Solution Found!
Solved: Using p ? v ? T DataThe following table lists
Chapter 3, Problem 7P(choose chapter or problem)
The following table lists temperatures and specific volumes of water vapor at two pressures:
Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate
(a) the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\), in \(\mathrm{m}^3 / \mathrm{kg}\).
(b) the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^3 / \mathrm{kg}\), in \(^{\circ} \mathrm{C}\).
(c) the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\), in \(\mathrm{m}^3 / \mathrm{kg}\).
Questions & Answers
QUESTION:
The following table lists temperatures and specific volumes of water vapor at two pressures:
Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate
(a) the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\), in \(\mathrm{m}^3 / \mathrm{kg}\).
(b) the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^3 / \mathrm{kg}\), in \(^{\circ} \mathrm{C}\).
(c) the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\), in \(\mathrm{m}^3 / \mathrm{kg}\).
ANSWER:Solution 7P:
Step 1 of 4:-
a) The specific volume at T = 240°C, P = 1.25 MPa, in m3/kg.
At this temperature, the pressure of comes between
and
.
The values of volume in 3rd row of the table are,
at
.
at
.
The slope of the slanted line will determine the specific volume at the particular pressure.
.
So, the specific volume at and
is:
.