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3 years ago

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How do you identify all sensors, functions, and where we can use them?

1 answer:

Alex17521 [72]3 years ago

7 0

Sensor/Detectors/Transducers are electrical, opto-electrical, or electronic devices composed of specialty electronics or otherwise sensitive materials, for determining if there is a presence of a particular entity or function. Many vehicles including cars, trains, buses etc. all use sensors to monitor oil temperature and pressure, throttle and steering systems and so many more aspects.

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Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi

VladimirAG [237]

Answer:

Source 2.

Explanation:

The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:

$\eta_{th} = 1 - \frac{T_{L}}{T_{H}}$

Where:

$T_{L}$ - Temperature of the cold reservoir, in K.

$T_{H}$ - Temperature of the hot reservoir, in K.

The thermal efficiencies are, respectively:

Source 1

$\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}$

$\eta_{th} = 0.466 \,(46.6\,\%)$

Source 2

$\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}$

$\eta_{th} = 0.0798 \,(7.98\,\%)$

The power produced by each device is presented below:

Source 1

$\dot W = (0.466)\cdot (11000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot W = 85.433\,kW$

Source 2

$\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot W = 146.3\,kW$

The source 2 produces the largest amount of power.

8 0

3 years ago

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe

fgiga [73]

Answer:

The rate of fuel required to drive the air conditioner $Q_h = 6.061 kW$

The flow rate of the cold air is $\r m = 0.30765 kg/s$

Explanation:

From this question we are told that

The efficiency is $\eta = 33$ % = 0.33

Temperature for the hot day is $T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)$

Temperature after cooling is $T_c = 5^oC = 278K$

The input power is $P_{in} = 2kW$

The rate of fuel required to drive the air conditioner can be mathematically represented as

$Q_h = \frac{P_{in}}{\eta}$

$= \frac{2}{0.33} = 6.061 kW$

From the question the air condition is assumed to be half as a Carnot refrigeration unit

This can be Mathematically interpreted in terms of COP(coefficient of performance) as

$\beta_{air} = 0.5 \beta$

where $\beta$ denotes COP and is mathematically represented as

$\beta = \frac{Q_c}{P_{in}}$

= > $Q_c = \beta P_{in}$

Where $Q_c$ is the rate of flue being burned for cold air to flow

Now if the COP of a Carnot refrigerator is having this value

$\beta_{Carnot } = \frac{T_c}{T_h - T_c}$

$= \frac{278}{308-278}$

$\beta_{Carnot} = 9.267\\$

Then

$\beta_{air} = 0.5 * 9.2667$

$= 4.6333$

Now substituting the value of $\beta$ to solve for $Q_c$

$Q_c = \beta P_{in}$

$= 4.6333 *2$

$9.2667kW$

The equation for the rate of fuel being burned for the cold air to flow

$Q_c = \r mc_p \Delta T$

Making the flow rate of the cold air

$\r m = \frac{Q_c}{c_p \Delta T}$

$= \frac{9.2667}{1.004}* (308 - 278)$

$= 0.30765 kg/s$

4 0

3 years ago

An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter

slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg , 1980.39 MJ/Kmol

Explanation:

molar mass of ethanol (C2H5OH) = 46 g/mol

molar mass of octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol = 0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg

HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0

3 years ago

What is the purpose of the graphic language?

solmaris [256]

Answer:

enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.

5 0

3 years ago

Please view the file attached below and help me answer all the questions!!

evablogger [386]

Answe whats the question i can help you

Explanation:

8 0

3 years ago