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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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People who are exceptionally tall (like seven feet and over) often have a disorder that causes them to keep growing past the age
AleksandrR [38]

Answer:

C is the answer

Explanation:

The pituitary gland also known as the Master gland controls and regulates secretion of most hormones, when taller people usually grow past expected height this is normally due to the Pituitary gland not properly regulating and it secretes too many growth hormones causing "gigantism."

6 0
2 years ago
A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0
3 years ago
A 706-N man stands in the middle of a frozen pond of radius 8.0 m. He is unable to get to the other side because of a lack of fr
mariarad [96]

Answer:

t=62 s

Explanation:

Applying the conservation of linear momentum formula:

(m1+m2)*v_{o1}=m1*v_{f1}+m2*v_{f2}

the initial velocity is zero, we can calculate the man's mass using the gravitational force formula:

F_g=m.g\\\\m=\frac{706N}{9.81}\\\\m=72.0kg

now:

m*v_{f1}=-m_b*v_b\\\\V_{f1}=-\frac{m_b*v_b}{m}\\\\V_{f1}=-\frac{1.2kg*8.0m/s}{72.0kg}\\\\V_f=-0.13m/s

That is 0.13m/s due south.

because there is no friction, the man will maintain a constant velocity, so:

d=v*t\\t=\frac{d}{v}\\t=\frac{8m}{0.13m/s}\\\\t=62s

8 0
3 years ago
A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper
vodka [1.7K]

Answer:

t = 402 years

Explanation:

To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:

I=nqv_dA  (1)

n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3

q: charge of the electron = 1.6*10^-19 C

vd: drift velocity of electron in the metal = ?

A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2

I: current in the wire = 1110 A

You solve the equation (1) for vd:

v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s

Next, you calculate the time by using the information about the length of the line transmission:

x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years

hence, the electrons will take aproximately 402 years in crossing the line of transmission

6 0
3 years ago
Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon
Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

T^2=\frac{4\pi}{GM}r^3

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min

the comet takes around 16.63min

8 0
3 years ago
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