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allsm [11]
3 years ago
13

The displacement volume of an internal combustion engine is 2.2 liters. The processes within each cylinder of the engine are mod

eled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by p1 = 1.6 bar, T1 = 325 K and V1 = 2.36 liters.
1) If the cycle is executed 2100 times per min, determine:
a) the compression ratio.
b) the net work per cycle, in kJ.
c) the maximum temperature, in K.
d) the power developed by the engine, in kW.
e) and the thermal efficiency.

Engineering
1 answer:
soldier1979 [14.2K]3 years ago
5 0

Answer:

A) 14.75

B) 3.36Kj

C) 2384.2k

D) 117.6kW

E) 57.69%

Explanation:

Attached is the full solutions.

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musickatia [10]

Answer:

The correct option is d ( Neither A nor B)

Explanation:

Technician A made 2 mistakes in his statement.Firstly the tire is self supporting not self sealing.

Secondly, this tire does not provide permanent sealing of punctured area option a is incorrect.

This self-supporting tire after being affected with complete air leakage can temporarily bear the load of the car and avoid rolling over a distance of 80 km at a maximum speed of 55 mph. Here is what technician B suggested incorrectly as the tire after being.Here the technician B suggested incorrectly as the tire after being affected with puncture can not travel at any speed so option B is wrong

Since option a and b are incorrect and c is invalid.

4 0
3 years ago
A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of
andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}

By putting the value, we get

⇒       =\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}

⇒       =3.7258 \ l

hence,

The heat loss will be:

⇒ Q=SF\times K(T_2-T_1)

       =3.7258\times 1\times 1.8\times (500-300)

       =3.7258\times 1.8\times (200)

       =1341.288 \ W/m

3 0
3 years ago
The rolling process is governed by the frictional force between the rollers and the workpiece. The frictional force at the entra
adell [148]

Answer:

b)false

Explanation:

Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling  is a metal forming process.

We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.

So given statement is wrong.

3 0
3 years ago
Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician
mariarad [96]

Answer:

Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

Explanation:

He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.

3 0
3 years ago
What is the composition of the two phases that form when a stream of 40% A, 39% B, and 21% C separates into two phases? Label th
irga5000 [103]

Answer:

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At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.

8 0
3 years ago
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