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Alex777 [14]
3 years ago
9

What is the minimum efficiency of a functioning current-model catalytic converter? a. 60% b. 75% c. 80% d. 90%

Engineering
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

d. 90%

Explanation:

As we know that internal combustion engine produce lot's of toxic gases to reduce these toxic gases in the environment a device is used and this device is know as current modeling converter.

Generally the efficiency of current model catalytic converter is more than 90%.But the minimum efficiency this converter is 90%.

So option d is correct.

d. 90%

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Why do need engineer and architect​
USPshnik [31]

Answer: By using the expertise of an engineer or architect, a project can be built more efficiently and economically since the homeowner's vision can evolve on paper rather than during the building process. They also will work diligently with your budget and select materials and builders within your price point.

Explanation:

5 0
3 years ago
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
You are preparing to work with Chemical A. You open the appropriate storage cabinet, and notice Chemical B, as well as Chemical
PtichkaEL [24]

The correct answer is; Stability and reactivity.

Further Explanation:

The stability and reactivity section of the SDS sheets is where to check for the possibility of hazardous reactions for the chemicals. This also lists the chemical stability of each chemical that people may be using. This can be found in section 10 of the OSHA Quick Card.

The SDS sheets has 16 sections for employees to use. Since 2015, the sections can be found in uniform format for easier and faster ways to find the section needed. The 16 sections for the SDS sheets are:

  1. Identification
  2. Hazard(s) identification
  3. Composition/information on ingredients
  4. First-aid measures
  5. First-aid measures
  6. First-aid measures
  7. Handling and storage
  8. Exposure controls/personal protection
  9. Physical and chemical properties
  10. Stability and reactivity
  11. Toxicological information
  12. Ecological information
  13. Disposal considerations
  14. Transport information
  15. Regulatory information
  16. Other information

Learn more about SDS sheets at brainly.com/question/9753408

#LearnwithBrainly

5 0
4 years ago
What gadgets are charge coupled devices used in?
Alinara [238K]

Answer:

They are used in imaging application gadgets such as video cameras,TV, surveillance cameras and document scanners

Explanation:

A charge couple device (CCDs) are highly capable in imagery detector.Its common application is in video and digital imaging.The quality of a charge couple device is determined by factors such as the dynamic range, dark charge level and the quantum efficiency.These devices serve the purpose of detecting optical images though some are installed with applications for data storage.

5 0
3 years ago
A compressed-air drill requires an air supply of 0.25 kg/s at gauge pressure of 650 kPa at the drill. The hose from the air comp
Klio2033 [76]

Answer:

L = 46.35 m

Explanation:

GIVEN DATA

\dot m  = 0.25 kg/s

D = 40 mm

P_1 = 690 kPa

P_2 = 650 kPa

T_1 = 40° = 313 K

head loss equation

[\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m

whereh_l = \frac{ flv^2}{2D}

h_m minor loss

density is constant

v_1 = v_2

head is same so,z_1 = z_2

curvature is constant so\alpha = constant

neglecting minor losses

\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}

we know\dot m is given as= \rho VA

\rho =\frac{P_1}{RT_1}

\rho =\frac{690 *10^3}{287*313} = 7.68 kg/m3

therefore

v = \frac{\dot m}{\rho A}

V =\frac{0.25}{7.68 \frac{\pi}{4} *(40*10^{-3})^2}

V = 25.90 m/s

Re = \frac{\rho VD}{\mu}

for T = 40 Degree, \mu = 1.91*10^{-5}

Re =\frac{7.68*25.90*40*10^{-3}}{1.91*10^{-5}}

Re = 4.16*10^5 > 2300 therefore turbulent flow

for Re =4.16*10^5 , f = 0.0134

Therefore

\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}

L = \frac{(P_1-P_2) 2D}{\rho f v^2}

L =\frac{(690-650)*`10^3* 2*40*10^{-3}}{7.68*0.0134*25.90^2}

L = 46.35 m

5 0
3 years ago
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