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meriva
3 years ago
15

Help..

Engineering
1 answer:
iren [92.7K]3 years ago
5 0

Answer:

7

Explanation:

A quotient is the answer to a division.For example,the quotient of 10 is 2 and 5 because 5÷10=2.

You might be interested in
Carbon dioxide (CO2) expands isothermally at steady state with no irreversibilities through a turbine from 10 bar, 500 K to 2 ba
xenn [34]

Answer:

The answer to the question above is = 152.02 KJ/Kg

Explanation:

Given:

Temperature at first state, (T1)= 500k

Temperature at second state, (T2)= 500k

The above explains an isothermal process as a thermodynamic process,in which the temperature of the system remains constant

Pressure at first state, (p1) = 10 bar

Pressure at second state, (p2) = 2 bar

The heat transfer=

Qrev/m= T x [s(T2) - s(T1) - R ㏑ (p2 ÷ p1)]

Isothermal means the temperature does not change, while Expansion means the volume has increased.

For the internal isothermal process:

Qrev/m=  T x [- R ㏑ (p2 ÷ p1)]

= 500 x  - (8.314 ÷ 44.01) x In (2 ÷ 10) = 152.02 KJ/Kg

Energy equation at turbine is for the internally reversible isothermal process is:

Q-w = m [( (V2²- V1²) ÷ 2) + g ( Z2 - Z1)]

where w= the most efficient work possible in  J

Neglecting the effect of both potential and kinetic energy

(w/ m) = Qrev/ m

= 152.02 KJ/Kg

4 0
3 years ago
g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th
xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }

\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }

if Gl≈El(l,5800)

\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt= 2*5800=11600 um-K, at this value, F=0.941

\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77

The hemispherical emissivity is equal to:

E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt=2*333=666 K, at this value, F=0

E=0+(1-0.7)(1)=0.3

The hemispherical absorptivity is equal to:

qe=EoTs^{4}+h(Ts-T\alpha  )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}

3 0
3 years ago
In highways the far left lane is usually the _____
Ivan
Fastest


(Known as the fast lane)
8 0
3 years ago
Read 2 more answers
A plot of land is an irregular trangle with a base of 122 feet and a height of 47 feet what is the area of the plot?
Reika [66]

Answer:

150 is the area

Explanation:

3 0
3 years ago
Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0
3 years ago
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