Solution Found!
Solved: Using u?h Data Using the tables for water,
Chapter 3, Problem 42P(choose chapter or problem)
Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state by hand on sketches of the p–v and T–v diagrams.
(a) At \(p=2 \ \mathrm{MPa}, T=300^{\circ} \mathrm{C}\). Find u, in \(\mathrm{kJ} / \mathrm{kg}\).
(b) At \(p=2.5 \ \mathrm{MPa}, T=200^{\circ} \mathrm{C}\). Find u, in \(\mathrm{kJ} / \mathrm{kg}\).
(c) At \(T=170 \mathrm{~F}\), x=50%. Find u, in Btu/lb.
(d) At \(p=100 \ \mathrm{lbf} / \mathrm{in}^2, T=300^{\circ} \mathrm{F}\). Find h, in Btu/lb.
(e) At \(p=1.5 \ \mathrm{MPa}, v=0.2095 \mathrm{~m}^3 / \mathrm{kg}\). Find h, in \(\mathrm{kJ} / \mathrm{kg}\).
Questions & Answers
QUESTION:
Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state by hand on sketches of the p–v and T–v diagrams.
(a) At \(p=2 \ \mathrm{MPa}, T=300^{\circ} \mathrm{C}\). Find u, in \(\mathrm{kJ} / \mathrm{kg}\).
(b) At \(p=2.5 \ \mathrm{MPa}, T=200^{\circ} \mathrm{C}\). Find u, in \(\mathrm{kJ} / \mathrm{kg}\).
(c) At \(T=170 \mathrm{~F}\), x=50%. Find u, in Btu/lb.
(d) At \(p=100 \ \mathrm{lbf} / \mathrm{in}^2, T=300^{\circ} \mathrm{F}\). Find h, in Btu/lb.
(e) At \(p=1.5 \ \mathrm{MPa}, v=0.2095 \mathrm{~m}^3 / \mathrm{kg}\). Find h, in \(\mathrm{kJ} / \mathrm{kg}\).
ANSWER:Step 1 of 5
(a)
Obtain the properties of water at p = 2 Mpa and T = 300 degrees Celsius from the table A-4, “superheated water tables”, we interpolate between T = 280 degrees Celsius and T = 320 degrees Celsius to find specific internal energy at T = 300 degrees Celsius:
\(u\left( {320^\circ \;{\rm{C}}} \right) = 2807.9\;{\rm{kJ/kg}}\)
\(u\left( {280^\circ \;{\rm{C}}} \right) = 2736.4\;{\rm{kJ/kg}}\)
And
\(\frac{{2807.9 - 2736.4}}{{320 - 280}} = \frac{{2807.9 - u}}{{320 - 300}}\)
\(1.7875 = \frac{{2807.9 - u}}{{20}}\)
\(u = 2772.15\;{\rm{kJ/kg}}\)
Therefore, the specific internal energy is 2772.15 kJ/kg.
Draw the T-v and p-v diagrams.