Question

what is the enthalpy value of steam at 1.9 MP and 290 C?

Answer 1

Note: The first step to solve this problem is to identify the condition of the steam at the specified temperature and pressure. Clearly, the condition of the steam is not saturated since for saturated steam, temperature and pressure are dependant properties. Saturated steam cannot exist at 1.9 MPa and 290^o C, hence the condition of the steam is superheated. (Ref: Steam Tables)

Solution

From superheated steam tables, there is no enthalpy value corresponding to 1.9 MPa and 290^oC. Therefore we need to do linear interpolation to find the required enthalpy value.

The following steps are to be followed

1. Interpolate between 250 ^oC and 300 ^oC to obtain enthalpy value at 290 ^oC corresponding to 1.8 MPa (3006.26 kJ/kg)

2. Again, Interpolate between 250 ^oC and 300 ^oC to obtain enthalpy value at 290 ^oC corresponding to 2 MPa. (3000.02 kJ/kg)

3. Lastly, intepolate the newly obatined enthalpy values (ie., 3006.26 kJ/kg and 3000.02 kJ/kg) to obtain the required enthalpy value at 1.9 MPa and 290 ^oC. (ie., 3003.14 kJ/kg)

Answer:

The required enthalpy value at 1.9 MPa and 290 ^oC is 3003.14 kJ/kg