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Mariana [72]
3 years ago
10

Air enters a compressor operating at steady state at a pressure of 1 bar, a temperature of 290 K, and a velocity of 6 m/s throug

h an inlet with an area of 0.1 m^2. At the exit, the pressure is 7 bars, the temperature is 450 K, and the velocity is 2 m/s. Heat transfer from the compressor to the surroundings occurs at a rate of 180 KJ/min, Assuming an ideal gas, find the power input to the compressor in kW.
Engineering
1 answer:
Svetach [21]3 years ago
4 0

Answer:

See attachment

Explanation:

Download pdf
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4. A person is standing 4 miles east from his house. He travels 2 miles east. a. Calculate the initial displacement. b. Calculat
Pani-rosa [81]

Answer:

4) a) s = +4\,mi, b) \Delta s = 2\,mi, 5) a) s = -2\,mi, b) \Delta s = 10\,mi

Explanation:

4) Let assume that distance with respect to origin is positive at east, while distance is negative at west.

a) The initial displacement is:

s = +4\,mi

b) The distance travelled is:

\Delta s = 2\,mi

\Delta s = 2\,mi

5) a) The final displacement is:

s = -2\,mi

b) The total distance travelled is:

\Delta s = 2\,mi + 6\,mi + 2\,mi

\Delta s = 10\,mi

5 0
4 years ago
Thrust is managed to maintain IAS, and glide slope is being flown. What characteristics should be observed when a headwind shear
SVEN [57.7K]

The characteristics that can be observed when a headwind shears to be a constant tailwind are:

  1. Pitch attitude: decreases.
  2. Required thrust: increased then reduced.
  3. Vertical Speed: increases.
  4. Indicated airspeed (IAS): decreases, then increases to approach speed.

<h3>What is thrust?</h3>

Thrust can be defined a force that moves an aircraft or a flying machine through the air, especially in the direction of the motion.

<h3>The importance of thrust.</h3>

Generally, thrust is used in aeronautic engineering to achieve the following:

  • To overcome the weight of a rocket.
  • To overcome the drag of an aircraft.
  • To maintain indicated airspeed (IAS).
  • To maintain glide slope being flown at.

Some of the characteristics that can be observed when a headwind shears to be a constant tailwind are:

  1. Pitch attitude: decreases.
  2. Required thrust: increased then reduced.
  3. Vertical Speed: increases.
  4. Indicated airspeed (IAS): decreases, then increases to approach speed.

Read more on thrust here: brainly.com/question/20068220

4 0
3 years ago
Heater control valves can not leak coolant<br><br> True <br> False
vladimir1956 [14]

Heater control valves can not leak coolant

it is true

3 0
2 years ago
Read 2 more answers
If 20 kg of iron, initially at 12 °C, is added to 30 kg of water, initially at 90 °C, what would be the final temperature of the
rjkz [21]

Answer:

final temperature of the combined system T = 84.78°C

Explanation:

Given data

mass of iron ( m1 )   = 20 kg

temperature iron ( t1 ) =  12 °C

mass of water ( m2 ) = 30 kg

temperature of water ( t2 )   =  90 °C

To find out

final temperature of the combined system

solution

we know the energy requirement formula to rise the temp

energy = mass × specific heat  × change in temperature  

we combine both system so both energy will be added

and

we know specific heat of iron ( c1 ) = 0.450 kJ/kg

and specific heat of water ( c2 ) = 4.186 kJ/kg

4.186 joule/gram °C

now combine both energy

energy = mass, m1 × specific heat, c1  × change in temperature, T - t1 + mass, 2 × specific heat, c2  × change in temperature, T - t2

energy = 20 × 0.450  × T - 12  + 30 × 4.186 × T -90

(20)(0.45)(T−12)=(30)(4.186)(90−T)

final temperature of the combined system T = 84.78°C

5 0
3 years ago
Calculate the energy in kJ added to 1 ft3 of water by heating it from 70° to 200°F.
Serga [27]

Answer:

The energy in kJ is 8558.16 kJ.

Explanation:

Data presented in the problem:

Water is heated from 70 (T1) to 200 °F (T2).

Volume (V) of the water is 1 ft3.

It is required for the specific heat of water(HW), which is 1 BTU/lb°F.

First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.

M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.

.After that, we can calculate the heat required (Q).

Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)

Q = 62.4 * 130 BTU = 8112 BTU.

Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ

Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.

Finally, the energy required is 8558.16 kJ.

3 0
3 years ago
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