 = ./(!12 ),
!
Where ℎ) are the distributed viscous losses (the minor ones are negligible according to the
problems assumptions) expressed as:
) #"
ℎ) = 𝑓 4 $5.
To solve the problem, we need to apply an iterative method. The final value of the velocity
is 𝑉 = 2.095 m/s.
The new velocity in the pipe is 𝑉 =
of pump power is:
$.78%
$
= 1.05 m/s. The energy balance expressed in terms
𝑃 = 𝜌𝑉𝐴(𝑔𝐻 + 𝑙( ).
The Reynolds number is:
𝑅𝑒 =
.(4
9
)
;
≃ 10% ⇒ 𝑓 = 4.4 × 10': ⇒ 𝑙( = 2𝑓𝑣 $ 4 = 2.91 <5.
The power of the pump is:
𝑃 = 𝜌𝑉𝐴(𝑔𝐻 + 𝑙( ) ≃ 2450 𝑊.
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Question 3
This is a simple problem of hydrostatic that is solved by applying the Stevino’s law.
We call M2 the point at the interface between the two fluids on the right tube; we call
M1 the point on the same horizontal line as M2, but on the left tube. We can write the
Stevino’s law on the point M1, as:
𝑝=" = 𝑝>?@ + 𝜌" 𝑔ℎ" ,
Where ℎ" = 0.22 𝑚, and 𝑝>?@ = 𝑃 − 𝑝>A, = 76 − 100 = −24 𝑘𝑃𝑎 is the gauge (or
gage) pressure of the air. Analogously, for M2 we can write:
𝑝=$ = 𝜌$ 𝑔ℎ$ ,
Where ℎ" = 0.33 𝑚.
Form the principle of communicating vessels, the pressure at M1 is equal to that at
M2, 𝑝=" = 𝑝=$ , therefore:
𝑝>?@ + 𝜌" 𝑔ℎ" = 𝜌$ 𝑔ℎ$ ⇒ 𝜌$ =
*#$%
52"
+
.& 2&
2"
.
The density of the fluid 1, is derived from the definition of the specific gravity SG, as:
𝜌" = 𝑆𝐺 × 𝜌B>AC@ = 13.55 × 1000 = 13550 𝐾𝑔/𝑚: .
The density of the fluid 2 is then evaluated as:
𝜌$ =
*#$%
52"
+
.& 2&
2"
= 1619.747 ⇒ 𝑆𝐺 = .
."
'#()%
≃ 16.2.
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