Exam 1 Prep Review

Problem Solving for Exam Preparation

Introduction

• The session focuses on solving problems in preparation for an upcoming exam.
• The speaker encourages students to interrupt for clarification if necessary.

Problem 1: Isothermal Compressibility and Work Done in Isobaric Process

• Initial Given Information:
  • Isothermal compressibility and volume expansivity values are provided.
  • The goal is to find the work needed to isobarically raise the temperature of a real fluid.
• Key Assumption:
  • Assume the process is reversible, simplifying calculations.
• Work Calculation Steps:
  • Work, $W = - ext{e} dB$, where $ ext{e}$ is the isothermal compressibility. In this case, $W = -P dV$ because it is an isobaric process.
  • The isobaric compressibility relationship provides:
    $\beta = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_V$
  • For volume changes, perform integration of the equation using separation of variables:
  • $\frac{dB}{V} = \beta dT$
  • Integrate from initial state to final state:
    $\ln \left( \frac{V<em>2}{V</em>1} \right) = \beta (T<em>2 - T</em>1)$
• Volume Change Calculation:
  • Rearranging yields:
    $V<em>2 = V</em>1 e^{\beta (T<em>2 - T</em>1)}$
  • Calculate $V_2$ as 0.506 liters/kg, given specific numerical values from the problem.
• Work Calculation Details:
  • Substitute into the work expression: $W = -PV<em>2 + PV</em>1$
  • Be cautious of units:
  • Convert bar to Pascal and liters to cubic meters to maintain SI units.
  • Completed calculations yield $W = 140.56$ joules.
• Critical Thinking on Work Sign:
  • Clarified potential confusion on sign error; the work should be negative due to increasing volume driven by temperature increase in isobaric conditions.

Problem 2: Turbine Work and Efficiency Calculation

• Scenario Overview:
  • An elevator powered by a steam turbine with energy transfer efficiency of 80%.
• Key Goals:
  • Determine work done by the turbine in two scenarios: adiabatic and non-adiabatic (losing heat).
• System Definition:
  • The turbine is the defined system, crucial for analyzing energy transfer.
  • Establish mass flow rate based on problem parameters: $\dot{m} = 5$ kg/s.
• Power Requirements:
  • Total power requirement for the elevator is 2,500 kW leading to:
    $W = -\frac{2500}{0.8} = -3125 kJ/s$
• Material and Energy Balance:
  • Apply the steady-state material balance:
  • Mass in = Mass out = 5 kg/s.
  • Energy balance (adiabatic):
  • Since $Q = 0$, the equation simplifies to:
    $\dot{m} (H<em>2 - H</em>1) = W$
• Enthalpy Determination:
  • Enthalpy at the inlet ($H_1$) is given from steam tables for superheated steam.
  • Substitute enthalpy values to solve for $H2$: $H</em>2 = W/\dot{m} + H_1$
  • Example values result in: $H_2 = 2823$ kJ/kg.
  • Determine temperature at outlet using interpolation between temperature and enthalpy tables. Results in: 172 °C from provided enthalpy values.
• Scenario B (non-adiabatic):
  • Update energy balance with heat loss, now $Q = 500 kJ/s$.
  • Rearranging yields:
    $H<em>2 = \frac{W + Q}{\dot{m}} + H</em>1$
  • Calculate new enthalpy as $H_2 = 2720$ kJ/kg, and interpolate for outlet temperature, resulting in 121.8 °C.

Problem 3: Ideal Gas Behavior during Heating, Compression, and Cooling

• Process Overview:
  • A cycle involving isobaric heating, adiabatic compression, and isochoric cooling of an ideal gas.
  • Initial conditions specified: 1 bar, 300 K to be heated to 750 K.
• Isobaric Heating:
  • Given that $Q = \Delta H = C<em>P (T</em>2 - T_1)$,
  • Calculate values for $\Delta U$ using $C_V$.
  • Net work is deduced from $\Delta U = Q + W$. Resulting in $W = -3741.3$ J/mol for this process.
• Adiabatic Process:
  • For adiabatic compression, establish conditions:
  • $Q = 0$, apply relevant enthalpy and internal energy equations. Results need validating through known relationships like $h$, $u$, or both.
• Isochoric Cooling:
  • Confirm $W = 0$ due to constant volume conditions.
  • Final checks on the overall cycle must confirm the cyclical nature where net changes should reach 0 for $\Delta U$ and $\Delta H$.
• Graphical Representation:
  • Outline of process expected to capture state changes effectively on PV diagrams, with critical consideration for behaviours of adiabatic processes (concave up forms as temperature drops).
• Summary Checks:
  • Validate overall energy and heat exchanges across the full cycle emphasizing the conservation first law. Conditions like $Q$ and $W$ should balance appropriately per the first law's implications.

Reflection on Problem Solving Strategy

• Keywords Identification:
  • For Problem 1: "isobaric" was crucial for simplifying assumptions.
  • For Problem 2: "80% efficiency" and "loses heat" dramatically shaped equations.
  • For Problem 3: "isobaric", "adiabatic", and "isochoric" clearly outlined steps and expectations for energy transfers.
• Preparation involves recognizing and isolating these key terms to facilitate clear logical pathways through problems.

Conclusion

• The session concludes with reminders on exam focus and availability for questions.
• Highlights thorough systematic approaches in handling thermodynamic problems and emphasizes critical thinking, careful reading of questions, and the significance of units throughout calculations.