Steam flows through a nozzle at mass flow rate of m = 0.1 kg/s with a heat loss of 5 kW. The enthalpies at inlet and exit are 2500 kJ/kg and 2350 kJ/kg, respectively. Assuming negligible velocity at inlet (C₁≈0) ,  the velocity (C₂)  of steam (in m/s) at the nozzle exit is _________ (correct to two decimal places) 

The correct answer is 447.21.

1. Given Parameters:
From the problem description and the nozzle schematic shown in the image, we have:
Inlet enthalpy of steam, $h_1=2500\text{kJ/kg}$
Exit enthalpy of steam, $h_2=2350\text{kJ/kg}$
Mass flow rate of steam, $\dot{m}=0.1\text{kg/s}$
Heat loss from the nozzle, $\dot{Q}=-5\text{kW}$ (negative since heat is lost from the system)
Inlet velocity, $C_1\approx 0\text{m/s}$

2. Steady Flow Energy Equation (SFEE):
Applying the steady flow energy equation to the nozzle control volume:

$\dot{m}(h_1+\frac{C_1^2}{2}+g{z}_1)+\dot{Q}=\dot{m}(h_2+\frac{C_2^2}{2}+g{z}_2)+\dot{W}$

For a nozzle, there is no work transfer ($\dot{W}=0$) and potential energy changes are negligible ($z_1=z_2$). Since the inlet velocity is also negligible ($C_1\approx 0$), the SFEE simplifies to:

$\dot{m}{h}_1+\dot{Q}=\dot{m}(h_2+\frac{C_2^2}{2})$

3. Calculation:
First, let us convert the enthalpies and heat transfer into standard SI units (Joules):
$h_1=2500\times {10}^3\text{J/kg}$
$h_2=2350\times {10}^3\text{J/kg}$
$\dot{Q}=-5\times {10}^3\text{W (or J/s)}$

Substitute these values into the simplified equation:

$0.1(2500\times {10}^3)-5\times {10}^3=0.1(2350\times {10}^3+\frac{C_2^2}{2})$

Divide the entire equation by $0.1$:

$2500\times {10}^3-50\times {10}^3=2350\times {10}^3+\frac{C_2^2}{2}$

$2450\times {10}^3=2350\times {10}^3+\frac{C_2^2}{2}$

Subtract $2350\times {10}^3$ from both sides:

$\frac{C_2^2}{2}=100\times {10}^3=100000\text{J/kg}$

$C_2^2=200000{\text{m}}^2/{\text{s}}^2$

$C_2=\sqrt{200000}\approx 447.2136\text{m/s}$

Rounding to two decimal places, we get the exit velocity:
$C_2=447.21\text{m/s}$