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2 years ago

Which of the following is true about modern hydraulic lifts?

Engineering

1 answer:

kaheart [24]2 years ago

6 0

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>

<em />

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

Learn more here:

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It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d

denis23 [38]

The exit temperature is 586.18K and compressor input power is 14973.53kW

Data;

Mass = 50kg/s
T = 288.2K
P1 = 1atm
P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\$

Let's substitute the values into the formula

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K$

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

$P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW$

The compressor input power is 14973.53kW

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3 years ago

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Zina [86]

Answer:

0.245 m^3/s

Explanation:

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Answer:

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