This document provides examples of problems from Chapter 2 of the textbook Principles of Hydraulics and Pneumatics. It includes 11 example problems that cover topics such as hydraulic force multiplication, velocity trade-offs, weight versus mass, viscosity, bulk modulus, basic flow rate, continuity equation, and Bernoulli's equation. The examples provide the relevant equations, given values, and step-by-step workings to calculate the requested output values for each problem.

1. ENGT 107: PRINCIPLES
OF HYDRAULICS AND
PNEUMATICS PROFESSOR JIM
EVANGELOS
Example ProblemsChapt. 2

2. Example Problems:
1 Example 2-4
2 Example 2-5
3 Example 2-6 & 2-6M
4 Example 2-7
5 Example 2-8
6 Example 2-9M
7 Example 2-12
8 Example 2-13
9 Example 2-17M
10 Example 2-19
11 Example 2-20

3. Example 2-4
Hydraulic Force Multiplication
 Figure 2-5 (pg. 17) shows an input
cylinder with a diameter of 1 in and an
output cylinder with a dia of 2.5 in. A
force of 250 lbs. is applied to the input
cylinder.
 What is the output force?
 How far would we need to move the input
cylinder to move the output cylinder 1 in?
𝐴𝑖𝑛 =
𝜋∙𝐷2
4
=
𝜋∙(1 𝑖𝑛)2
4
= 0.7854 𝑖𝑛2
𝐴 𝑜𝑢𝑡 =
𝜋∙𝐷2
4
=
𝜋∙(2.5 𝑖𝑛)2
4
= 4.909 𝑖𝑛2
From equation 2-3:
𝐹𝑜𝑢𝑡
𝐹 𝑖𝑛
=
𝐴 𝑜𝑢𝑡
𝐴 𝑖𝑛
𝐹𝑜𝑢𝑡 =
𝐴 𝑜𝑢𝑡
𝐴 𝑖𝑛
∙ 𝐹𝑖𝑛
𝐹𝑜𝑢𝑡 =
4.909 𝑖𝑛2
0.7854 𝑖𝑛2 ∙ 250 𝑙𝑏𝑠 = 1563 𝑙𝑏𝑠
From equation 2-5:
𝑑𝑖𝑛
𝑑 𝑜𝑢𝑡
=
𝐴 𝑜𝑢𝑡
𝐴 𝑖𝑛
𝑑𝑖𝑛 =
𝐴 𝑜𝑢𝑡
𝐴 𝑖𝑛
∙ 𝑑 𝑜𝑢𝑡
𝑑𝑖𝑛 =
4.909 𝑖𝑛2
0.7854 𝑖𝑛2 ∙ 1 𝑖𝑛 =
𝐹𝑜𝑢𝑡 =
4.909 𝑖𝑛2
0.7854 𝑖𝑛2 ∙ 250 𝑙𝑏𝑠 = 1563 𝑙𝑏𝑠
𝑑𝑖𝑛 =
4.909 𝑖𝑛2
0.7854 𝑖𝑛2 ∙ 1 𝑖𝑛 = 6.25 in

4. Example 2-5
Velocity trade off
 The output cylinder in
the previous example is
required to move at 4
in/s.
 At what speed must the
input cylinder move?
𝑣𝑖𝑛 =
4.909 𝑖𝑛2
0.7854 𝑖𝑛2 ∙
4 𝑖𝑛
𝑠
= 250 in/s
From equation 2-6:
𝑣 𝑖𝑛
𝑣 𝑜𝑢𝑡
=
𝐴 𝑜𝑢𝑡
𝐴 𝑖𝑛
𝑣𝑖𝑛 =
𝐴 𝑜𝑢𝑡
𝐴 𝑖𝑛
∙ 𝑣 𝑜𝑢𝑡
𝑣𝑖𝑛 =
4.909 𝑖𝑛2
0.7854 𝑖𝑛2 ∙
4 𝑖𝑛
𝑠
= 250 in/s

5. Example 2-6 & 2-6M
Weight versus Mass
 An object weighs 1000
lbs.
 What is the mass?
 What is the mass if the
weight is 200 N?
𝑤 = 𝑚 ∙ 𝑔
𝑚 =
𝑤
𝑔
𝑚 =
1000 𝑙𝑏𝑠
32. 2𝑓𝑡/𝑠2
= 31.06
𝑙𝑏 ∙ 𝑠2
𝑓𝑡
= 31.06 𝑠𝑙𝑢𝑔𝑠
𝑚 =
200 𝑁
9.81 𝑚/𝑠2 = 203.9
𝑁 ∙ 𝑠2
𝑚
= 31.06 𝐾𝑔

6. Example 2-7 Viscosity
 The top plate shown has a wetted
area of 0.5 ft2 and is moving at
velocity of 10 ft/s. The force
required to maintain this speed is 5
lbs. The fluid film thickness is 0.05
in.
 What is the viscosity of the fluid?
From equation 2−6:
𝜇 =
𝐹 ∙𝑦
𝑣 ∙𝐴
𝜇 =
5 𝑙𝑏 ∙(0.00417 𝑓𝑡)
10 𝑓𝑡/𝑠 ∙(0.5 𝑓𝑡2)
= 0.00417
𝑙𝑏 ∙𝑠
𝑓𝑡2
0.05 𝑖𝑛 ∙
1 𝑓𝑡
12 𝑖𝑛
= 0.00417 𝑓𝑡
𝜇 =
5 𝑙𝑏 ∙(0.00417 𝑓𝑡)
10 𝑓𝑡/𝑠 ∙(0.5 𝑓𝑡2)
= 0.00417
𝑙𝑏 ∙𝑠
𝑓𝑡2

7. Example 2-8 Bulk Modulus
 In a system using hydraulic oil
(B = 250,000 psi), the pressure is
3000 psi.
 By what percentage is the oil being
compressed relative to the
unpressurized state (p = 0 psi)?
From equation 2−13: B =
−∆𝑝
∆𝑉
𝑉
calculate the change in pressure:
∆𝑝 = 3000 𝑝𝑠𝑖 − 0 𝑝𝑠𝑖 = 3000 𝑝𝑠𝑖
calculate the proportional change in
volume by manipulating 2−13:
∆𝑉
𝑉
=
−∆𝑝
𝐵
=
−3000 𝑝𝑠𝑖
250,000 𝑝𝑠𝑖
=
−3000 𝑝𝑠𝑖
250,000 𝑝𝑠𝑖
= −0.012
Next, we multiply by 100 to obtain % change
= - 1.2%

8. Example 2-9M Basic Flow rate
 A fluid flows at a velocity of 25 m/min through a
conduit with an ID of 30 mm. Determine the flow
rate.Calculate the conduit area: A =
𝜋∙𝐷2
4
=
𝜋∙(.030 𝑚)2
4
= .0007070 m2
Calculate the flow rate: Q = 𝑣 ∙ 𝐴 =25
𝑚
𝑚𝑖𝑛
∙ .0007070 m2 = .017675 m3/min
Flow rates are commonly in lpm, 1m3 =
1000 l
.017675 m3/min ∙ 1000 l/m3 = 17.67 𝑙/𝑚𝑖𝑛.017675 m3/min 1000 l/m3

9. Example 2-12 Continuity
Equation
 A fluid flows at a velocity of 120 in/min at point 1 in
the system shown. The diameter at point 1 is 2 in
and the diameter at point 2 is 1.5 in.
 Determine the flow velocity at point 2.
 Also determine the flow rate in gpm.
𝐴1 =
𝜋∙(𝐷1 )2
4
=
𝜋∙(2 𝑖𝑛)2
4
= 3.142 𝑖𝑛2
𝐴2 =
𝜋∙(𝐷2 )2
4
=
𝜋∙(1.5 𝑖𝑛)2
4
= 1.767 𝑖𝑛2
From equation 2−15: 𝑣1 𝐴1
= 𝑣2 𝐴2
𝑅𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑖𝑛𝑔: 𝑣2 = 𝑣1 ∙
𝐴1
𝐴2
=
𝑄 = 𝑣1 𝐴1
120 𝑖𝑛/𝑚𝑖𝑛 ∙
3.142 𝑖𝑛2
1.767 𝑖𝑛2120 𝑖𝑛/𝑚𝑖𝑛 ∙
3.142 𝑖𝑛2
1,767 𝑖𝑛2
= 213.4
𝑖𝑛
𝑚𝑖𝑛
= 120
𝑖𝑛
𝑚𝑖𝑛
∙ (3.142 𝑖𝑛2
)
= 377.0
𝑖𝑛3
𝑚𝑖𝑛
377.0
𝑖𝑛3
𝑚𝑖𝑛
∙
1 𝑔𝑎𝑙
231 𝑖𝑛3377.0
𝑖𝑛3
𝑚𝑖𝑛
∙
1 𝑔𝑎𝑙
231 𝑖𝑛3
= 1.632
𝑔𝑎𝑙
𝑚𝑖𝑛

10. Example 2-13 Bernoulli’s
Equation
 A fluid (𝛾 = 0.0324 𝑙𝑏𝑠/𝑖𝑛3
) flows at a constant flow rate
of 150
𝑖𝑛3
𝑠
through the system shown in the figure. The
areas at points 1 & 2 are equal. The pressure at point1
is 100 psi and h = 200 in.
 Determine the pressure at point 2.
From equation 2−16:
ℎ1 +
𝑝1
𝛾
+
𝑣1
2
2𝑔
= ℎ2 +
𝑝2
𝛾
+
𝑣2
2
2𝑔
But the areas at points 1 and 2 are exactly the same, and
this means the velocity at both places are the same.
This cancels the velocity terms in the above equation:
ℎ1 +
𝑝1
𝛾
= ℎ2 +
𝑝2
𝛾
𝑝2 = 𝑝1 − 𝛾(ℎ2 − ℎ1)
𝑝2 = 100
𝑙𝑏𝑠
𝑖𝑛2
− 0.0324
𝑙𝑏𝑠
𝑖𝑛3
(200 𝑖𝑛)𝑝2 = 100
𝑙𝑏𝑠
𝑖𝑛2
− 0.0324
𝑙𝑏𝑠
𝑖𝑛2
(200 𝑖𝑛)
= 93.5
𝑙𝑏𝑠
𝑖𝑛2

11. Example 2-17M
 .

12. END OF LECTURE