Respuesta :

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

h₁ = 763.088 kJ/kg

s₁ = 4.287 kJ/(kg·K)

Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

h₂ = 79.572 kJ/kg

The saturation temperature at the given

T₂ = 79°C

The rate of work output [tex]\dot W[/tex] = [tex]\dot m[/tex]×[tex]c_p[/tex]×(T₂ - T₁)

Where;

[tex]c_p[/tex] = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)

[tex]\dot m[/tex] =  The mass flow rate = 2.0 kg/s

Substituting the values, we have;

[tex]\dot W[/tex] = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

[tex]\dot W[/tex] = -396.17 kJ/s

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