Question

A liquid comes out of the reactor of a chemical plant at 250 °C temperature with a flow rate of 100 LPM. The ambient temperature is 25 °C. Density and specific heat of the liquid remain constant at 1000 kg/m³ and 4.18 kJ/kg K, respectively, for the given range of temperature. The rate of exergy associated with the hot liquid stream at the exit of the plant is ________ kW (rounded off to 2 decimal places).

Hint:

Use the flow exergy formula for a liquid stream, combining the enthalpy drop with the entropy-based unavailable part.

Answer 1

Correct Answer: 399.55

Step 1: Get the mass flow rate first.
$100$ LPM $= 100/1000/60 = 0.001667$ m$^3$/s, so with $\rho=1000$ kg/m$^3$, $\dot m = 1.667$ kg/s.

Step 2: Compute the enthalpy and entropy changes separately, both measured from the dead state $T_0$.
For an incompressible liquid, $\Delta h = c_p(T-T_0)$ and $\Delta s = c_p\ln(T/T_0)$. With $T=523.15$ K and $T_0=298.15$ K: $\Delta h = 4.18(225) = 940.5$ kJ/kg, and $\Delta s = 4.18\ln(1.7547) = 4.18(0.5623) = 2.3504$ kJ/(kg K).

Step 3: Combine using the exergy definition $\psi=\Delta h-T_0\Delta s$.
$T_0\Delta s = 298.15(2.3504) = 700.9$ kJ/kg, so $\psi = 940.5-700.9 = 239.6$ kJ/kg.

Step 4: Multiply by the mass flow rate to get the exergy rate.
$\dot X = \dot m \psi = 1.667(239.6) = 399.5$ kW.

Final Answer:
Splitting the exergy into an enthalpy part and an entropy penalty gives the same rate as the combined formula. \[ \boxed{\dot X = 399.55 \ \text{kW}} \]