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3 years ago

Which of the following describes the most direct and attainable way to reduce pollution in the environment?

Engineering

2 answers:

vampirchik [111]3 years ago

8 0

Answer:

making practical decisions about using natural resources and energy

Explanation:

kirill115 [55]3 years ago

5 0

Answer:

i think its A. increasing research to find alternative natural resources for the future

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____ grinders are used to grind diameters, shoulders, and faces much like the lathe is used for turning, facing, and boring oper

skelet666 [1.2K]

Answer:

Cylindrical

Explanation:

A cylindrical grinder is a tool for shaping the exterior of an item. Although cylindrical grinders may produce a wide range of forms, the item must have a central axis of rotation. Shapes such as cylinders, ellipses, cams, and crankshafts are examples of this. Cylindrical grinding machines are specialized grinding machines that are used to process cylinders, rods, and similar workpieces. The cylinders revolve in one direction between two centers, while the grinding wheel or wheels are close together and rotate in the other direction.

8 0

2 years ago

Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressur

Paha777 [63]

Answer:

Explanation:

Answer:

Explanation:

Answer:

Explanation:

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.

(a). here we are asked to determine the Temperature and Pressure.

Given that the properties of Air;

ha = 230.02 KJ/Kg

Ta = 230 K

Pra = 0.5477

From the energy balance equation for a diffuser;

ha + Va²/2 = h₁ + V₁²/2

h₁ = ha + Va²/2 (where V₁²/2 = 0)

h₁ = 230.02 + 220²/2 ˣ 1/10³

h₁ = 254.22 KJ/Kg

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg

from this we have;

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]

Pr₁ = 0.77759

therefore T₁ = 254.15K

P₁ = (Pr₁/Pra)Pa

= 0.77759/0.5477 ˣ 26

P₁ = 36.91 kPa

now we calculate Pr₂

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349

⇒ now we obtain properties of air at

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg

calculating the enthalpy of air at state 2

ηc = h₁ - h₂ / h₁ - h₂

0.85 = 254.22 - 505.387 / 254.22 - h₂

h₂ = 549.71 KJ/Kg

to obtain the properties of air at h₂ = 549.71 KJ/Kg

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃ i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄ + 0 = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

6 0

3 years ago

Match the word with the definition:

aksik [14]

1. Renewable Resources = (Renewable means you can keep making it) = resources that can be replenished (such as trees)

2. Nonrenewable Resources = ( Nonrenewable means it can't be made once it is used up) = resources that are gone once they are used (such as fossil fuels)

3. Producer = ( produces something) = person who makes goods or provides services

4. Consumer = ( uses something) = person whose wants are satisfied by using goods and services

5. Allocate = ( put someplace) = distribute

6. Choice = option

7 0

3 years ago

Read 2 more answers

Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin

Vladimir79 [104]

Answer:

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 = 1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 = 1.013 μm

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 = 0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.243 μm

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 = 0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.608 μm

6 0

3 years ago

⦁ An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and

Anon25 [30]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx

8 0

3 years ago