An inventor tries to sell you his new heat engine that takes in 40 J of heat at 87°C on each cycle, expels 30 J at 27°C, and doe
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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
To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are
Real power in 3 phase
Now the Phase Voltage is,
The current phase would be,
Rearranging,
Replacing,
Therefore the current per phase is 2.26kA