TFS · Thermodynamics · Problem 16PDFSolution in PDF ↓
TFS · Thermodynamics · Problem 16
Problem & Solution
Video Synthesis
- Problem: 80 degree atmospheric air enters a compressor at a velocity of 10 feet per second through a 50 square inch inlet and exits out of a velocity of 5 feet...
- Approach: So to start out here, we're going to use the equation that you'll find in the reference handbook by searching steady dash flow systems.
- Key step: Entry p for the exit plus velocity at the exit squared over 2 plus g times the height at the exit plus u dot n minus w dot out equals 0.
- Watch out: But in terms of its actual contribution to the change in energy, it's not worth the time to factor that in for a small change in velocity.
- Result: So now we're ready to find Qn and I'll put that term out front W out minus 106 plus the mass flow rate point 255 pound per second times the difference...
- ✅ Answer: B
Office Hours
3
Student questions asked in live office hours about this problem
OH 58
Q: In the compressor problem, you used m·ΔH with exit and inlet enthalpy switched compared to my approach, and I got answer A using it the other way—did I get my figures backwards?
A: You're right to catch that—I made an algebra error in the solution video. When solving for q_in, the correct formula should be m·(h_e - h_i) + w_out, not m·(h_i - h_e), because the enthalpy terms change sign when moving across the equation. Make sure you're careful with the algebraic manipulation when isolating q_in.
OH 91
Q: How did you know at the beginning that the velocity difference would be insignificant in the compressor energy balance problem?
A: From experience, I've learned velocity terms often end up negligible, so I struck it out early. But you don't have to take my word for it—the velocity data was given, so you can include it to prove to yourself it doesn't change the answer much. Over time, as you work similar problems repeatedly, you'll develop judgment about when terms matter and when they don't.
OH 112
Q: Why is velocity canceled in the first law equation for a compressor when it contributes 20,000-30,000 BTU/hr? Could this omission cause you to miss a test answer, and why do my final numbers differ from yours by 10,000 BTU/hr?
A: You caught a computational error on my end—your number is correct. The more interesting point is about units: enthalpy naturally gives BTU/lb, but velocity and elevation terms give ft²/sec², which aren't directly comparable to BTU/lb, so they need unit conversion before you can actually add them to the enthalpy term in the first law equation.