Math Problem Statement

The rigid tank illustrated in the following figure has a volume of 0.06 m³ and initially contains a two-phase liquid–vapor mixture of H₂O at a pressure of 15 bar and a quality of 20%. As the tank contents are heated, a pressure-regulating valve keeps the pressure constant in the tank by allowing saturated vapor to escape. Neglecting kinetic and potential energy effects. (a) Determine the total mass in the tank, in kg, and the amount of heat transfer, in kJ, if heating continues until the final quality is x = 0.5. (b) Plot the total mass in the tank, in kg, and the amount of heat transfer, in kJ, versus the final quality x ranging from 0.2 to 1.0.

Solution

To solve this problem, we need to analyze the properties of the water (H₂O) in the tank using thermodynamic principles. Let's break down the steps for each part:

Part (a) Determining the Total Mass and Heat Transfer:

  1. Initial Conditions:

    • Volume V=0.06 m3V = 0.06 \ \text{m}^3
    • Pressure p=15 barp = 15 \ \text{bar}
    • Initial quality xinitial=0.2x_{initial} = 0.2

    At p=15 barp = 15 \ \text{bar}, we need to determine the specific volumes of saturated liquid (vfv_f) and vapor (vgv_g) from steam tables:

    • vfv_f: Specific volume of saturated liquid at 15 bar.
    • vgv_g: Specific volume of saturated vapor at 15 bar.
  2. Calculate Initial Specific Volume: The specific volume of the mixture can be calculated using the quality: vinitial=vf+xinitial(vgvf)v_{initial} = v_f + x_{initial} (v_g - v_f)

  3. Calculate Total Mass: Using the specific volume and the volume of the tank, we can find the mass: m=Vvinitialm = \frac{V}{v_{initial}}

  4. Final Conditions:

    • Final quality xfinal=0.5x_{final} = 0.5
    • The pressure remains constant at 15 bar.

    Use the same formula for the specific volume at the final state: vfinal=vf+xfinal(vgvf)v_{final} = v_f + x_{final} (v_g - v_f)

    Since the volume of the tank is constant, any vapor that is released through the pressure-regulating valve will affect the mass of water remaining in the tank.

  5. Heat Transfer Calculation: Use the energy balance: Q=m(ufinaluinitial)Q = m \cdot (u_{final} - u_{initial}) where:

    • uinitial=uf+xinitial(uguf)u_{initial} = u_f + x_{initial} (u_g - u_f)
    • ufinal=uf+xfinal(uguf)u_{final} = u_f + x_{final} (u_g - u_f)
    • ufu_f and ugu_g: Specific internal energies of saturated liquid and vapor at 15 bar.

    This calculation will give us the heat transfer required to achieve the final quality.


Part (b) Plotting Total Mass and Heat Transfer:

For this part, we need to determine the total mass and heat transfer for a range of qualities from x=0.2x = 0.2 to x=1.0x = 1.0:

  1. Range of Qualities: xx from 0.2 to 1.0.

  2. Calculate Specific Volume and Mass for Each xx: v(x)=vf+x(vgvf)v(x) = v_f + x(v_g - v_f) m(x)=Vv(x)m(x) = \frac{V}{v(x)}

  3. Calculate Heat Transfer for Each xx: Use the same energy balance: Q(x)=m(x)(u(x)uinitial)Q(x) = m(x) \cdot (u(x) - u_{initial}) where u(x)=uf+x(uguf)u(x) = u_f + x(u_g - u_f).

  4. Plotting:

    • Create two plots:
      • Total mass mm vs. quality xx.
      • Heat transfer QQ vs. quality xx.

Let me know if you need help with specific values from the steam tables or with the calculation steps in detail! Would you like a detailed explanation for part (a) or help with plotting for part (b)? Here are some follow-up questions to consider:

  1. What are the specific volume values (vfv_f and vgv_g) for H₂O at 15 bar?
  2. How is quality (xx) used in calculating specific volumes of a mixture?
  3. What is the significance of a pressure-regulating valve in this problem?
  4. How can we find the internal energy (uu) values from steam tables?
  5. What are the implications of neglecting kinetic and potential energy effects in energy balance?

Tip: When using steam tables, be sure to match the pressure with the saturation conditions to find the correct values of vfv_f, vgv_g, ufu_f, and ugu_g.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Quality (vapor-liquid ratio)
Energy balance

Formulas

v_initial = v_f + x_initial (v_g - v_f)
m = V / v_initial
Q = m * (u_final - u_initial)
v(x) = v_f + x(v_g - v_f)
m(x) = V / v(x)
Q(x) = m(x) * (u(x) - u_initial)

Theorems

First Law of Thermodynamics
Ideal gas law approximation (for thermodynamic properties)

Suitable Grade Level

College-level thermodynamics