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2 years ago

Subject of the study about vacuum cleaner

Engineering

1 answer:

3241004551 [841]2 years ago

3 0

The subject of the study about vacuum cleaners is Cyclonic Technology.

<h3>Which generation is applied in vacuum cleaner?</h3>

Cyclonic Technology. In smooth words, your vacuum cleaner works similar to the way your washing tool extracts stains from the clothes. Cyclonic generation allows the vacuum cleaners to put off dirt and dust from the air it collects on the identical time as cleaning the space.

The vacuum cleaners art work because of Bernoulli's Principle, which states that because the price of air increases, the pressure decreases. Air will commonly flow from a high-pressure place to a low-pressure place, to attempt to balance out the pressure.
In smooth words, your vacuum cleaner works similar to the way your washing tool extracts stains from the clothes. Cyclonic generation allows the vacuum cleaners to put off dirt and dust from the air it collects on the identical time as cleaning the space.

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A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature

mafiozo [28]

Answer:

a)31%

b)34MW

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

For this problem, we will first find the enthalpies in all states

h1=3231kJ/Kg

h2=2310kJ/Kg

h3=h4=272kJ/Kg

A) using the eficiency ecuation

Efficiency = (h1-h2) / (h1-h4)

Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>

b)using ecuation for Wout

Wout = m (h1-h2)

Wout=37(3231-2310)=34077KW=<u>34.077MW</u>

6 0

4 years ago

Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe

kow [346]

Answer:

a. 81 kj/kg

b. 420.625K

c. 101.24kj/kg

Explanation:

$\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }$

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

$\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }$

$\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347$

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

= 80995 J

= 81KJ/kg

b. $n=\frac{t2-t1}{t'2-t1}$

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

$0.80 =\frac{48.5}{t'2-360}$

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

8 0

3 years ago

Your boss asks you to classify some of the components in a temperature measurement system. The system consists of a thermocouple

lianna [129]

Answer:

Explanation:

1) Resolution and uncertainty of both ADC ranges

a) Resolution (for +/- 5 mV range) = 5 mV / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 mV / 256 = 0.0195 mV ..................1

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*0.0195 mV = 0.00975 mV ..........2

b) Resolution (for +/- 5 V range) = 5 V / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 V / 256 = 0.0195 V = 19.5 mV ..................3

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*19.5 mV = 9.75 mV ..........4

2)Thermocouple sensitivity = ( Maximum output voltage - Minimum Output voltage) / (Maximum Temperature - Minimum Temperature)

Thermocouple sensitivity = (3.649mv - 0 ) / (70 - 0) = 0.0521 mV / Deg.C ............5

This is the required Thermocouple sensitivity

3) Water bath temperature is given as 57 deg.C

Hence voltage read by Thermocouple = Sensitivity*57 = 0.0521*57 mV = 2.9697 mV ........6

4)We need to use ADC with a range of +/- 5 mV range as ADC with +/- 5 V range can not do measurement as it's resolution is higher than output voltage.

ADC will measure voltage as 2.9695 mV ......................7

8 0

3 years ago

A 2-lane highway is to be constructed across a 6-ft diameter metal culvert which is oriented perpendicular to the highway center

kvv77 [185]

Answer:

Dude just like forget the highway and drive on the interstate bruh

Explanation:

5 0

4 years ago

Supercharging is the process of (a) Supplying the intake of an engine with air at a density greater than the density of the surr

iVinArrow [24]

Answer:

a)supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere

Explanation:

Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere.

By doing this , it increases the power out put and increases the brake thermal efficiency of the engine.It also increases the volumetric efficiency of the engine.

So the our option a is right.

4 0

3 years ago