TFS · Thermodynamics · Problem 7PDFSolution in PDF ↓
TFS · Thermodynamics · Problem 7
Problem & Solution
Video Synthesis
- Problem: Equal parts of 50 degree water and 250 degree steam each at 5 psi g are added to a mixing chamber and thoroughly mixed.
- Key step: We don't know what that's going to be probably going to be a saturated mixture but we don't know that yet we'll prove that during the problem and ever
- Watch out: We don't know what that's going to be probably going to be a saturated mixture but we don't know that yet we'll prove that during the problem and ever...
- Result: And that gives us 0.413, which is closest to answer choice A.
- ✅ Answer: A
Office Hours
2
Student questions asked in live office hours about this problem
OH 106
Q: Are there scenarios where it's appropriate to assume that the enthalpy of a compressed liquid equals hf? I used h = hf and got the wrong answer — I want to understand when this assumption is valid.
A: The h ≈ hf assumption is acceptable when pressure is near the saturation pressure at that temperature, but it breaks down when the liquid is significantly sub-cooled — meaning pressure is well above saturation. The more accurate formula for compressed liquid enthalpy is Δh = Cp × ΔT from the saturated state, which is what we used in the solution. Use h = hf as a rough sanity check or when Cp × ΔT is negligibly small; for this problem the pressure was high enough that the subcooling correction mattered.
OH 107
Q: Can you just look up the enthalpy of water at 50°F in the saturated steam tables and use hf = 18 BTU/lb?
A: No — the state is fully defined by 50°F and ~20 psia (5 psig), and when you look up water at 50°F in the saturation table, you'll find that the saturation pressure at 50°F is only about 0.18 psia, which is far below 20 psia. That means the water is a sub-cooled (compressed) liquid, not saturated liquid, so hf from the saturation table doesn't apply. For sub-cooled liquid, use h = hf(T) + Cp × (P – Psat) × v, or the simplified h ≈ hf + Cp × ΔT approach.