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2 years ago

What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?

Engineering

1 answer:

irga5000 [103]2 years ago

7 0

Answer:

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency increase.

Now from T-S diagram

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

$W_{net}=(h_3-h_4)-(h_2-h_1)$

$Q_{in}=h_3-h_5$

Due to generation $(h_5-h_2)$ amount of energy has been saved.

$Q_{generation}=Q_{saved}$

So efficiency of cycle $\eta =\frac{W_{net}}{Q_{in}}$

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Effectiveness of re-generator

$\varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}$

So the efficiency of regenerative cycle

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

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3 years ago

What motivated software engineers to move from the waterfall model to the incremental or spiral model

scoray [572]

Answer:

1. They needed to develop multiple components in software programs.

2. The ability to overlap the development to be more evolutionary in nature.

3. The need to be more risk-averse or the unwillingness to take risks led to the use of a spiral model.

Explanation:

Software development life cycle (SDLC) can be defined as a strategic process or methodology that defines the key steps or stages for creating and implementing high quality software applications.

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<em>What motivated software engineers to move from the waterfall model to the incremental or spiral model is actually due to the following fact;</em>

They needed to develop multiple components in software programs.

The ability to overlap the development to be more evolutionary in nature.

The need to be more risk-averse or the unwillingness to take risks led to the use of a spiral model.

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3 years ago

Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo

AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

= ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

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2 years ago

Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at l

Tema [17]

Answer:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

$m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\$

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

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2 years ago