Circulates air through the engine to keep engine components cool

What steps might one take to make a decision or solve a problem

marusya05 [52]

you must think and plan out what you want to do

8 0

3 years ago

simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl

Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

a) Air temperature at turbine exit

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) $(\frac{783.05-778.18}{800.13-778.18} )$ = 764.45K

b) The net work output

first we determine the actual work input to compressor

Wc = h2 - h1 ( calculated values )

= 626.57 - 295.17 = 331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4 ( calculated values )

= 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

= 541.88 - 331.4 = 210.48 kJ/kg

c) determine thermal efficiency

лth = Wnet / qin ------ ( 1 )

where ; qin = h3 - h2

equation 1 becomes

лth = Wnet / ( h3 - h2 )

= 210.48 / ( 1324.93 - 626.57 )

= 0.3014 = 30.14%

6 0

3 years ago

Help please its due today will mark you brainliest

Tems11 [23]

Answer:

launch- The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit

powered ascent-The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit

coasting flight-

When the rocket runs out of fuel, it enters a coasting flight. The vehicle slows down under the action of the weight and drag since there is no longer any thrust present. The rocket eventually reaches some maximum altitude which you can measure using some simple length and angle measurements and trigonometry.

ejection charge-At the end of the delay charge, an ejection charge is ignited which pressurizes the body tube, blows the nose cap off, and deploys the parachute. The rocket then begins a slow descent under parachute to a recovery. The forces at work here are the weight of the vehicle and the drag of the parachute.

slow decent- slow downs (i guess)

recovery-A recovery period is typically characterized by abnormally high levels of growth in real gross domestic product, employment, corporate profits, and other indicators. This is a turning point from contraction to expansion and often results in an increase in consumer confidence

Explanation:

6 0

4 years ago

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

jok3333 [9.3K]

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$ .

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

We know that

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where Q is volume flow rate

L is length of tube

$d_i$ is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$ .

6 0

3 years ago

Increase vehicle control through lane position and speed adjustments is not necessary risk management, just common sense.

Vesnalui [34]

True

If it was wrong I’m sorry

6 0

3 years ago

Circulates air through the engine to keep engine components cool​

Circulates air through the engine to keep engine components cool