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3 years ago

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Problem 9.49: Air enters the turbine of a gas turbine at 1200 kPa, 1200 K, and expands to 100 kPa in two stages. Between the sta

ges, the air is reheated at a constant pressure of 350 kPa to 1200 K. The expansion through each turbine stage is isentropic. Determine, in kJ per kg of air flowing, (a.) the work developed by each stage. (b.) the heat transfer for the reheat process. (c.) the increase in net work as compared to a single stage of expansion with no reheat.

1 answer:

Nitella [24]3 years ago

8 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

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Which of the following is not sublimable? *

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An electron (e = 1.6 x 10-19 C) is traveling at 4.00 x 107 m/s due North in a horizontal plane through a point where the earth’s

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Answer:

Explanation:

Force on the electron due to magnetic field 's north component will be zero because both velocity of electron and direction of magnetic field are same .

Force due to downward component of magnetic field

= B q v where B is magnetic field , q is charge moving and v is velocity of charge

F = 5 .00 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 4 x 10⁷

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The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m

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Answer:

$v=(6ti+6k)\ m/s$

Explanation:

Given that,

The position of a particle is given by :

$r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m$

Let us assume we need to find its velocity.

We know that,

$v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s$

So, the velocity of the particle is $(6ti+6k)\ m/s$ .

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3 years ago

A car traveling with constant speed travels 150 km in 7200s what is the speed of the car

Oksana_A [137]

Velocity is distance/time

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3 years ago

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

2 years ago

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