Chapter 4 (Lecture 9)

ENERGY ANALYSIS OF CLOSED SYSTEMS: MAE 2131, LECTURE 9

3–3 Phase-Change Processes of Pure Substances

• Figures:

  • Figure 3–10: T-v diagram for the heating process of water at constant pressure.

  • Access the text alternative for slide images. © McGraw Hill LLC

Classification of States for Water:

• Saturated Liquid: Refer to tables A-4 or A-5

• Saturated Vapor: Refer to tables A-4 or A-5

• Mixture of Saturated Liquid and Vapor:

  • Refer to tables A-4 or A-5 and associated quality (x)

• Superheated Vapor: Refer to table A-6

• Compressed Liquid: Refer to table A-7 or table A-4

Property Equations:

• Specific Volume:
  $v = (1 - x)v<em>f + x v</em>g = v<em>f + x(v</em>g - v_f)$ (Eq. 3.2)

• Internal Energy:
  $u = (1 - x)u<em>f + x u</em>g = u<em>f + x(u</em>g - u_f)$ (Eq. 3.6)

• Enthalpy:
  $h = (1 - x)h<em>f + x h</em>g = h<em>f + x(h</em>g - h_f)$ (Eq. 3.7)

• Approximation for Quality:
  $y ext{ is approximately } y_f ext{ at } T$

Example 4-11: Enthalpy of Compressed Liquid

• Objective: Determine the enthalpy of liquid water at 100°C and 15 MPa by multiple methods:

  • (a) Using compressed liquid tables

  • (b) Approximating as a saturated liquid

  • (c) Using correction from Eq. 4-38

Solution:

• Given:

  • At 100°C, saturation pressure of water is 101.42 kPa

  • Since pressure (P) > saturation pressure (P_saturated), the water exists as a compressed liquid.

• (a) From compressed liquid tables:

  • Pressure (P): 15 MPa

  • Enthalpy (h): 430.39 kJ/kg

  • Temperature (T): 100°C (Exact value)

• (b) Approximating as a saturated liquid at 100°C:

  • Resulting enthalpy is in error by about 2.6%.

• (c) From Eq. 4-38:

  • $h ext{ approx } = h_y @ 100°C = 419.17 kJ/kg + (0.001 m³/kg) [(15000 - 101.42) kPa]$

  • $= 434.07 kJ/kg$ (Using Table A-7)

Discussion:

• The correction term reduces the error from 2.6% to approximately 1%. The improvement in accuracy may not always justify the effort.

Example 3-6: Cooling of Superheated Water Vapor

• Problem Statement: One pound-mass of water fills a 2.29 ft³ rigid container at an initial pressure of 250 psia, which is then cooled to 100°F. Determine initial temperature and final pressure of the water.

Solution:

• Initial Specific Volume:
  $V = rac{2.29ft³}{1 lbm} = 2.29 ft³/lbm$

• At 250 psia,:

  • Specific Volume of Saturated Vapor (v_g): 1.8440 ft³/lbm (Table A-5E)

  • Initial state is superheated vapor, given that internal energy (U₁) is greater than U₂.

• State Parameters:

  • $P_1 = 250 psia$

  • $T_1 = 550°F$

  • $v_1 = 2.29 ft³/lbm$

• Final Conditions:

  • Final Temperature (T₂): 100°F

  • Final Specific Volume (V₂): Equal to initial volume, hence $V₂ = V₁ = 2.29 ft³/lbm$

  • Final Pressure (P₂): Saturation pressure at final temperature 100°F $P₂ = P @ 100°F = 0.9505 psia$

Discussion:

• A closed rigid tank maintains constant specific volume, represented as a vertical line in the P-U diagram.

First Law of Thermodynamics for a Closed System

• Energy Change of a System:
  $ext{For fixed boundaries: } E<em>2 - E</em>1 = Q - W$
  ext{For moving boundaries: } \
  \Delta E = \Delta KE + \Delta PE + \Delta U = Q - W

4–1 Moving Boundary Work

• Definition: Moving boundary work (PdV work) represents expansion and compression work in a piston-cylinder device.

Process Descriptions:

• Quasi-equilibrium Process:

  • A process during which the system is nearly in equilibrium at all times.

• Differential Work Done: $\delta W = P \cdot ds$

  • $W<em>b = \int</em>{V<em>1}^{V</em>2} PdV$

• Signs of Work:

  • Work done during expansion: $W_b > 0$

  • Work done during compression: $W_b < 0$

4–2 Energy Balance for Closed Systems

Equation Summary:

• General energy balance equation: $Q - W - W_b = \Delta U$

  • Specific Form:
    $Q - W = \Delta H = m imes (h<em>2 - h</em>1)$
    $Q - W = \Delta U + W<em>b$ $W</em>b = mP<em>0 (v</em>2 - v_1)$

Example 4-1: Boundary Work for a Constant-Volume Process

• Scenario: A rigid tank containing air at 500 kPa and 150°C undergoes cooling, reducing temperature and pressure to 65°C and 400 kPa.

• Analysis:

  • Work done during this process is determined from:
    $W = ext{PdV} = 0$

  • No boundary work is done as dV = 0 for a rigid tank.

Discussion:

• Boundary work during a constant-volume process is always zero, which is indicated by no area under the process curve in the P-V diagram.

Example 4-2: Boundary Work for a Constant-Pressure Process

• Scenario: A frictionless piston-cylinder device contains 10 lbm of steam at 60 psia and 320°F. Heat is added until the temperature reaches 400°F.

Analysis:

• Assumptions: Quasi-equilibrium expansion process.

• Work Done Calculation:

  • $W<em>1 = m P</em>0 (v<em>2 - v</em>1)$

  • Specific volumes at states determined from the superheated vapor table:

  • $v_1 = 7.4863 ft³/lbm$ at 60 psia, 320°F

  • $v_2 = 8.3548 ft³/lbm$ at 60 psia, 400°F

  • Substituting values yields:
    $W = (10 lbm)(60 psia)(8.3548 - 7.4863) ft³/lbm = 96.4 Btu$

Discussion:

• The positive work sign indicates work done by the system; steam utilized 96.4 Btu to perform this work.

Example 4-5: Electric Heating of Water Vapor at Constant Pressure

• Problem: In a piston-cylinder device containing 25 g of saturated water vapor at 300 kPa, an electric current of 0.2 A for 5 min is passed. Heat loss is 3.7 kJ. Determine the final temperature of the steam.

Analysis:

• Need to assess energy input against losses to find final steam temperature.