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Anna11 [10]
1 year ago
5

What happens when you run liquid air or nitrogen through a gas turbine

Chemistry
1 answer:
lidiya [134]1 year ago
4 0

Answer:

When a low cost fuel is available, internal combustion drivers surpass all others in compactness and low cost of installation and operation. For example, gas compression on a large scale has long been done with integral engine compressors. 

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If you repeat an experiment and the result are very different from the result you got the first time the next step would be to w
Bess [88]
You would want to make sure that you have controlled the variables properly, and if you determine that you did then you would repeat the experiment to be sure of the results.
4 0
3 years ago
A radioactive substance of mass 768g has a half life of 3 years. After how many years does it leave only 6g undecayed?​
vagabundo [1.1K]

Answer:

The answer is 21 years.

Explanation:

7×3

8 0
2 years ago
Read 2 more answers
How many quantum numbers are used to describe the properties of electrons in atomic orbitals?
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4 Quantum numbers are used

1.Azimuthal
2.Principal
3.Spin
4.Magnetic
6 0
3 years ago
What can we tell from the column that an element is in?
Paul [167]

Answer:

The number of electrons in the outer energy level

Explanation:

6 0
3 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
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