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Lilit [14]
1 year ago
11

Why not just put all the set up steps within each step? it is because we want to keep our code __ ? (3 letters)

Engineering
1 answer:
andrezito [222]1 year ago
8 0

Why not simply include each step's setup instructions within it? is it due to our desire to maintain DRY (Don't Repeat Yourself) code?

<h3>What is the purpose of unit testing?</h3>

Program testing is known as "unit testing" involves testing individual software components. When developing an application, unit testing is done on the software product. An individual component could be a technique or a specific function.

Unit testing's primary goal is to separate written code for testing to see if it functions as intended. Unit testing is a crucial stage in the development process because, when done properly, it can aid in finding early code issues that could be more challenging to identify in subsequent testing phases.

The core of the testing process consists of unit testing and functional testing. The primary distinction between the two is that during the development cycle, the developer conducts unit testing. The tester does functional testing at the system testing level.

To learn more about unit testing refers to:

brainly.com/question/22900395

#SPJ4

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Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

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3 years ago
Sort the following alphabets using MergeSort and give required steps. [2 Marks]
OlgaM077 [116]

Answer:

I'm afraid i can't visualise it to you but visit the site below to help you out <3

Explanation:

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