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taurus [48]
3 years ago
11

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.

Physics
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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Somebody please help me with this question..!
Rudik [331]
8.1) Here, Pressure increases with depth in the ocean at higher rate then that of increase in altitude in atmosphere. So, the rate of change of pressure is different in these journeys.So, your most correct answer would be option D.

8.2) Difference in the shape of lines is due to different density of air & water. So, that physical property and your answer would be option A

So, in Short Answers of your questions are:
8.1) - Option D
8.2) - Option A

Hope this helps!


3 0
3 years ago
calculate the work done by a girl of mass 40 kg when she climbs a stair case of 20 steps each of height 10 cm and acceleration i
Volgvan

Answer:

i think its a 800x² hope you like it

5 0
3 years ago
Read 2 more answers
The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
Rainbow [258]

Answer: 0.0353\ s^{-1}

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in 2.59\times 10\ s

Sample at any time is given by

N=N_oe^{-\lambda t}

where, \lambda=\text{decay constant}

Put values

\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10

Taking natural logarithm both side

\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}

8 0
3 years ago
Americium-241 is used in smoke detectors. It undergoes alpha decay with a half life of 432 years. The alpha particles ionize som
babymother [125]

Answer:

860.6 years.

Explanation:

The parameters given are;

Initial detector activity = 370000 alpha decays per second

Final detector activity = 93000 alpha decays per second

Formula for time to change in activity is given by the following relation;

t_{93000} = \dfrac{-ln\dfrac{A}{A_0} }{\lambda} =  \dfrac{-ln\dfrac{93000}{370000} }{5.08 \times 10^{-11}} = 2.72 \times 10^{10} \, seconds

t₉₃₀₀₀ = 2.72 × 10¹⁰ seconds = 860.6 years.

3 0
3 years ago
The magnetic field of a magnet is strongest
Talja [164]
The answer is C !!!!
4 0
3 years ago
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