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Lerok [7]
4 years ago
13

2. Air at a temperature of 20 ºC passes through a pipe with a constant velocity of 40 m/s. The pipe goes through a heat exchange

r in which hot gases outside the pipe heat up the air to 820 ºC. It then enters a turbine with a velocity of 40 m/s and expands till the temperature falls to 620 ºC. The air stream loses 4.3 kW heat in the turbine. If the air flow rate is 2.5 kg/s, calculate (a) How much heat is transferred to the air in the heat exchanger. (b) The power output of the turbine.
Physics
1 answer:
frozen [14]4 years ago
7 0

Answer:

a) Q = 1436 kW

b) P ≈ 776 kW

Explanation:

Let's begin by listing out the given parameters:

T1 = 20 °C, u = 40 m/s, T2 = 820 °C, P = 4.3 kW, m = 2.5 kg/s, T3 = 510 °C, V1 = 40 m/s,

V2 = 40 m/s, V3 = 55 m/s, ṁ = 2.5 kg/s

To solve the question, we make this assumption that the size of the pipe is constant

a) No change in velocity implies that heat added is isochoric

Q = m * C * ΔT

Cv of air at 300 K(≈20 °C) = 0.718

Q = 2.5 * 0.718 * (820 − 20)

Q = 1436 kW

b) P = ṁ * Cp * ΔT + ṁ * (V2² - V3²) ÷ 2000] - Ql

V2² - V3² = 55² - 40² = 1425

ΔT = T2 - T3 = 820 - 510 = 310 °C

Cp of air at 300 K(≠20 °C) = 1.005 kJ/kgK

Ql = 4.3 kW

P = 2.5 * (1.005 * 310) + 2.5 * (1425 ÷ 2000) - 4.3

P = 778.875 + 1.78125 - 4.3 = 776.35625

P ≈ 776 kW

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A stone is thrown vertically upward with a speed of 18 m/s. (a) How long does it take the stone to reach a height of 11 m? (b) h
bagirrra123 [75]

Answer:

a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.

b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.

Please, see the attached figures and the explanation for a description of the figures.

Explanation:

Hi there!

The equations for the height and velocity of the stone are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time t

a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:

y = y0 + v0 · t + 1/2 · g · t²        

11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²    

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m

Solving the quadratic equation:

t = 0.7743 s and t = 2.899 s

(Notice that I have used more significant figures to avoid error by rounding)

The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down  (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of  11 m for the first time.

b)  Let´s use the equation of velocity:

v = v0 + g · t

at t = 0.77443 s

v = 18 m/s - 9.8 m/s² · 0.77443 s

v = 10.41 m/s

at t = 2.899 s

v = 18 m/s - 9.8 m/s² · 2.899 s

v = - 10.41 m/s

(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.

Figures

The functions to plot are the following:

height in function of time (figure 1, x-axis: time. y-axis: height)

y = -4.9t² + 18t

velocity in function of time (figure 2, x-axis: time. y-axis velocity)

v = -9.8t + 18

Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)

a = -9.8

5 0
4 years ago
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