Problem: Steam supply to an engine is made of two streams that mix before entering the engine. Stream 1 flows at a $.01\frac{kg}{s}$ with an enthalpy of $2952\frac{kJ}{kg}$ and a velocity of $20\frac{m}{s}$. The other stream is supplied at a rate of $.01\frac{kg}{s}$ with an enthalpy of $2569\frac{kJ}{kg}$ and a velocity of $20\frac{m}{s}$. The fluid leaves the engine as two streams at the exit, one of water at a rate of $.001\frac{kg}{s}$ with an enthalpy of $420\frac{kJ}{kg}$, and the other of steam. The fluid velocities are negligible. The engine develops a shaft power of 25kW. Heat transfer is negligible. Find the enthalpy of the second exit stream.
My attempt: I determined that I could use the steady flow energy equation where potential energy effects are negligble to solve this problem.
$$\dot{Q}-\dot{W}=\sum_{out} \dot{m}(h+\frac{V^{2}}{2})-\sum_{in} \dot{m}(h+\frac{V^{2}}{2})$$ $$-\dot{W}=\dot{m_{4}}h_{4}+\dot{m_{3}}h_{3}-\dot{m_{1}}(h_{1}+\frac{V_{1}^{2}}{2})-\dot{m_{2}}(h_{2}+\frac{V_{2}^{2}}{2})$$
The values of $\dot{W}$, $\dot{m_{1}}$, $V_{1}$, $h_{1}$, $\dot{m_{2}}$, $h_{2}$, $V_{2}$, $\dot{m_{3}}$, $h_{3}$ are given in the problem statement. Since this is a steady flow and mass is conserved throughout, then: $$\dot{m_{in}}=\dot{m_{out}}$$ $$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}+\dot{m_{4}}$$ Therefore, $\dot{m_{4}}=.109\frac{kg}{s}$ I should be able to put all these values in and solve for $h_{4}$. I am getting an answer of $2860\frac{kJ}{kg}$ and my teacher's answer is $2401\frac{kJ}{kg}$. Am I doing anything wrong?
Edits: The second stream is supplied at a rate of $.1\frac{kg}{s}$, not $.01\frac{kg}{s}$. Also I changed the steady flow energy equations and received the correct answer. Thanks for the help guys!
