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3 years ago

Harlin is designing a new car engine that does not create pollution. Which technological design factor is probably the most

Engineering

1 answer:

kari74 [83]3 years ago

8 0

Answer: environmental impact

Explanation:

From the question, we are informed that Harlin is designing a new car engine that does not create pollution.

The technological design factor which is probably the most important for the design is the impact on the environment.

Pollution has a negative effect on the environment a d doing this shows that emphasis is placed on how the environmental issue of pollution can be tackled.

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The welding method that requires the operator to observe and only make corrections is

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Which regulations are related to guard rail height and dimensions and uniformity of stairs?

galina1969 [7]

Answer:

structural safety

Explanation:

Guards protecting floor surfaces must be 36 inches in height, while guards for stairs must be 34 inches in height measured vertically from the tread nosing. A guard may also serve as the required handrail (34 to 38 inches high) provided the top rail meets the requirements for grip size.

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An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces

Allushta [10]

Answer:

(a) The amount of heat transferred to the air, $q_{out}$ is 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is 308.07 kJ/kg

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

$c_p$ = 1.005 kJ/(kg·K), $c_v$ = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

$T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}} \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K$

From;

$\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }$

We have;

$p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa$

Process 2 to 3 is reversible constant volume heating, therefore;

$\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}$

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

$T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86 \times 729.21}{2190.43} = 1458.42 \, K$

Process 3 to 4 is isentropic expansion, therefore;

$T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}} \right )^{k-1}$

$1458.42= T_{4} \times \left (9.2 \right )^{0.4}$

$T_4 = \dfrac{1458.42}{(9.2)^{0.4}} = 600.3 \, K$

$q_{out} = m \times c_v \times (T_4 - T_1) = 0.718 \times (600.3 - 300.15) = 215.5077 \, kJ/kg$

The amount of heat transferred to the air, $q_{out}$ = 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is found as follows;

$W_{net} = q_{in} - q_{out}$

$q_{in} = m \times c_v \times (T_3 - T_2) = 0.718 \times (1458.42 - 729.21) = 523.574 \, kJ/kg$

$\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg$

$\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100= \dfrac{308.07}{523.574} \times 100= 58.8\%$

(d) From the general gas equation, we have;

$V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg$

The Mean Effective Pressure, MEP, is given as follows;

$MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa$

The Mean Effective Pressure, MEP = 393.209 kPa.

3 0

3 years ago

Which of the following sentences uses the active voice

MrRa [10]

Answer: Last week, Nate and I counted all the inventory.

Explanation: all other choices are passive voices

this sentence follows a clear subject + verb + object construct that's why it is an active voice. In fact, sentences constructed in the active voice add impact to your writing. but on the other hand With passive voice, the subject is acted upon by the verb.

Ape x

7 0

3 years ago

What limits the practical realization of higher efficiencies in the Otto cycle?

AlekseyPX

Answer:

We know that all petrol engines are works on Otto cycle.Otto cycle have four process out of four two are constant volume process and others two are isentropic processes.

There are lots of limitations for practical Otto cycle these are as follows

1.In practical cycle heat can not add at constant volume.

2.In practical cycle there is a gap between combustion of two set of fuel.

3.Lots of heat is dissipated by cylinders.

4.Valve opening and closing is not a sudden process it requires some time.

5.There is a limitations of cylinder material ,it means that temperature of cycle can not rise after a specified limit of material.

Due to these above limitations practical cycles have low efficiency as compare to ideal cycle.

5 0

3 years ago