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satela [25.4K]
3 years ago
7

A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi

nd the pressure ratio and the specific heat addition by the combustion for this cycle. (Borgnakke, 02/2019, p. P-93) Borgnakke, C., Sonntag, R. E. (02/2019). Fundamentals of Thermodynamics, Enhanced eText, 10th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781119494966 Always check citation for accuracy before use.
Physics
1 answer:
motikmotik3 years ago
3 0

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

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Answer:

  • on the moon, they will fall at the time
  • on earth, the coin will fall faster to the ground

Explanation:

A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.

If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).

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3 years ago
143°C = _____<br><br> 416 K<br> -130 K<br> 0 K<br> 143 K
dezoksy [38]
The answer is 0k because 143c equals nothing
5 0
2 years ago
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The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

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4 0
3 years ago
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A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

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B. Because I say so
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