Transcript Document

MECH1300
Basic Principles of Hydraulics
Topics
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Pascal’s Law
Transmission of Force
Properties of Hydraulic Fluids
Liquid Flow
Static Head Pressure
Power
Hydraulic Systems
Chapter 2
MECH1300
Basic Principles of Hydraulics
Hydraulic Systems usually operate from 500 to 5000 lbs per square inch.
They are precise with speed and positioning.
Smooth transitions.
Have more power.
MECH1300
Pascal’s Law
Pressure is the intensity of a force, it’s created when a force from one object
acts over an area of another object.
Pascal’s Law – The pressure exerted on a confined fluid at rest is
transmitted undiminished in all directions and acts at right angles to the
containing surfaces.
Pressure transmits force and energy, not the fluid.
Most industrial hydraulic systems use pressurized fluids
and are considered hydrostatic
MECH1300
Pressure
𝐹𝑜𝑟𝑐𝑒
𝑁
𝑙𝑏𝑠
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
= 2= 2
𝐴𝑟𝑒𝑎
𝑚
𝑖𝑛
𝐹
𝑝=
𝐴
One kPa is equal to 1000 Pascals, and
6.9 kPa equals 1 psi
1 bar = 100KPascal
𝑙𝑏𝑠
𝑖𝑛2
is abbreviated as psi (U.S. Units)
𝑁
𝑚2
𝑖𝑠 𝑎𝑏𝑏𝑟𝑒𝑣𝑖𝑎𝑡𝑒𝑑 𝑎𝑠 𝑎 𝑝𝑎𝑠𝑐𝑎𝑙𝑠 (𝑃𝐴) (S.I. Units)
𝑝=
5
= 2 𝑝𝑎𝑠𝑐𝑎𝑙𝑠
2.5
𝑝=
5
= 5 𝑝𝑎𝑠𝑐𝑎𝑙𝑠
1
MECH1300
Pressure
Ex: What is the amount of pressure in the given containier?
10
𝑝=
= 100 𝑝𝑎𝑠𝑐𝑎𝑙𝑠
0.1
MECH1300
Pressure
Ex: The piston has a diameter of 1.5in. We don’t want to exceed a pressure of 500psi. What is the
maximum force this system can withstand?
𝜋 ∙ 𝐷2 3.14159 ∗ (1.5)2
𝐴=
=
= 1.767𝑖𝑛2
4
4
𝐹 = 𝑝 ∙ 𝐴 = 500
𝑙𝑏𝑠
∙ 1.767𝑖𝑛2 = 884 𝑙𝑏𝑠
2
𝑖𝑛
MECH1300
Pressure
Ex: The piston is required to support a force of 10kN (10,000N). We do not want to exceed a
pressure of 70 bar. What size cylinder is required?
𝑁
70 𝑏𝑎𝑟 = 7,000,000 𝑃𝑎 = 7,000,000 2
𝑚
𝐴=
𝐹
𝑝
=
10,000𝑁
𝑁
7,000,000
𝑚
Solve A =
𝐷=
4∙𝐴
𝜋
= 0.001429𝑚2
𝜋∙𝑑2
4
for D
=
4∙(0.001429)
3.14159
= 0.04265m
MECH1300
Pressure
Ex: The piston has a diameter of 40mm. We don’t want to exceed a pressure of 3500 kPa. What is
the maximum force this system can withstand?
𝑁
3500𝑘𝑃𝐴 = 3,500,000 𝑃𝑎 = 3,500,000 2
𝑚
𝜋 ∙ 𝐷2 3.142 ∙ 0.04𝑚
𝐴=
=
4
4
2
= 0.001257𝑚2
𝑁
𝐹 = 𝑝 ∙ 𝐴 = 3500000 2 ∙ 0.001257𝑚2 = 4399𝑁
𝑚
MECH1300
Multiplying Force
𝑊𝑜𝑟𝑘 = 𝐹𝑜𝑟𝑐𝑒 ∙ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑊 = 𝐹 ∙ 𝑑
𝐹𝑜𝑢𝑡
𝐴𝑟𝑒𝑎𝑜𝑢𝑡
=
∙ 𝐹𝑖𝑛
𝐴𝑟𝑒𝑎𝑖𝑛
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑖𝑛 =
𝐴𝑟𝑒𝑎𝑜𝑢𝑡
∙ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑜𝑢𝑡
𝐴𝑟𝑒𝑎𝑖𝑛
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦𝑖𝑛 =
𝐴𝑟𝑒𝑎𝑜𝑢𝑡
∙ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦𝑜𝑢𝑡
𝐴𝑟𝑒𝑎𝑖𝑛
This is a closed system – Fluid must remain constant.
Whatever amount of fluid flows out of the input piston
must flow into the output piston.
MECH1300
Multiplying Force
Ex: If we have an input cylinder with a diameter of 1 in and an
output cylinder with a diameter 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?
Input piston area
𝜋 ∙ 𝐷2 3.142 ∙ 1𝑖𝑛 2
𝐴=
=
= 0.7854𝑖𝑛2
4
4
Output piston area
𝜋 ∙ 𝐷2 3.142 ∙ 2.5𝑖𝑛 2
𝐴=
=
= 4.909𝑖𝑛2
4
4
Output Force
𝐴𝑟𝑒𝑎𝑜𝑢𝑡
4.909𝑖𝑛2
𝐹𝑜𝑢𝑡 =
∙ 𝐹𝑖𝑛 =
∙ 250𝑙𝑏𝑠 = 1563𝑙𝑏𝑠
𝐴𝑟𝑒𝑎𝑖𝑛
0.7854𝑖𝑛2
Input distance
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑖𝑛
𝐴𝑟𝑒𝑎𝑜𝑢𝑡
4.909𝑖𝑛2
=
∙ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑜𝑢𝑡 =
∙ 1𝑖𝑛 = 6.25𝑖𝑛
𝐴𝑟𝑒𝑎𝑖𝑛
0.7854𝑖𝑛2
MECH1300
Basic Properties of Hydraulic Fluids
Mass – How much matter an object contains. (Slugs, Kg)
Weight – The measure of how much force an objects mass creates due to gravity. (lbs, N)
Gravity (Earth) = 32.2 𝑓𝑡
𝑠𝑠
= 9.81 𝑚
𝑠2
𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑎𝑠𝑠 ∙ 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
EX: An object weighs 500 lbs. What is its mass?
𝑤 500𝑙𝑏𝑠
𝑙𝑏 ∙ 𝑠 2
𝑚= =
= 15.53
= 15.53 𝑠𝑙𝑢𝑔𝑠
𝑓𝑡
𝑔
𝑓𝑡
32.2 2
𝑠
EX: An object weighs 1000N. What is its mass?
𝑤 1000 𝑁
𝑁 ∙ 𝑠2
𝑚= =
= 101.94
= 101.94 𝐾𝑔
𝑔 9.81 𝑚
𝑚
𝑠2
𝑤 = 𝑚𝑔
MECH1300
Basic Properties of Hydraulic Fluids
Density – A substance’s mass per unit volume. (𝑠𝑙𝑢𝑔𝑠
𝑓𝑡 3
, 𝐾𝑔
𝑚𝑎𝑠𝑠
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑣𝑜𝑙𝑢𝑚𝑒
𝑚3 )
𝑚
𝜌=
𝑉
𝑤𝑒𝑖𝑔ℎ𝑡
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑊𝑒𝑖𝑔ℎ𝑡 =
𝑉𝑜𝑙𝑢𝑚𝑒
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 =
𝑤
𝛾=
𝑉
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑙𝑢𝑖𝑑
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑤𝑎𝑡𝑒𝑟
𝛾𝑤𝑎𝑡𝑒𝑟 = 62.4 𝑙𝑏𝑠
𝑠𝑔𝑥 =
𝛾𝑥
𝛾𝑤𝑎𝑡𝑒𝑟
𝑓𝑡 3
Specific gravity does not have units associated with it.
Specific weight of Petroleum based oils is 56 𝑙𝑏𝑠
𝑓𝑡 3
𝑜𝑟 8800 𝑁
𝑚3
MECH1300
Basic Properties of Hydraulic Fluids
EX: Calculate the specific gravity of oil.
𝑠𝑔𝑜𝑖𝑙 =
𝛾𝑜𝑖𝑙
𝛾𝑤𝑎𝑡𝑒𝑟
=
56 𝑙𝑏𝑠
62.4 𝑙𝑏𝑠
𝑓𝑡 3
𝑓𝑡 3
= 0.9
MECH1300
Basic Properties of Hydraulic Fluids
Viscosity – The measure of a fluid’s thickness or resistance to flow.
𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 =
𝐹∙𝑦
𝑣∙𝐴
𝐹 = 𝑓𝑜𝑟𝑐𝑒 𝑙𝑏𝑠, 𝑁
𝑦 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 (𝑓𝑡, 𝑚)
𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑓𝑡 𝑠 , 𝑚 𝑠)
𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑣𝑖𝑛𝑔 𝑝𝑙𝑎𝑡𝑒 (𝑓𝑡 2 , 𝑚2 ) (wetted area)
𝑙𝑏∙𝑠 𝑁∙𝑠
,
𝑓𝑡 2 𝑚2
MECH1300
Basic Properties of Hydraulic Fluids
EX: The top plate shown has a wetted area of 0.5𝑓𝑡 2 and is moving at a velocity of 10 𝑓𝑡 𝑠. The
force required to maintain this speed is found to be 5 lbs. The fluid film thickness is 0.05 in.
What is the viscosity of the fluid?
𝜇=
𝐹∙𝑦
𝑣∙𝐴
0.05 𝑖𝑛 ∙
𝜇=
1 𝑓𝑡
12 𝑖𝑛
= 0.00417 𝑓𝑡
5 𝑙𝑏𝑠∙(0.00417𝑓𝑡)
10𝑓𝑡 𝑠∙ 0.5𝑓𝑡 2
= 0.00417
𝑙𝑏∙𝑠
𝑓𝑡 2
MECH1300
Basic Properties of Hydraulic Fluids
Kinematic viscosity – "momentum diffusivity“- The dynamic viscosity
divided by the density.
𝑣=
𝜇
𝜌
𝑙𝑏∙𝑠
Kinematic viscosity of oil =
0.00417 2
𝑓𝑡
1.74
𝑙𝑏∙𝑠2
𝑓𝑡4
= 0.00240 𝑓𝑡
2
𝑁∙𝑠
=
0.0833 2
𝑚
𝑁∙𝑠2
897 4
𝑚
= 0.0000929 𝑚
𝑠
2
𝑠
MECH1300
Basic Properties of Hydraulic Fluids
Viscosity is determined using a Saybolt viscometer. It measures the time for a 60-milliliter sample of
oil to drain through a standard orifice, resulting in saybolt seconds universal (SSU or SUS).
Viscosity decreases with increasing temperature.
Viscosity Index (V.I.) - a scale from 1-100. The higher the number, the more stable the fluids viscosity
with varying temperature.
If viscosity is too high, then it becomes more difficult for fluid to flow through a system.
If viscosity is too low, it creates excessive wear because the oil is not thick enough to provide good
lubrication. It can also cause leakage.
MECH1300
Bulk Modulus
The Bulk Modulus of a liquid is a measure of it’s stiffness or incompressibility.
𝐵=
𝐵𝑢𝑙𝑘 𝑆𝑡𝑟𝑒𝑠𝑠
𝐵𝑢𝑙𝑘 𝑆𝑡𝑟𝑎𝑖𝑛
=
𝐹𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎
∆𝑉 𝑉
=
−∆𝑝
∆𝑉 𝑉
𝐵 = 𝑏𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 (𝑝𝑠𝑖, 𝑃𝑎)
∆𝑝 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑝𝑠𝑖, 𝑃𝑎)
∆𝑉 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑖𝑛3 , 𝑚3 )
𝑇ℎ𝑒 ∆𝑉 𝑉 𝑡𝑒𝑟𝑚 𝑖𝑠 𝑎 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒
MECH1300
Bulk Modulus
EX: In a system using hydraulic oil (B = 250,000 psi), the pressure is 4000 psi.
By what percentage is the oil being compressed relative to the unpressurized
state (𝑝 = 0 𝑝𝑠𝑖)?
Calculate the change in pressure:
∆𝑝 = 4000 𝑝𝑠𝑖 − 0 𝑝𝑠𝑖 = 4000 𝑝𝑠𝑖
Calculate the proportional change in
volume by solving for ∆𝑉 𝑉
−∆𝑝
−4000 𝑝𝑠𝑖
∆𝑉 𝑉 =
=
= −0.016
𝐵
250,000 𝑝𝑠𝑖
Multiply by 100 to get the % change
−0.016 × 100 = −1.6%
At 4000 psi the oil is only 1.6%
of its original volume. This
justifies that oil is relatively
incompressible.
MECH1300
Liquid flow
Flow Rate – The volume of fluid that moves through a system in a
given period of time. It determines the speed at which the output device
will operate.
Flow velocity – Is the distance the fluid travels in a given period of
time.
𝑄 =𝑣∙𝐴
𝑖𝑛3 𝑚3
𝑄 = 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
,
𝑚𝑖𝑛 𝑚𝑖𝑛
𝑖𝑛
𝑚
𝑣 = 𝑓𝑙𝑜𝑤 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
,
𝑚𝑖𝑛 𝑚𝑖𝑛
2
2
𝐴 = 𝑎𝑟𝑒𝑎 𝑖𝑛 , 𝑚
MECH1300
Liquid flow
𝑖𝑛
500
𝑚𝑖𝑛
EX: A fluid flows at a velocity of
through a conduit with an ID
(inner diameter) of 3 in. Determine the flow rate.
Calculate the conduit area:
𝜋 ∙ 𝐷2 3.142 ∙ 3 𝑖𝑛
𝐴=
=
4
4
2
= 7.07 𝑖𝑛2
Calculate the flow rate:
3
𝑖𝑛
𝑖𝑛
𝑄 = 𝑣 ∙ 𝐴 = 500
∙ 7.07𝑖𝑛2 = 3534.75
𝑚𝑖𝑛
𝑚𝑖𝑛
Convert to gpm:
𝑖𝑛3
1 𝑔𝑎𝑙
𝑔𝑎𝑙
3534.75
∙
= 15.30
𝑚𝑖𝑛 231𝑖𝑛3
𝑚𝑖𝑛
MECH1300
Liquid flow
𝑚
50
𝑚𝑖𝑛
EX: A fluid flows at a velocity of
through a conduit with an ID
(inner diameter) of 20 mm. Determine the flow rate.
Calculate the conduit area:
𝜋 ∙ 𝐷2 3.142 ∙ 0.020 𝑚
𝐴=
=
4
4
2
= 0.000314 𝑚2
Calculate the flow rate:
3
𝑚
𝑚
𝑄 = 𝑣 ∙ 𝐴 = 50
∙ 0.000314𝑚2 = 0.0157
𝑚𝑖𝑛
𝑚𝑖𝑛
Convert to lpm:
𝑚3
1000 𝑙
𝑙
0.0157
∙
= 15.71
𝑚𝑖𝑛
1𝑚3
𝑚𝑖𝑛
MECH1300
Liquid flow
EX: A fluid flows at a rate of 20 𝑙𝑝𝑚 through a conduit with an ID (inner
diameter) of 30 mm. Determine the flow velocity in m/s.
Calculate the conduit area:
𝜋 ∙ 𝐷2 3.142 ∙ 0.030 𝑚
𝐴=
=
4
4
Convert to 𝑚
3
2
= 0.000707 𝑚2
𝑚𝑖𝑛:
𝑙
1𝑚3
𝑚3
20
∙
= 0.02
𝑚𝑖𝑛 1000 𝑙
𝑚𝑖𝑛
Calculate the flow velocity:
𝑚3
0.02 𝑚𝑖𝑛
𝑄
𝑚
𝑣= =
=
28.29
𝐴 0.000707𝑚2
𝑚𝑖𝑛
Convert to m/s:
𝑚
1 𝑚𝑖𝑛
𝑚
28.29
∙
= 0.4715
𝑚𝑖𝑛
60 𝑠
𝑠
MECH1300
Continuity Equation
Steady Flow – The volume of fluid that flows through one section of a
system must flow through any other section.
This occurs if a pump is producing a constant flow rate so that we can
assume that the fluid is incompressible.
MECH1300
Continuity Equation
𝑄1 = 𝑄2
𝑣1 ∙ 𝐴1 = 𝑣2 ∙ 𝐴2
This is a system with a steady flow rate, with a reduction in
area. The reduction in area will correspond to an increase
in flow velocity.
MECH1300
Continuity Equation
Ex: A fluid flows at a velocity of 100 in/min at point 1. Its diameter at point 1 is 3 in
and the diameter at point 2 is 2 in. Determine the flow velocity at point 2.
Determine the flow rate in gpm.
Calculate the conduit areas:
𝜋 ∙ 𝐷2 3.142 ∙ 3 𝑖𝑛 2
𝐴1 =
=
= 7.07 𝑖𝑛2
4
4
2
𝜋∙𝐷
3.142 ∙ 2 𝑖𝑛 2
𝐴2 =
=
= 3.142 𝑖𝑛2
4
4
Calculate the flow velocity:
𝐴1
𝑖𝑛 7.07𝑖𝑛2
𝑖𝑛
𝑣2 = 𝑣1 ∙
= 100
∙
= 207.21
𝐴2
𝑚𝑖𝑛 3.142𝑖𝑛2
𝑚𝑖𝑛
Calculate the flow rate:
3
𝑖𝑛
𝑖𝑛
𝑄 = 𝑣1 ∙ 𝐴1 = 100
∙ 7.07𝑖𝑛2 = 707
𝑚𝑖𝑛
𝑚𝑖𝑛
Convert to gpm:
𝑖𝑛3
1 𝑔𝑎𝑙
𝑔𝑎𝑙
707
∙
=
3.06
𝑚𝑖𝑛 231𝑖𝑛3
𝑚𝑖𝑛
MECH1300
Bernoulli’s Equation
Bernoulli’s equation describes the total energy of an incompressible fluid.
1. Potential Energy (due to elevation and gravity) = 𝑤 ∙ ℎ
𝑝
2. Pressure Energy = w ∙
𝛾
3. Kinetic Energy (due to velocity) = 𝑤
𝑤 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑙𝑏𝑠, 𝑁
ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 (𝑖𝑛, 𝑚)
𝑝 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒(𝑙𝑏𝑠 𝑖𝑛2 , 𝑁 𝑚2 )
𝛾 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡(𝑙𝑏𝑠 𝑖𝑛3 , 𝑁 𝑚3 )
𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑖𝑛 𝑠 , 𝑚 𝑠)
𝑔 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 396.4 𝑖𝑛 𝑠2 = 9.81 𝑚
𝑠2
𝑣2
∙
2∙𝑔
MECH1300
Bernoulli’s Equation
Potential Energy 1+Pressure Energy 1+Kinetic Energy 1 = Potential Energy 2+Pressure Energy 2+Kinetic Energy 2
𝑤 ∙ ℎ1 + 𝑤
𝑝
ℎ1 + 1
𝛾
+
𝑝1
∙
𝛾
𝑣1 2
2∙𝑔
+𝑤
=
𝑣1 2
∙
2∙𝑔
𝑝
ℎ2 + 2
𝛾
= 𝑤 ∙ ℎ2 + 𝑤 ∙
𝑝2
𝛾
+𝑤∙
𝑣2 2
2∙𝑔
𝑣2 2
+
2∙𝑔
It tells us that an increase in energy in any one of the three areas must be balanced by a reduction in one or more of the other
two areas of an equal amount.
Elevation Head + Pressure Head + Velocity Head = total head
Head is a measure of energy a fluid posses per pound.
MECH1300
Bernoulli’s Equation
3
EX: A fluid 𝛾 = 0.0324 𝑙𝑏𝑠 𝑖𝑛3 flows at a constant rate of 150 𝑖𝑛 𝑠. The areas at point 1 and point 2
are equal. The pressure at point 1 is 100 psi and h=200 in. Determine the pressure at point 2.
𝑝1 𝑣1 2
𝑝2 𝑣2 2
ℎ1 + +
= ℎ2 + +
𝛾 2∙𝑔
𝛾 2∙𝑔
Since the two areas are equal and the flow rate is
constant, we can ignore the velocity terms because
they cancel out, so the equation reduces to:
𝑝1
𝑝2
ℎ1 + = ℎ2 +
𝛾
𝛾
Solve for 𝑝2
𝑝2 = 𝑝1 − 𝛾 ℎ2 − ℎ1
Substitute known values
𝑙𝑏𝑠
𝑙𝑏𝑠
𝑙𝑏𝑠
𝑝2 = 100 2 − 0.0324 3 200𝑖𝑛 = 93.5 2
𝑖𝑛
𝑖𝑛
𝑖𝑛
MECH1300
Bernoulli’s Equation
3
EX: A fluid 𝛾 = 8000 𝑁 𝑚3 flows at a constant flow rate of 0.004 𝑚 𝑠. The areas are 𝐴1 = 0.003𝑚2
and 𝐴2 = 0.002𝑚2 . If the pressure at point 1 is 500kPa, determine the pressure at point 2.
Calculate the flow velocities:
𝑣1 =
𝑣2 =
𝑄
𝐴1
𝑄
𝐴2
𝑚3
=
=
0.004 𝑠
0.003𝑚2
𝑚3
𝑠
0.002𝑚2
0.004
= 1.33
=2
𝑚
𝑠
𝑚
𝑠
Because ℎ1 = ℎ2 the elevation terms drop out.
Solve for 𝑝2 :
𝛾
𝑝2 = 𝑝1 +
𝑣1 2 − 𝑣2 2
2∙𝑔
Reducing the size of the conduit causes and increase in
velocity and a decrease in pressure.
Substitute known values:
𝑁
𝑁
3
𝑚
𝑝2 = 500,000 2 +
𝑚
𝑚
2 ∙ 9.81 2
𝑠
8000
𝑚
1.33
𝑠
2
𝑚
− 2
𝑠
2
𝑁
= 99090 2 = 99.09𝑘𝑃𝑎
𝑚
MECH1300
Bernoulli’s Equation
Venturi – reducing a conduits size to intentionally reduce pressure. The reduced diameter section is
called the throat.
The heights are equal and the elevation terms drop out. Bernoulli’s equation becomes:
𝑝1
𝛾
𝑣1 2
+ 2∙𝑔
=
𝑝2
𝛾
+
𝑣2 2
2∙𝑔
Then solve for 𝑣2
𝐴
using the continuity equation plug in 𝑣1 = 𝑣2 ∙ 𝐴2
1
𝑣2 =
2 ∙ 𝑔 ∙ 𝑝1 − 𝑝2
𝐴
𝛾 ∙ 1 − 𝐴2
1
MECH1300
Torricelli’s Theorem
A special case of Bernoulli’s equation that applies to a liquid draining from a tank.
In this case the pressure at points 1 and 2 are both 0 psi and the velocity at point 1 is zero because it
is going to be very small compared to the velocity at point 2. Bernoulli’s equation is reduced to:
ℎ1 = ℎ2 +
𝑣2 2
2∙𝑔
We solve for 𝑣2
𝑣2 =
2 ∙ 𝑔 ℎ1 − ℎ2
However, ℎ1 − ℎ2 is simply just h, so
𝑣2 =
2∙𝑔∙ℎ
MECH1300
Torricelli’s Theorem
EX: The tank is fill with liquid to a height of 4 feet, determine the velocity at the outlet.
𝑣2 =
2∙𝑔∙ℎ
𝑣2 =
2 ∙ 32.2
𝑓𝑡
𝑠2
∙ 4𝑓𝑡 = 16.04
𝑓𝑡
𝑠
EX: The tank is fill with liquid to a height of 2 meters, determine the
velocity at the outlet.
𝑣2 =
2∙𝑔∙ℎ
𝑣2 =
2 ∙ 9.81 𝑠2 ∙ 2𝑚 = 6.26
𝑚
𝑚
𝑠
MECH1300
Laminar and Turbulent Flow
Laminar Flow – A thin layer of fluid adheres to the conduit wall, as you move away from the conduit
wall, the velocity increases at each layer until it is at it’s maximum in the center of the conduit.
If the average velocity increase
the flow can transition from
Laminar to Turbulent
Turbulent Flow – Fluid flows in a chaotic manner with parcels of fluid flowing in all
directions, as a type of “churning” action
MECH1300
Laminar and Turbulent Flow
We can predict the type of flow by calculating the Reynolds number.
𝑅𝑒 =
𝑣∙𝐷
v
𝑅𝑒 = 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑛𝑜 𝑢𝑛𝑖𝑡𝑠
𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑡 𝑠2 , 𝑚 𝑠2
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑓𝑡, 𝑚
𝑓𝑡 2 𝑚2
v = 𝑘𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦
𝑠,
𝑠
Laminar flow 𝑅𝑒 < 2000
2000 < 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛𝑎𝑙 < 4000
Laminar
Turbulent Flow 4000 < 𝑅𝑒
Type of flow is difficult to predict in the transitional range
MECH1300
Laminar and Turbulent Flow
2
EX: A fluid with a kinematic viscosity of 0.0045 𝑓𝑡 𝑠 is flowing through a 3-in diameter conduit at a
flow rate of 400 gpm. Determine if the flow will be laminar or turbulent.
3
Convert gpm to 𝑖𝑛 𝑠:
𝑔𝑎𝑙 231𝑖𝑛3
1 𝑚𝑖𝑛
𝑖𝑛3
𝑄 = 400
∙
∙
= 1540
𝑚𝑖𝑛
1 𝑔𝑎𝑙
60 𝑠
𝑠
Calculate the area:
𝜋 ∙ 𝐷2 𝜋 ∙ 3𝑖𝑛
𝐴=
=
4
4
2
= 7.069𝑖𝑛2
Calculate the flow velocity:
𝑖𝑛3
𝑄 1540 𝑠
𝑖𝑛
𝑣= =
=
217.87
𝐴 7.069𝑖𝑛 𝑠
𝑠
Convert to 𝑓𝑡 𝑠:
𝑖𝑛
1 𝑓𝑡
𝑓𝑡
𝑣 = 217.87 ∙
= 18.16
𝑠
12 𝑖𝑛
𝑠
Convert diameter into feet:
1 𝑓𝑡
𝐷 = 3𝑖𝑛 ∙
= 0.25 𝑓𝑡
12 𝑖𝑛
Calculate 𝑅𝑒 :
𝑓𝑡
𝑣 ∙ 𝐷 18.16 𝑠 ∙ 0.25 𝑓𝑡
𝑅𝑒 =
=
= 1008.89
𝑓𝑡 2
v
0.0045
𝑠
Laminar
MECH1300
Static Head Pressure
Static Head Pressure – The deeper that you submerge an object underwater it is subject to
pressure. The pressure is do to the weight of the water that is above it, pressing down on it.
Its magnitude is determined by:
𝑝𝐻 = 𝛾 ∙ ℎ
𝑝𝐻 = ℎ𝑒𝑎𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝛾 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡
ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 (𝑖𝑛, 𝑚)
𝑙𝑏𝑠
𝑙𝑏𝑠
𝑖𝑛3
𝑖𝑛3
,𝑁
,𝑁
𝑚3
𝑚3
MECH1300
Static Head Pressure
EX: A tank is fill with oil with a specific weight of 8000 𝑁
at the bottom of the tank?
𝑚3
to a depth of 8 m. What is the pressure
𝑁
𝑁
𝑝𝐻 = 𝛾 ∙ ℎ = 8000 𝑚3 ∙ 8𝑚 = 64000 2 = 64𝑘𝑃𝑎
𝑚
MECH1300
Pressure loss
• When there is a flowing fluid, pressure drops as the
fluid flows through the system due to energy loss (due
to viscosity of the fluid resisting flow).
• When flow stops, pressure equalizes again.
• Lost energy is transformed into heat, raising the
temperature of the fluid.
• As temperature rises, viscosity goes down.
• It’s ability to lubricate weakens, causing wear and tear
to components
MECH1300
Power Review
𝑤
𝑡
=
𝐻𝑃 =
𝐹∙𝑣
550
𝑃=
1 ℎ𝑝 = 550
𝑓𝑡∙𝑙𝑏𝑠
𝑠
𝐹∙𝑑
𝑡
𝑣=
𝑑
𝑡
so 𝑃 = 𝐹 ∙ 𝑣
HP means power is in the horsepower
𝑘𝑊 =
𝐹∙𝑣
1000
EX: A 20,000 N load must be moved at a velocity of 2 m/s. How much power is required?
𝑘𝑊 =
20000𝑁 ∙ 2
1000
𝑚
𝑠 = 40𝑘𝑊
MECH1300
Power In Hydraulics
Power in terms of pressure
𝑤
𝐹∙𝑑
𝑑
𝑃= =
𝑣=
so 𝑃 = 𝐹 ∙ 𝑣 and 𝐹 = 𝑝 ∙ 𝐴
𝑡
𝑡
𝑃 =𝑝∙𝐴∙𝑣
𝑡
𝑝𝑜𝑤𝑒𝑟 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∙ 𝑎𝑟𝑒𝑎 ∙ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑄 = 𝐴 ∙ 𝑣
𝑃 =𝑝∙𝑄
𝐻𝑃𝐻 =
𝑝∙𝑄
1714
𝑘𝑊𝐻 =
𝑝∙𝑄
60000
MECH1300
Power In Hydraulics
EX: Oil is flowing through a hydraulic system at 15 gpm and 2700 psi.
What is the hydraulic power of this system?
𝑝 ∙ 𝑄 2700 𝑝𝑠𝑖 ∙ 10 𝑔𝑝𝑚
𝐻𝑃𝐻 =
=
= 15.75 ℎ𝑝
1714
1714
MECH1300
Hydraulic Systems
Three main segments:
1. Power supply – supplies flow to the system. The pump takes the
mechanical power from the prime mover (motor or engine) and
converts it to fluid power
2. Control – Controls the direction of the fluid with valves, allowing a
cylinder to extend or retract.
3. Output – Output force is controller by the pressure of the fluid.