Find the mass of 0.50m^3 of wet steam at a pressure of 4 bar and dryness fraction of 0.8. Also determine the enthalpy of 1m^3 of steam

To find the mass of 0.50 m³ of wet steam at a pressure of 4 bar with a dryness fraction of 0.8, we can use the properties of steam and the concept of specific volume. The dryness fraction indicates the proportion of steam in the mixture, with the rest being water. Let's break this down step by step.

Understanding the Properties of Wet Steam

Wet steam is a mixture of liquid water and steam, and its properties can be determined using steam tables. The key parameters we need are:

• Pressure (P): 4 bar
• Dryness Fraction (x): 0.8
• Volume (V): 0.50 m³

Step 1: Finding Specific Volume

Specific volume (v) is defined as the volume occupied by a unit mass of steam. For wet steam, the specific volume can be calculated using the formula:

v = v_f + x(v_g - v_f)

Where:

• v_f = specific volume of saturated liquid (water)
• v_g = specific volume of saturated vapor (steam)
• x = dryness fraction

Step 2: Using Steam Tables

At 4 bar, we can refer to steam tables to find the values of v_f and v_g:

• v_f (specific volume of water) ≈ 0.00113 m³/kg
• v_g (specific volume of steam) ≈ 0.025 m³/kg

Now, substituting these values into the specific volume formula:

v = 0.00113 + 0.8(0.025 - 0.00113)

v = 0.00113 + 0.8(0.02387)

v = 0.00113 + 0.019096

v ≈ 0.020226 m³/kg

Step 3: Calculating Mass

Now that we have the specific volume, we can find the mass (m) of the wet steam using the formula:

m = V / v

Substituting the values:

m = 0.50 m³ / 0.020226 m³/kg

m ≈ 24.7 kg

Determining the Enthalpy of 1 m³ of Steam

To find the enthalpy (h) of 1 m³ of steam at the same conditions, we can use the formula:

h = h_f + x(h_g - h_f)

Where:

• h_f = enthalpy of saturated liquid
• h_g = enthalpy of saturated vapor

From the steam tables at 4 bar:

• h_f ≈ 340 kJ/kg
• h_g ≈ 2676 kJ/kg

Now substituting these values into the enthalpy formula:

h = 340 + 0.8(2676 - 340)

h = 340 + 0.8(2336)

h = 340 + 1868.8

h ≈ 2208.8 kJ/kg

Final Results

To summarize:

• The mass of 0.50 m³ of wet steam at 4 bar with a dryness fraction of 0.8 is approximately 24.7 kg.
• The enthalpy of 1 m³ of steam under the same conditions is approximately 2208.8 kJ/kg.

This approach illustrates how to utilize steam tables effectively to derive important thermodynamic properties of steam in practical applications.