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3 years ago

What is 39483048^349374*3948048/3i4u4

Engineering

1 answer:

Verizon [17]3 years ago

3 0

Answer:

1. $3.81813506×10^2^9$

2. $1.71479428×10^6^5$

3. $9.38483383×10^2^6$

4. $1.150847×10^2^9$

Explanation:

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The best saw for cutting miter joints is the

ZanzabumX [31]

Answer:

The best saw for cutting miter joints is the backsaw.

Add-on:

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6 0

3 years ago

Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati

Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0

3 years ago

Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new

Marina86 [1]

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

$C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$ where i=1 to N-1 and N=6

$C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$

A area of the polygon can be found by the following formula $A=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i})$ where i=1 to N-1

$A=\frac{1}{2}[ (x_{1} y_{2} -x_{2} y_{1})+ (x_{2} y_{3} -x_{3} y_{2})+(x_{3} y_{4} -x_{4} y_{3})+(x_{4} y_{5} -x_{5} y_{4})+(x_{5} y_{6} -x_{6} y_{5})]$

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

$C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(x_{2}+x_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(x_{3}+x_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(x_{4}+x_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(x_{5}+x_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{x} =15.36$

For Cy

$C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(y_{2}+y_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(y_{3}+y_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(y_{4}+y_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(y_{5}+y_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{y} =22.55$

So coordinates for the fire station should be (15.36,22.55)

5 0

3 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

$R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm$

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })$

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })$

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

$V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V$

$w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15} } } =3.35x10^{-5} cm=0.335um$

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

$Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm$

7 0

3 years ago

What does a peak flow meter allow you to assess?

Alex Ar [27]

Answer:

peak flow and any engineering considerations related thereto

Explanation:

It should be no surprise that a peak flow meter will report peak flow, sometimes with important maximum-value, time-constant, or bandwidth limitations. There are many engineering issues related to flow rates. A peak flow meter can allow you to assess those issues with respect to the flows actually encountered.

Peak flow can allow you to assess adequacy of flow and whether there may be blockages or impediments to flow that reduce peak levels below expected values. An appropriate peak flow meter can help you assess the length of time that peak flow can be maintained, and whether that delivers sufficient volume.

It can also allow you to assess whether appropriate accommodation is made for unexpectedly high flow rates. (Are buffers or overflow tanks of sufficient size? Is there adequate protection against possible erosion? Is there adequate support where flow changes direction?)

3 0

4 years ago