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Andru [333]
3 years ago
15

These are the most widely used tools and most often abuse tool​

Engineering
2 answers:
Mars2501 [29]3 years ago
4 0
Where is the picture in this problem? How am I supposed to answer if I can’t see any footage taken from this problem.
mojhsa [17]3 years ago
3 0
Probably a screw driver
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Who is web developer?????​<br>plez answer..
vredina [299]

Web developers design and create websites. They are responsible for the look of the site. They are also responsible for the site's technical aspects, such as its performance and capacity, which are measures of a website's speed and how much traffic the site can handle. In addition, web developers may create content for the site

8 0
3 years ago
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Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.
Rufina [12.5K]

Answer:

a) \dot m = 16.168\,\frac{kg}{s}, b) v_{out} = 680.590\,\frac{m}{s}, c) \dot W_{out} = 18276.307\,kW

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0

Energy Balance

-q_{loss} - w_{out} + h_{in} - h_{out} = 0

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

\nu_{in} = 0.055665\,\frac{m^{3}}{kg}

h_{in} = 3650.6\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mix)

\nu_{out} = 5.89328\,\frac{m^{3}}{kg}

h_{out} = 2500.2\,\frac{kJ}{kg}

a) The mass flow rate of the steam is:

\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

\dot m = 16.168\,\frac{kg}{s}

b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

v_{out} = 680.590\,\frac{m}{s}

c) The power output of the steam turbine is:

\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

\dot W_{out} = 18276.307\,kW

6 0
2 years ago
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati
snow_tiger [21]

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

8 0
3 years ago
A disk of radius 2.1 cm has a surface charge density of 5.6 µC/m2 on its upper face. What is the magnitude of the electric field
Assoli18 [71]

Answer:

=6.3*10^3 N/C

Explanation:

solution:

from this below equation (1)

E=σ/2εo(1-\frac{z}{\sqrt{z^2-R^2} } )...........(1)

we obtain:

=5.6*10^-6 \frac{c}{m^2} /2(8.85*10^-12\frac{c^2}{N.m^2} ).(1-\frac{9.5 cm}{\sqrt{9.5^2-2.1^2} } )

=6.3*10^3 N/C

8 0
3 years ago
Fire extinguishers need to be checked _______ to make sure they are working properly.
Zepler [3.9K]

Answer:

Answer D

Explanation:

They need to be checked monthly

3 0
2 years ago
Read 2 more answers
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