Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such that the specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by 0.007 kJ/kg.K. The power input to the electric heater is 2.50 kW. There is no other work or heat interaction between the system and the surroundings. Assuming an ambient temperature of 300 K, the irreversibility rate of the system is _________kW (round off to two decimal places).

The correct answer is 2.10.

Step-by-Step Explanation:

To find the irreversibility rate of the system, we can use the Gouy-Stodola theorem, which states that the rate of irreversibility ($\dot{I}$) is directly proportional to the rate of entropy generation (${\dot{S}}_{\mathrm{gen}}$) in the system:

$\dot{I}=T_0{\dot{S}}_{\mathrm{gen}}$

where:
- $T_0$ is the ambient (surrounding) temperature in Kelvin ($300\text{K}$).
- ${\dot{S}}_{\mathrm{gen}}$ is the rate of entropy generation within the control volume.

1. Identify the given parameters:
- Mass flow rate of water, $\dot{m}=1\text{kg/s}$
- Increase in specific enthalpy, $h_2-h_1=2.50\text{kJ/kg}$
- Increase in specific entropy, $s_2-s_1=0.007\text{kJ/kg}·\text{K}$
- Power input to the electric heater, ${\dot{W}}_{\text{in}}=2.50\text{kW}$
- Ambient temperature, $T_0=300\text{K}$

2. Perform an entropy balance on the system:
For a steady-flow system, the entropy balance is written as:

${\dot{S}}_{\mathrm{in}}+{\dot{S}}_{\mathrm{gen}}={\dot{S}}_{\mathrm{out}}$

Since the electric heater provides work input and there is no heat interaction between the system and its surroundings ($\dot{Q}=0$), the only entropy entering and leaving the control volume is carried by the fluid stream. Electrical work input does not carry any entropy. Thus, the entropy balance simplifies to:

$\dot{m}{s}_1+{\dot{S}}_{\mathrm{gen}}=\dot{m}{s}_2$

Solving for the rate of entropy generation (${\dot{S}}_{\mathrm{gen}}$):

${\dot{S}}_{\mathrm{gen}}=\dot{m}(s_2-s_1)$

Substitute the given values into the equation:

${\dot{S}}_{\mathrm{gen}}=1\text{kg/s}\times 0.007\text{kJ/kg}·\text{K}=0.007\text{kW/K}$

3. Calculate the irreversibility rate ($\dot{I}$):
Using the Gouy-Stodola relationship:

$\dot{I}=T_0{\dot{S}}_{\mathrm{gen}}$

$\dot{I}=300\text{K}\times 0.007\text{kW/K}=2.10\text{kW}$

Therefore, the irreversibility rate of the system is 2.10 kW.