Basic Thermodynamics Sample Problem 3

Question

Consider 1 kg of water that isheated at constant pressure (0.50 MPa) from a wet state (x=85%) to 400°C. Determine (a) the change of enthalpy and (b) the change of entropy.

Answer 1

To determine the change of enthalpy and the change of entropy for the given process, we can use the steam tables. Here are the steps:

(a) The change of enthalpy can be calculated as follows:

From the steam tables, we can find that the enthalpy of saturated liquid water at 0.50 MPa is 683.5 kJ/kg, and the enthalpy of saturated steam at the same pressure is 3075.7 kJ/kg.

At x=85%, the water is in a wet state, which means it is a mixture of saturated liquid and saturated vapor. We can find the specific enthalpy of the wet water using the formula:

h = (1-x) * h_f + x * h_g

where h_f is the specific enthalpy of saturated liquid water, h_g is the specific enthalpy of saturated steam, and x is the quality (i.e., the fraction of the mixture that is in the vapor phase).

Plugging in the values, we get:

h = (1-0.85) * 683.5 + 0.85 * 3075.7 = 2626.63 kJ/kg

The enthalpy of water at 400°C and 0.50 MPa is 3341.5 kJ/kg.

The change of enthalpy can be calculated as:

Δh = h₂ - h₁ = 3341.5 - 2626.63 = 714.87 kJ/kg

Therefore, the change of enthalpy is 714.87 kJ/kg.

(b) The change of entropy can be calculated using the formula:

Δs = s₂ - s₁ = s_g,2 - s_f,1

where s_f,1 is the specific entropy of saturated liquid water at the initial state, and s_g,2 is the specific entropy of saturated steam at the final state.

From the steam tables, we can find that the specific entropy of saturated liquid water at 0.50 MPa is 1.3059 kJ/kg·K, and the specific entropy of saturated steam at the same pressure and temperature of 400°C is 7.0273 kJ/kg·K.

The change of entropy can be calculated as:

Δs = 7.0273 - 1.3059 = 5.7214 kJ/kg·K

Therefore, the change of entropy is 5.7214 kJ/kg·K.