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A nozzle in a horizontal orientation is designed to have steady flowing steam exit it with a velocity of 250 m/s. If the outlet specific enthalpy of the steam is 1,986 kJ/kg, what is the required inlet specific enthalpy? Assume that heat transfer to the surroundings and the inlet steam velocity are negligible.

Respuesta :

Answer:

[tex]h_{1}[/tex] = 2017.25 kJ/kg

Explanation:

GIVEN DATA:

Exit velocity [tex]v_{2}[/tex] = 250 m/s

outlet enthalpy [tex]h_{2}[/tex]= 1986 kJ/kg

inlet velocity  [tex]v_{1}[/tex]= 0

heat transfer Q = 0

from steady flow energy equation(SFEE) between inlet and exit point

[tex]h_{1}+\frac{v_{1}^{2}}{2}=h_{2}+\frac{v_{2}^{2}}{2}+ Q[/tex]

[tex]h_{1}=h_{2}+\frac{v_{2}^{2}}{2}[/tex]

[tex]h_{1}[/tex]=2017.25 kJ/kg

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