All categories

3 years ago

She spent her time reading stories about people who fought in the Revolution.

Engineering

1 answer:

Sonbull [250]3 years ago

4 0

She spent her time reading stories about people who fought in the Revolution.

You might be interested in

Where did fossil fuels come from

Levart [38]

Any fuel such as coal, petroleum (crude oil) or natural gas that has developed within the Earth over millions of years from the decayed remains of bacteria, plants or animals

6 0

3 years ago

Read 2 more answers

An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat

IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature $T_L$ = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature $T_H$ = 99° F = ( 99 + 460 )R = 559 R

Cooling Load $Q_L$ = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

Coefficient of performance of the air conditioner;

$COP_{air-condition$ = Cooling Load $Q_L$ / power P

we substitute

$COP_{air-condition$ = 30000 Btu/h / 8396.619 Btu/h

$COP_{air-condition$ = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

- Power input required ( in hp )

$Q_L$ / $P_{required$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

30000 Btu/h / $P_{required$ = 530 R / ( 559 R - 530 R )

30000 Btu/h / $P_{required$ = 530 R / 29 R

we solve for $P_{required$

$P_{required$ = ( 30000 Btu/h × 29 R ) / 530 R

$P_{required$ = ( 870000 Btu/h / 530 )

$P_{required$ = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

$P_{required$ = ( 1641.5094 / 2544.43 ) hp

$P_{required$ = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

$COP_{rev-air-condition$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

$COP_{rev-air-condition$ = 530 R / ( 559 R - 530 R )

$COP_{rev-air-condition$ = 530 R / 29 R

$COP_{rev-air-condition$ = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0

3 years ago

A pressure gage and a manometer are connected to a compressed air tank to measure its pressure. If the reading on the pressure g

mixer [17]

Answer:

h=1.122652m

Explanation:

Assuming density of air = 1.2kg/m³

the differential pressure is given by:

$h^{i} =h(\frac{density of manometer}{density of flowing air}-1)\\h^{i} =h(\frac{1000}{1.2}-1)\\ h^{i}=832.33h...(1)\\\\but\\ h^{i} =\frac{change in pressure}{air density*g} \\\\h^{i} =\frac{11*10^3}{1.2*9.81}\\\\h^{i}= 934.42...(2)\\\\equating, \\\\934.42=832.33h\\\\h=1.122652m$

7 0

4 years ago

On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch

rewona [7]

Answer:

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

Explanation:

Distance of earth from sun = $R_{2} = 147.4 \times 106 Km$

Spacecraft perihelion = $R_{2} = 120\times106Km$

gravitational parameters are now given as

$\mu_{sun} = 132.7\times 10^{9}$

$\mu_{earth} = 398600$

radius of earth = 6378 Km

Heliocentric spacecraft velocity at earth sphere of influence =

$V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }$

$V_{D}^{v} =28.43\frac{km}{s}$

Heliocentric velocity of earth = $v_{earth} = 30.06\frac{km}{sec}$

$V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}$

assume

$r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km$

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

$v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}$

geometric space craft velocity in its circular parking orbit

$v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } } = 7.784 \frac{km}{s}$

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

7 0

4 years ago

Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressur

Veseljchak [2.6K]

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

Initial temperature T1 = 27°C

Initial pressure P1 = 100 kPa

final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

Total work done W is determined where there is volume change which from point 2 to 3.

W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

7 0

4 years ago