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AlekseyPX
3 years ago
11

Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as

a result of heat transfer to the surrounding medium at 20°C. Determine the rate of entropy change of the air.
Engineering
1 answer:
adell [148]3 years ago
4 0

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

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What is the minimum recommended safe distance from an X-ray source?
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2 years ago
Read 2 more answers
1.0•10^-10 standard form
Drupady [299]

Answer:

1.0 * 10^{-10} = 0.0000000001

Explanation:

Given

1.0 * 10^{-10}

Required

Convert to standard form

1.0 * 10^{-10}

From laws of indices

a^{-x} = \frac{1}{a^x}

So, 1.0 * 10^{-10} is equivalent to

1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}

1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}

1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}

1.0 * 10^{-10} = 1.0 * 0.0000000001

1.0 * 10^{-10} = 0.0000000001

Hence, the standard form of 1.0 * 10^{-10} is 0.0000000001

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