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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!

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You seem to have made some typos in the question, but my first guess anyways is that you've made a mistake with the units on $\tfrac{v^2}{2}$: $1 \tfrac{\text{m}^2}{\text{s}^2}$=1 J/kg, so if you add it to enthalpies in kJ/kg you'll get a greater outlet enthalpy. – rdhs Nov 24 '11 at 15:46
That's exactly the equations that's in the book. What typos do you see in the question though? And in my calculations, I made sure to change the enthalpies to J/kg instead of kJ/kg. – Greg Harrington Nov 24 '11 at 15:55
Oh. Nevermind, then. But if you have two streams coming in at .01 kg/s, you can't have an exit stream at .109 kg/s. – rdhs Nov 24 '11 at 16:02
@rdhs Check the edits, I figured it out. I did write one of the flow rates wrong in the description but I was using the right mass flow rate in the problem. Turns out the problem lied in algebraically rewriting the steady flow energy equation. Algebraic mishaps are such middle school – Greg Harrington Nov 24 '11 at 16:47
@rdhs do you think you could go look at my question about using the non-steady flow energy equation. I have most of the problem done but i just need some advice for the last portion – Greg Harrington Nov 27 '11 at 0:12

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