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 =
×
×(T₂ - T₁)
Where;
= The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)
= The mass flow rate = 2.0 kg/s
Substituting the values, we have;
= 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s
= -396.17 kJ/s