12. Never spray brakes with a high-pressure stream of water or air because it could blow asbestos fibers into the air.

Consider a Pitot static tube mounted on the nose of an experimental airplane. A Pitot tube measures the total pressure at the ti

kotykmax [81]

Answer:

M∞ = 0.53

M∞ = 1.5

M∞ = 3.1

Explanation:

Find: For each case the free stream Mach number.

-Pitot pressure=1.22×10^5N/m2 , static pressure=1.01 × 10^5N/m2 .

Solution:

- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:

P = 1.01 × 10^5 .. static pressure

Po = 1.22×10^5 ... pitot pressure ( hydrodynamic )

- Take the ratio:

P / Po = (1.01 × 10^5) / (1.22×10^5) = 0.8264.

- Use Table A.13 and look up the ratio P/Po = 0.8264 for Mach number M∞.

M∞ = 0.53

Find:

-Pitot pressure=7222 lb/ft^2 , static pressure=2116 lb/ft^2

Solution:

- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:

P = 2116 .. static pressure

Po = 7222 ... pitot pressure ( hydrodynamic )

- Take the ratio:

P / Po = (2116) / (7222) = 0.2930.

- However, since this is supersonic, a normal shock sits in front of the Pitot tube. Hence, Po is now the total pressure behind a normal shock wave. Thus, we have to use Table A.14.

P1 = 2116 .. static pressure

Po2 = 7222 ... pitot pressure ( hydrodynamic )

- Take the ratio:

Po2 / P1 = (7222) / (2116) = 3.412.

- Use Table A.14 and look up the ratio Po2/P1 = 3.412 for Mach number M∞.

M∞ = 1.5

Find:

-Pitot pressure=13197 lb/f^t2 , static pressure=1020 lb/ft^2

Solution:

- The free stream Mach number is a function of static to hydrodynamic pressures. So for this case we have:

P = 1020 .. static pressure

Po = 13197 ... pitot pressure ( hydrodynamic )

- Take the ratio:

P / Po = (1020) / (13197) = 0.0772.

- Again, since this is supersonic, a normal shock sits in front of the Pitot tube. Hence, Po is now the total pressure behind a normal shock wave. Thus, we have to use Table A.14.

P1 = 1020 .. static pressure

Po2 = 13197 ... pitot pressure ( hydrodynamic )

- Take the ratio:

Po2 / P1 = (13197) / (1020) = 12.85.

- Use Table A.14 and look up the ratio Po2/P1 = 12.85 for Mach number M∞.

M∞ = 3.1

3 0

3 years ago

When all conductors of the same alternating-current circuit are in the same raceway, the magnetic flux around each conductor flu

LuckyWell [14K]

Answer:

B. Cancels out

8 0

4 years ago

Read 2 more answers

A train travels 650 meters in 25 seconds. What is the train's velocity?

frosja888 [35]

The train is traveling 26 meters A second .

3 0

3 years ago

Which of these cars traveled faster during time interval <br> please show solution

Mazyrski [523]

Answer:

I think D is correct

Explanation:

C is decreasing function, probably worst

A is arctan -> in radian, the rate of increasing is very slow-> second worst

B(14) = ln(9*14) = 4.8

D(14) = sqrt(8+14^2)=14.2

3 0

3 years ago

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

b)

- 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

a)

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

b)

- 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