A mass of 2.5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate the work done by the steam during this process. Use steam tables.

the work done by the steam during this process = 119.575 kJ

Explanation:

From the information given:

At state 1

Let obtain the specific volume of the saturated vapor from the Saturated water- Pressure table A-5 at the pressure of 100 kPa

where is the specific volume of the saturated vapor at state 1.

At state 2:

From the tables A-6 of Superheated water at the pressure of 100 kPa or 0.1 MPa and at the temperature of 200°C, the specific volume

where is the specific volume of the superheated water at state 2.

The workdone by the steam during the process can be expressed by the formula:

where;

m = mass of the saturated water vapor

P = pressure of the saturated water vapor

volume of the superheated water at state 2

volume of the saturated water at state 1

Replacing our values ;

W = 2.5 (100) ( 2.1724 -1.6941)

W = 250(0.4783 )

W = 119.575 kPa.m³

W = 119.575 kJ

∴

the work done by the steam during this process = 119.575 kJ

i think if it were me i whould divide 300 dide 15

step-by-step explanation: