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nekit [7.7K]
3 years ago
7

A gas-turbine power plant operating on the simple Brayton cycle has a pressure ratio of 7. Air enters the compressor at 0°C and

100 kPa. The maximum cycle temperature is 1500 K. The compressor has an isentropic efficiency of 80 percent, and the turbine has an isentropic efficiency of 90 percent. Assume constant properties for air at 300 K with cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.a) sketch the T-s diagram for the cycle.
b) if the net power output is 150 MW, determine the volume flow rate of the air into the compressor, in m^3/s.
c) for a fixed compressor inlet velocity and flow area, explain the effect of increasing compressor inlet temperature (i.e., summertime operation versus wintertime operation) on the inlet mass flow rate and the net power output with all other parameters of the problem being the same.

Physics
1 answer:
Ivahew [28]3 years ago
4 0

Answer:

Answers of the Both parts  are in the following attachment

Explanation:

         

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A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
IrinaVladis [17]

The coefficient of friction between the road and the car's tire is determined as 0.78.

<h3>Acceleration of the car</h3>

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

  • u is the initial velocity = 97 km/h = 26.94 m/s

a = (26.94)²/(2 x 47)

a = 7.72 m/s²

<h3>Coefficient of friction</h3>

μ = a/g

μ = (7.72)/9.8

μ = 0.78

Learn more about coefficient of friction here: brainly.com/question/14121363

#SPJ1

5 0
2 years ago
4A. How high is a 12 kg monkey in a tree if it has 509 J of gravitational potential Energy?
True [87]

4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)

509 J = 12x10xH

509 J = 120xH

H = 509/120

H = 4.24 m

Hope u got the answer....pls rate the answer if it is helpful for u....and I'm sorry I could not understand B part so I didn't do it.

Thank you

8 0
3 years ago
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
I am Lyosha [343]

Answer:

The density of this object is approximately 1.36\; {\rm kg \cdot L^{-1}}.

The density of the oil in this question is approximately 0.600\; {\rm kg \cdot L^{-1}}.

(Assumption: the gravitational field strength is g =9.806\; {\rm N \cdot kg^{-1}})

Explanation:

When the gravitational field strength is g, the weight (\text{weight}) of an object of mass m would be m\, g.

Conversely, if the weight of an object is (\text{weight}) in a gravitational field of strength g, the mass m of that object would be m = (\text{weight}) / g.

Assuming that g =9.806\; {\rm N \cdot kg^{-1}}. The mass of this 63.8\; {\rm N}-object would be:

\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}.

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was (63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}. Hence, the weight of water that this object displaced would be 47.0 \; {\rm N}.

The mass of water displaced would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}.

The volume of that much water (which this object had displaced) would be:

\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}.

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately 4.793\; {\rm L}.

The mass of this object is 6.50\; {\rm kg}. Hence, the density of this object would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately 4.793\; {\rm L}.

The weight of oil displaced would be equal to the magnitude of the buoyancy force: 63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}.

The mass of that much oil would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}.

Hence, the density of the oil in this question would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

7 0
2 years ago
A cylindrical metal can is to have no lid. It is to have a volume of 125π in3. What height minimizes the amount of metal used?
Alex Ar [27]

Answer:

Minimum height of metal = 5 inches

Explanation:

Volume of the cylindrical metal = πR²H = 125π

cancelling out π on both sides

R²H = 125

Hence it can be deduced that R² = 25 and H = 5

Hence minimum height of metal = 5 inches

6 0
3 years ago
TRUE OR FALSE. if an object does not change its position at a given time interval, then it is at rest or its speed is zero or no
Andrew [12]

Answer:

true

Explanation:

a body can only be accelerating and have speed if it's in motion

3 0
3 years ago
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