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4 years ago

I don't understand this question whoever helps thank you!

Physics

1 answer:

GenaCL600 [577]4 years ago

8 0

I am pretty sure it is C.

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If a cup of coffee is at 90°C and a person with a body temperature of 36'C touches it,how will heat flow between them

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the heat always transfers from high temperature to low temperature body without aid of any external energy to this law the heat transfers from cup of coffee to the person body until both bodies reaches to the equilibrium state

Explanation:

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Air enters the diffuser of a jet engine operating at steady state at 18 kPa, 216 K and a velocity of 265 m/s, all data correspon

earnstyle [38]

Answer:

45.44m/s

Explanation:

To solve this problem it is necessary to go back to the concepts related to the first law of thermodynamics,

in which it deepens on the conservation of the Energy.

The first law of Thermodynamics is given by the equation:

$0 = \dot{Q}-\dot{W}+\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})+\dot{m}g(z_1-z_2)$

Where,

$\dot{Q}=$ Heat transfer

$\dot{W} =$ Work

$\dot{m} =$ Flow mass

$V_i =$ Velocity

$h_i =$ Specific Enthalpy

$g =$ Gravity

$z_i =$ Height

From this equation we have that there is not Heat transfer, Work and changes in Height. Then,

Then our equation would be,

$0=\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})$

Solving for $V_2,$

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

From the tables of ideal gas (air) at 216K we have,

$h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})$

$h_1 = 215.97kJ/kg$

From the tables at 250K, we have that

$h_2 = 250.05kJ/kg$

The velocity was previously given, then

$V_1 =265m/s$

Replacing in the equation:

$V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}$

$V_2 = 45.44m/s$

Therefore the velocity of the air at the diffuser exit is 45.44m/s

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4 years ago