The modern hydraulic lifts make use of biodegradable fluid to transmit hydraulic power
<em>Question: The options are left out in the question. The details and facts about the modern hydraulic lift are presented here</em>
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Details about the modern hydraulic lifts include;
The development of the modern hydraulic occurred in the Industrial Revolution to perform task done previously by steam powered elevators
The power of the hydraulic lift come from the hydraulic cylinder known as the actuator, which in turn is powered by pressurized hydraulic fluid such as oil
The hydraulic fluid is pushed by a piston rod through which energy is capable of being transferred, such that the applied force is multiplied, to provide more power for lifting
<u>Facts about the modern hydraulic lifts include;</u>
- The dry motor in the modern hydraulic lift is more efficient and consumes 20% less energy
- It comprises of valves that are controlled electronically such that the response is much rapid and the energy consumption is reduced by a further 20%
- The cars used in the modern lift are lighter, as well as the slings, which reduces the power usage by 20%
- It makes use of chemicals which are environmentally friendly as hydraulic fluid
- The flash point of the fluid used is higher, as well as it posses 50% lower compressibility as well elasticity
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Answer:
Both Technician A and Technician B are correct
Explanation:
Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.
Technician B is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.
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Answer:
The fluid property responsible for the development of velocity boundary layer is majorly the fluid's viscosity.
For non-viscous fluids (in theory, because no fluid is entirely non-viscous), there will be no velocity boundary layer.
Explanation:
The velocity boundary layer is the thin layer of viscous fluid that is in direct contact with the pipe surface. The velocity of fluid in this layer is 0 as fluid doesn't move in this layer.
This phenomenon is due to the viscosity of the fluid. Viscosity of the fluid refers to the internal friction that exists between fluid layers, so, the layer of fluid in contact with non-moving, static surface of the pipe experiences friction that causes this layer to not move, causing the fluid velocity to vary from 0 at this surface to the maximum value at the centre of the pipe, before the velocity begins to drop again until it reaches 0 at the other end of the circular pipe.
Since viscosity is the primary cause of this, non-viscous or inviscid fluids are saved from this phenomenon as their flows do not have the velocity boundary layer.
Although, a completely non-viscous or inciscid fluid is an idealized concept because all fluids will experience some sort of viscosity (no matter how small) between their fluid layers. Hence, a velocity boundary layer, no matter how thin (or of minute thickness), will exist in the flow of real fluids.
But, an idealized non-viscous or inviscid fluid will not have a velocity boundary layer.
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Answer:
the compressor power is -223.12 hP { -ve indicate work done }
Volumetric flow rate at exit = 262.74 ft³/s
Explanation:
Given the data in the question;
p₁ = 14.2 psi = 0.978 bar
p₂ = 120 psi = 8.268 bar
T₁ = 60°F = 288.706 K
T₂ = 500°F = 533.15 K
Q₁ = 1200 ft³/min = 20ft³/s = 0.566 m³/s
δ₁ = p₁/RT₁ = 0.978 × 10⁵ / ( 287 × 288.706) = 1.18 kg/m³
δ₂ = P₂/RT₂ = 8.268 × 10⁵ / ( 287 × 533.15 ) = 5.4 kg/m³
so
ω₁ = δ₁Q₁ = 1.18 × 0.566 = 0.668 kg/s
we know that; ω₁ = ω₂ { in a steady flow }
ω₂ = δ₂Q₂
Q₂ = ω₂/δ₂
Q₂ = 0.668 / 5.4 = 0.1237 m³/s
Hence Volumetric flow rate at exit = 262.74 ft³/s
from the steady state energy equation;
ω( h₁ - h₂ ) = dW/dt
{ where h = CpT)
( Cp = 1.005 )
dW/dt = ωCp( T₁ - T₂ )
we substitute
dW/dt = 0.668 × 1.005( 288.706 - 533.15 )
dW/dt = 0.67134 × -244.444
dW/dt = -164.105 kW = -223.12 hP
Hence, the compressor power is -223.12 hP { -ve indicate work done }