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2 years ago

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Which option should the engineers focus on as they develop the train in the following scenario?

1 answer:

pav-90 [236]2 years ago

5 0

Answer:

Engineers can design a train with a regenerative braking system

Explanation:

Assuming the point of the question is that the engineers want to focus on using energy efficiently when starting and stopping, they would likely want to consider a regenerative braking system. Such a system can store energy during braking so that it can be used during starting, reducing the amount of energy that must be supplied by an outside power source.

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Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres

natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

$h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}$

$\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}$

$\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}$

$G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}$

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

$P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))$

The rate of heat:

$Q=G_{m}(h_{3}-h_{2})$

The thermal efficiency:

$\mu_{ther}=\frac{P_{net}}{Q}$

3 0

2 years ago

Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at

makkiz [27]

Answer:

$\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}$

Explanation:

The Young's module is:

$E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }$

$E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}$

Let assume that both specimens have the same geometry and load rate. Then:

$E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}$

The displacement rate for steel is:

$\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}$

$\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )$

$\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}$

7 0

2 years ago

Read 2 more answers

Two identical 3 in. major-diameter power screws (single-threaded) with modified square threads are used to raise and lower a 50-

sp2606 [1]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

6 0

2 years ago

In a team, a person’s efforts are less identifiable than when that person works independently. Because the person’s efforts are

Eduardwww [97]

The answer is deindividuation - a psychological state in which a person does not feel individual responsibility.

3 0

2 years ago

The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort < 0. Fort 2 0 they areV =75 ~75e-10

masya89 [10]

Answer:

maximum value of the power delivered to the circuit =3.75W

energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

Explanation:

V =75 - 75e-1000t V

l = 50e -IOOOt mA

power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)

=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)

=

maximum value of the power delivered to the circuit =3.75W

the total energy delivered to the element = $\int\limits^t_0 {3.75(e^{ -IOOOt} - e ^{-2OOOt} )} , dx \\\\$

$3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750$

5 0

2 years ago