A well-designed product will increase?

Consider a Carnot heat-engine cycle executed in a closed system using 0.028 kg of steam as the working fluid. It is known that t

Paha777 [63]

Answer:

T=138 °C

Explanation:

Given that

m = 0.028 kg

Net work output W= 60 KJ

T₂=2T₁

As we know that efficiency of Carnot heat engine given as

$\eta=1-\dfrac{T_1}{T_2}$

$\eta=1-\dfrac{T_1}{2T_1}$

η = 0.5

We know that

$\eta=\dfrac{W}{Q_a}$

Qa=heat addition

W= net work output

$\eta=\dfrac{W}{Q_a}$

$0.5=\dfrac{60}{Q_a}$

Qa= 120 KJ

From first law

Qa= W+ Qr

Qr= 120 - 60

Qr= 60 KJ

Qr Is the heat rejection.

Heat rejection per unit mass

Qr=60 / 0.028 = 2142.85 KJ/kg

Qr= 2142.85 KJ/kg

Temperature at which latent heat of steam is 2142.85 KJ/kg will be our answer.

T=138 °C

The temperature corresponding to 2142.85 KJ/kg will be 138 °C.

T=138 °C

8 0

3 years ago

What is clearance? What is backlash? What is interference? Explain briefly.

Anton [14]

Explanation:

Clearance:

For easy matching and dis matching of hole and shaft we use size of hole little bit more than the size of shaft and this difference in size is called clearance.

Backlash:

It is the clearance between the two mating gear to avoids failure of gears.Actually when temperature of gears increases then at the same time the size of gear also increases ,due this there is a possibility foe jamming of gears so to avoids this backlash is provides.

Interference:

When two gears are matting then addendum of one gear inters into the deddendum of another gear and due to this gears get jam .This phenomenon is called interference.

5 0

3 years ago

Find the phasor form of:

mixer [17]

Answer:

$I=24.598\angle 50.377$

Explanation:

A tension or current expressed in cosine form and with a positive sign can be converted directly into a phasor. This is done by indicating the tension and the offset angle:

$Acos(10\omega t +\phi)=A\angle \phi$

So:

$i(t)=10cos(10t+63)+15cos(10t-42)=10\angle 63 + 15\angle42=I$

You can sum the phasors simply using a calculator, however, let's do it manually:

Let's find the rectangular form of each phasor using the next formulas:

$A=\sqrt{a^2+b^2} \\\phi=arctan(\frac{b}{a})$

For $10\angle 63$

$63=arctan(\frac{b}{a} )\\\\tan(63)=\frac{b}{a} \\\\b=a*tan(63)$

$10=\sqrt{a^2+(a*tan(63))^2} \\\\10^2=a^2+a^2*(tan(63))^2\\\\Solving\hspace{3}for\hspace{3}a\\\\a=\sqrt{\frac{100}{1+tan(63)^2} } =4.539904997\\\\and\hspace{3}b\\b=\sqrt{100-a^2} =8.910065242$

So:

$Z_1=a+bj=4.539904997+8.910065242j$

For $15\angle 42$

$42=arctan(\frac{b_2}{a_2} )\\\\tan(42)=\frac{b_2}{a_2} \\\\b_2=a_2*tan(42)$

$15=\sqrt{a_2^2+(a_2*tan(42))^2} \\\\15^2=a_2^2+a_2^2*(tan(42))^2\\\\Solving\hspace{3}for\hspace{3}a_2\\\\a_2=\sqrt{\frac{225}{1+tan(42)^2} } =11.14717238\\\\and\hspace{3}b_2\\\\b_2=\sqrt{225-a^2} =10.0369591$

So:

$Z_2=a_2+b_2j=11.14717238+10.0369591j$

Hence:

$Z_T=Z_1+Z_2=(4.539904997+11.14717238)+(8.910065242+10.0369591)j\\Z_T=15.68707738+18.94702434j$

Finally:

$I=\sqrt{15.68707738^2+18.94702434^2} \angle arctan (\frac{18.94702434}{15.68707738} )=24.598\angle 50.377$

7 0

3 years ago

I'll give a free brainliest

spayn [35]

All I need is one more.
Thx!!
<333333

5 0

3 years ago

Read 2 more answers

A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re

Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

A.

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

B.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

4 0

3 years ago

A well-designed product will increase?​

A well-designed product will increase?