Note that common skills are listed toward the top, and less common skills are listed toward the bottom.

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

slamgirl [31]

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%

7 0

2 years ago

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$

but Tinitial = 0

$\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001$

$\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}$

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0

2 years ago

If the density of states function in the conduction band of a particular semiconductor is a constant equal to K, derive the expr

s2008m [1.1K]

Answer:

full details of the answer is attached

5 0

2 years ago

As the driver of the red SUV, what are your options for navigating this roundabout?

IgorC [24]

Your options for navigating this roundabout would include driving in a counter-clockwise direction and then making your exit to the right lane.

<h3>What is a roundabout? </h3>

A roundabout is a circular intersection that make junctions or intersections of roads safer, because it is designed and developed to physically direct drivers and pedestrians to move in a counter-clockwise direction.

In this scenario, your options for navigating this roundabout as the driver of the red SUV would include driving in a counter-clockwise direction and then making your exit to the right lane.

Read more on roundabout here: brainly.com/question/22580476

#SPJ1

3 0

1 year ago

A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti

Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = $\frac{0.03194}{0.315}$

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m

here h is heat required and m is mass flow rate

H = 2680.36 × 0.10139

H = 271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = $\frac{H}{0.90}$

net boiler heat = $\frac{271.75}{0.90}$

net boiler heat = 301.94 kW

5 0

2 years ago