Using p − v − T Data

The following table lists temperatures and specific volumes of water vapor at two pressures:

p= 1.0 MPa |
p=1.5 MPa |
||

T (°C) |
v(m3/kg) |
T (°C) |
v(m3/kg) |

200 |
0.2060 |
200 |
0.1325 |

240 |
0.2275 |
240 |
0.1483 |

280 |
0.2480 |
280 |
0.1627 |

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°C, p = 1.25 MPa, in m3/kg.

(b) the temperature at p = 1.5 MPa, v = 0.1555 m3/kg, in°C.

(c) the specific volume at T = 220°C, p = 1.4 MPa, in m3/kg.

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: .

Step 2 of 4:-

b) Here we need to determine the temperature at P = 1.5 MPa, V = 0.1555 m3/kg, in°C.

We know the pressure is, .

The volume of will be surely in between the temperatures and . this is evident from the given table.

Diagrammatically,

Here also the slope of the slanted line will provide the required temperature.

So,

.

So, the temperature at P = 1.5 MPa, V = 0.1555 m3/kg is: .