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: Convert temperatures to Kelvin and get the mass flow rate.
\(T = 250+273.15 = 523.15\) K, \(T_0 = 25+273.15=298.15\) K. The volumetric flow is \(100\) L/min \(= 0.1/60 = 0.001667\) m\(^3\)/s, so \(\dot m = \rho \dot V = 1000(0.001667)=1.667\) kg/s.

Step 2: Write the flow exergy formula for a liquid stream.
Ignoring kinetic, potential and pressure effects, the specific flow exergy relative to the dead state is \(\psi = c_p\left[(T-T_0) - T_0\ln\dfrac{T}{T_0}\right]\), which accounts for both the available temperature drop and the unavailable part tied to entropy generation on cooling to \(T_0\).

Step 3: Plug in the numbers.
\(T-T_0 = 523.15-298.15=225\) K. \(T/T_0 = 523.15/298.15 = 1.7547\), so \(\ln(T/T_0)=0.5623\). \(T_0\ln(T/T_0)=298.15(0.5623)=167.65\) K. \((T-T_0)-T_0\ln(T/T_0)=225-167.65=57.35\) K.

Step 4: Multiply by \(\dot m c_p\).
\(\dot m c_p = 1.667(4.18)=6.967\) kW/K. \(\dot X = \dot m c_p\left[(T-T_0)-T_0\ln(T/T_0)\right]=6.967(57.35)=399.55\) kW.

Final Answer:
This is the maximum useful work still recoverable from the hot stream before it is dumped at ambient conditions. \[ \boxed{\dot X = 399.55 \ \text{kW}} \]