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Rainbow [258]
3 years ago
13

1. Looking at the case study provided under the Companion Material section, what is the main problem that is addressed in this c

ase study? Maintenance workers painted over the pipes. Engineers lost all of their work because of the flood. The water pipes broke and flooded the space center. There was no paperwork for the maintenance work done on the pipes.
Engineering
1 answer:
Fiesta28 [93]3 years ago
6 0

Incomplete question. However, I provided information that could assist you in identifying the main problem or issue addressed in any case study.

Explanation:

First, note that a case study is simply a learning aid that allows one to learn from a real-life scenario.

To determine the main problems of a case study one needs to:

  • Read the case as many times as possible to become familiar with the message been expressed. For example,<em> by highlighting or underlining the most important facts </em>it can help you to discover the main problem or issue.
  • Check for any facts provided in the case study, by so doing you can identify the most important problems.

Thus, by taking these few steps you may be able to determine the main problem in that case study.

You might be interested in
Air is drawn from the atmosphere into a turbomachine. At the exit, conditions are 500 kPa (gage) and 130oC. The exit speed is 10
finlep [7]

Answer:

P=- 88.41 KW

Negative sign indicates that power is given to the system.

Explanation:

Given that

P₂=500 KPa

T₂=130°C

V₂=100 m/s

mass flow rate ,m= 0.8 kg/s

Lets take inlet condition for air

T₁=25°C

P₁=100 KPa

V₁=0 m/s

We know that

Heat capacity for air Cp=1.005 KJ/kg.k

We know that for air change in enthalpy only depends only on temperature

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}+W

1.005\times 298+\dfrac{0^2}{2000}=1.005\times 403+\dfrac{100^2}{2000}+W

W=1.005\times 298-1.005\times 403-\dfrac{100^2}{2000}

W=-110.52 KJ/kg

Shaft power P = m .W

P = -110.52 x 0.8

P=- 88.41 KW

Negative sign indicates that power is given to the system.

 

3 0
4 years ago
The C language allows developers to pass functions as parameters to other functions. Provide a declaration for a function named
MariettaO [177]

Answer:

answer is given below

Explanation:

 Declaration for garnet  

  • void garnet(char,int (*f)(int));
  • void - garnet return nothing

garnet accepts two parameters

  1. 'Char'
  2. function

Function as a parameter

  • int (*f) (int))

and here

first int refers return type of function

  • (*f)  function as pointer variable

second int

  • arguments that accept by inner function- function as a parameter

so here Garnet is a function that accepts two parameters. One parameter that contains the character and another parameter that takes an argument (here I am taking an integer) and returns an integer. Function The function returns nothing. Zero is mentioned here.

5 0
4 years ago
5.<br> Which of the following is considered ideal conditions?
dlinn [17]

Answer:

umm what

Explanation:

5 0
3 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

4 0
3 years ago
What did the ancient Greeks use simple machines for?
nalin [4]
Native Americans who used spears to hunt were using wedges. In the third century BC, the Greek scientist Archimedes invented a way to lift water, called the Archimedes screw. It was used to water crops and to move water out of ships.
6 0
3 years ago
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