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Iteru [2.4K]
2 years ago
5

Incremental software development could be very effectively used for customers who do not have a clear idea about the systems nee

ded for their operations. Discuss..
​
Engineering
1 answer:
avanturin [10]2 years ago
5 0

<u>Software Development and Client Needs</u>

In Incremental method of software development customers who do not have a basic idea of the development process are being carried along on like other methods that will relegate them to the background until a product is ready.

With this model and structure in place, when softwares/ products are built from several stages e.g prototype, testing, and when new features are added customers are always carried along with their valuable feedback and suggested greatly considered to achieve the customers satisfactions

This model will work well for the customers/clients who does not have a clear idea on the systems needed for their operations.

In summary the incremental model combines features from the waterfall and prototyping model.

For more information on soft ware development process kindly visit

brainly.com/question/20369682

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Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

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Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

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Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

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Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

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B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

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W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

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Answer:

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solving the above equation,

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