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Air at 80 kPa, 27 degrees C and 220 m/s eneters a diffuser at a rate of 2.5 kg/s and leaves at 42 degrees C. the exit area of the diffuser is 400 cm2. the air is estimated to lose heat at a rate of 18 kJ/s during this process.
Determine
a.The exit velocity.
b.The exit pressure of the air.

Respuesta :

davidlcastiblanco

Answer:

a).[tex]v_{2}=62.04\frac{m}{s}[/tex]

b).[tex]P_{2}=91.1kPa[/tex]

Explanation:

The outlet velocity is calculated from energy balance relation using the specific heat at constant pressure so

[tex]E_{in}=E_{out}[/tex]

[tex]m'*(h_{1}+\frac{v_{1}^2}{2})=m'*((h_{1}+\frac{v_{2}^2}{2})+Q'[/tex]

[tex]m'*(\frac{v_{1}^2-v_{2}^2}{2})=m'*Cp*T+Q'[/tex]

a).

[tex]v_{2}^2=v_{1}^2-2*Cp*T-\frac{2*Q'}{m'}[/tex]

[tex]v_{2}=\sqrt{v_{1}^2-2*Cp*T-\frac{2*Q'}{m'}}[/tex]

[tex]v_{2}=\sqrt{220^2-2*1.005x10^3*15-\frac{2*18x10^3}{2.5}}[/tex]

[tex]v_{2}=62.04\frac{m}{s}[/tex]

b).

[tex]P_{2}=\frac{m'*R*T_{2}}{A_{2}*v_{2}}[/tex]

[tex]P_{2}=\frac{2.5*287*315}{400x10^{-4}*62.04}[/tex]

[tex]P_{2}=91.1kPa[/tex]