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aksik [14]
3 years ago
8

determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004

Engineering
1 answer:
Afina-wow [57]3 years ago
6 0

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 ,  To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

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a sprue is 12 in long and has a diameter of 5 in at the top. The molten metal level in the pouring basing is taken to be 3 in fr
vampirchik [111]

Answer:

See explaination

Explanation:

We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.

See the attached file for detailed solution of the given problem.

8 0
3 years ago
Read 2 more answers
What is a shearing stress? Is there a force resulting from two solids in contact to which is it similar?
Luba_88 [7]

Answer:

Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.

Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.

Mathematically in a plane AB the shearing stresses are given by

\tau =\frac{Fcos(\theta )}{A}

Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.  

7 0
4 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
Ultimate tensile strength is: (a) The stress at 0.2% strain (b) The stress at the onset of plastic deformation (c) The stress at
MissTica

Answer:

By definition the ultimate tensile strength is the maximum stress in the stress-strain deformation. The stress at 0.2% strain, the stress at the onset of plastic deformation, the stress at the end of the elastic deformation and the stress at the fracture correspond, by definition, to other points of the stress-strain curve.

Explanation:

4 0
3 years ago
A single Gates Rubber V belt is to be selected to deliver engine power to the wheel-drive transmission of a riding tractor. A 5-
beks73 [17]

Answer:

Hello there, please check step by step explanations to get answers.

Explanation:

Given that:

5-hp single-cylinder engine is used. At most, the belt transmits 60 percent of this power. The driving sheave has a diameter of 6.2 in. and the driven, 12.0 in. The belt selected should be as close to 92 in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt, and specify it using the standard designation.

See attached documents for clearity and step by step procedure to answer

7 0
4 years ago
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