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sdas [7]
2 years ago
14

Please help for our stem project :(

Engineering
1 answer:
Darina [25.2K]2 years ago
4 0

Answer:

Do you have some a picture

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

Otto cycle T-S diagram

T₂ = 288.706*6.25^{0.393} = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

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

T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0
3 years ago
What do you guys like in engineering
Drupady [299]

Answer:

building lol and actually workin

Explanation:

3 0
3 years ago
Read 2 more answers
For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho
ad-work [718]

Answer:d

Explanation:

Given

Temperature=200^{\circ}\approc 473 K

Also \gamma for air=1.4

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.=\frac{V}{\sqrt{\gamma RT}}

(a)1000 km/h\approx 277.78 m/s

Mach no.=\frac{277.78}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.63

(b)500 km/h\approx 138.89 m/s

Mach no.=\frac{138.89}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.31

(c)2000 km/h\approx 555.55 m/s

Mach no.=\frac{555.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=1.27

(d)200 km/h\approx 55.55 m/s

Mach no.=\frac{55.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0
3 years ago
A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the
alekssr [168]

Answer:

answer

Explanation:

4 0
2 years ago
A list is sorted in ascending order if it is empty or each item except the last one is less than or equal to its successor. HERE
Free_Kalibri [48]

Using the knowledge of computational language in python it is possible to write a code that writes a list and defines the arrange.

<h3>Writing code in python:</h3>

<em>def isSorted(lyst):</em>

<em>if len(lyst) >= 0 and len(lyst) < 2:</em>

<em>return True</em>

<em>else:</em>

<em>for i in range(len(lyst)-1):</em>

<em>if lyst[i] > lyst[i+1]:</em>

<em>return False</em>

<em>return True</em>

<em>def main():</em>

<em>lyst = []</em>

<em>print(isSorted(lyst))</em>

<em>lyst = [1]</em>

<em>print(isSorted(lyst))</em>

<em>lyst = list(range(10))</em>

<em>print(isSorted(lyst))</em>

<em>lyst[9] = 3</em>

<em>print(isSorted(lyst))</em>

<em>main()</em>

See more about python at brainly.com/question/18502436

#SPJ1

7 0
2 years ago
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