Thermodynamics: Work and the First Law

Lecture Learning Outcomes

• Recall the evaluation of expansion work in a closed system.
• State the first law of thermodynamics.
• Apply the first law of thermodynamics to obtain net work done or net heat transfer in processes.
• Apply the energy balance rate to steady and transient state operations.

Worked Example: Work in Process

• Evaluate the work in kJ for a two-step process consisting of an expansion with n = 1.0 from an initial pressure of 3 bar and an initial volume of 0.1 m³ to a final volume of 0.15 m³.
• The process is followed by another expansion with n = 0 from an initial volume of 0.15 m³ to a final volume of 0.2 m³.
• First work process formula: 𝑝1 𝑉1¹ = 𝑝2 𝑉2¹ = 𝑐; 𝑝2 = 𝑝1 (𝑉1/𝑉2)¹ = 2 bar
• Work is determined as 𝑊 = ∫ 𝑝 𝑑𝑉 = ∫ 𝑐/𝑉ⁿ 𝑑𝑉 = ∫ 𝑐/𝑉 𝑑𝑉 = 𝑐 ln 𝑉₂/𝑉₁ = 𝑝₁𝑉₁ ln 𝑉₂/𝑉₁ = +12.16 𝑘𝐽
• Second work process formula: 𝑝2 𝑉2⁰ = 𝑝3 𝑉3⁰ = 𝑐; 𝑝2 = 𝑝3 = 2 bar
• Work is determined with the formula 𝑊 = ∫ 𝑝 𝑑𝑉 = 𝑝 ∫ 𝑑𝑉 = 𝑝(𝑉3 − 𝑉2) = +10 𝑘𝐽
• Total work equals + 22.16 𝑘𝐽

Change in Total Energy

• The change in the total energy of a system in engineering thermodynamics includes three macroscopic contributions.
• The macroscopic contributions include changes in kinetic energy, gravitational potential energy, and internal energy.
• The change in total energy is expressed as 𝐸2 − 𝐸1 = (𝑈2 − 𝑈1) + (𝐾𝐸2 − 𝐾𝐸1) + (𝑃𝐸2 − 𝑃𝐸1)
• The change in total energy can also be written as ∆𝐸 = ∆𝑈 + ∆𝐾𝐸 + ∆𝑃𝐸

Energy Transfer by Heat

• Closed systems interact with their surroundings in ways that cannot be categorized as work.
• When gas in a rigid container interacts with a hot plate, the energy of the gas increases, even though no work is done.
• Energy transfer by heat defines this type of interaction.
• The variable Q denotes the amount of energy transferred across a system's boundary during heat interaction with the surroundings.

Heat Transfer: Sign Convention

• The sign convention for heat transfer is the reverse from the one adopted for work.
• When heat transfers into a system, it is positive, but the reverse is negative.
• Q > 0 defines heat transfer to the system.
• Q < 0 defines heat transfer from the system.

Heat Transfer Details

• The value of Q depends on the interactions between the system and surroundings during a process, not just the end states.
• Heat is not a property of the system or surroundings.
• The amount of energy transfer by heat is given by the integral, where the limits mean 'from state 1 to 2'.
• Limits do not refer to the values of heat at the state and each state does not have a value of heat.

Heat Transfer Rate

• The net rate of heat transfer is denoted by 𝑄.
• Energy transfer by heat during a period of time is found by integrating from time 𝑡1 to 𝑡2, giving us 𝑄 = ∫ 𝑄𝑑𝑡
• A process is called adiabatic if a system undergoes a process involving no heat transfer with its surroundings.

Energy Balance for Closed Systems

• Conservation defines the energy concept.
• This is expressed in the first law of thermodynamics.
• The first law of thermodynamics states that within a system the change in energy equals the net amount of energy transferred in across the system boundary by heat transfer during the time interval minus the net amount of energy transferred out across the system boundary by work during the time interval.

Accounting Balance for Energy

• The first law of thermodynamics serves as an accounting balance for energy.
• In any process of a closed system, the energy of the system either increases or decreases by an amount equal to the net amount of energy transferred across the boundary.
• The first law can be written as 𝐸2 − 𝐸1 = 𝑄 − 𝑊
• The first law is also expressed as ∆𝐾𝐸 + ∆𝑃𝐸 + ∆𝑈 = 𝑄 − 𝑊.

Differential Energy Balance

• In its differential form, the energy balance can be given as 𝑑𝐸 = 𝛿𝑄 − 𝛿𝑊.
• The instantaneous time rate form is expressed as 𝑑𝐸/𝑑𝑡 = 𝑄 − 𝑊.
• This rate form can also be written as 𝑑𝐾𝐸/𝑑𝑡 + 𝑑𝑃𝐸/𝑑𝑡 + 𝑑𝑈/𝑑𝑡 = 𝑄 − 𝑊.
• Important things to not about the energy balance:
  • Be careful about signs and units.
  • Recognize the location of the system boundary.

Cooling Gas in a Piston-Cylinder Assembly

• Four-tenths of a kilogram of a certain gas is contained within a piston-cylinder assembly.
• The gas undergoes a process for which the pressure-volume relationship is 𝑝𝑉1.5 = 𝑐.
• The initial pressure is 3 bar, the initial volume is 0.1𝑚3, and the final volume is 0.2𝑚3.
• The change in specific internal energy of the gas in the process is 𝑢2 − 𝑢1 = −55 kJ/kg.
• There are no significant changes in kinetic or potential energy.
• Engineering model details in the example:
  • The gas is a closed system.
  • The process is described by 𝑝𝑉1.5 = 𝑐
  • There is no change in KE and PE of the system.
• The model follows the equations ∆𝐾𝐸 + ∆𝑃𝐸 + ∆𝑈 = 𝑄 − 𝑊, ∆𝑈 = 𝑄 − 𝑊, and ∆𝑈 = 𝑚(𝑢2 − 𝑢1) = −55𝑘𝐽/𝑘𝑔.
• For the model 𝑊 = +17.6 𝑘𝐽
• The energy balance can be expressed as 𝑄 = ∆𝑈 + 𝑊, 𝑄 = 𝑚(𝑢2 − 𝑢1) + 𝑊 = 0.4(−55) + 17.6, and 𝑄 = −4.4 𝑘𝐽.
• Heat transfers out of the system to the surroundings.

Energy Balance Examples

• Class Exercise 1: If the gas undergoes a process for which 𝑝𝑉 = 𝑐 and ∆𝑢 = 0, determine the heat transfer in kJ keeping the initial pressure and volume fixed.
  • Answer: + 20.79 𝑘𝐽
• Class Exercise 2: A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per kg, an elevation decrease of 50m, and an increase in velocity from 15 m/s to 30 m/s. The specific internal energy decreases by 5 kJ/kg, and the acceleration due to gravity is 9.7m/s2. Determine the heat transfer for the process in kJ.

Steady State Energy Transfer

• A system is at steady state if none of its properties change with time.
• In real applications, devices are assumed to be at steady state when property variations with time are small enough to ignore.

Energy Transfer Rates of a Gearbox at Steady State

• During steady state operation, a gearbox receives 60 kW through the input shaft and delivers power through the output shaft.
• An engineering model example:
  • The gearbox is a closed system at steady state.
  • Convection is the dominant heat transfer.
  • Energy by heat transfers out of the system.
• The rate of energy convection heat transfer is given by 𝑄 = ℎ𝐴(𝑇𝑏 − 𝑇𝑓).
• Given ℎ = 0.171 kW/m2.K as the heat transfer coefficient and 𝐴 = 1.0𝑚2 as the outer surface of the gearbox.
• The temperature of the outer surface is 𝑇𝑏 = 300 𝐾 (27℃), and the surrounding air temperature is 𝑇𝑓 = 293 𝐾 (20℃).
• The steady state energy rate balance reduces to 0 = 𝑄 − 𝑊, or 𝑄 = 𝑊.
• Given that 𝑄 = ℎ𝐴(𝑇𝑏 − 𝑇𝑓) = 0.171 × 1 × (300 − 293) = 1.2 𝑘𝑊, heat is transferred out of the system and 𝑄 = −1.2 𝑘𝑊.
• Solved via the formula −1.2 = −60 + 𝑊2 , 𝑊2 = +58.8 𝑘𝑊
• Evaluate the heat transfer rate and power delivered through the output shaft in kW.

Transient State Operation

• Many devices undergo periods of transient operation where the state changes with time.
• A transient state occurs during startup and shutdown periods
• An example is provided with a motor: the rate of heat transfer between an electric motor and its surroundings varies with time as 𝑄 = −0.2[1 − 𝑒(−0.05𝑡)], measured in kW, where t is time in seconds.
• The shaft of the motor rotates at a constant speed of 𝜔 = 100 𝑟𝑎𝑑/𝑠, applying a torque of ℑ = 18 𝑁. 𝑚.
• The motor draws a constant electrical power input of 2.0 kW.
• Key details:
  • Time-varying rate of heat transfer between motor and surroundings is given.
  • Changes are tracked in the rate of heat transfer, power, and energy as time changes.
• Formula 𝑊𝑠ℎ𝑎𝑓𝑡 = ℑ𝜔 = +1.8 𝑘𝑊
• Formula 𝑊 = −𝑊𝑒𝑙𝑒𝑐 + 𝑊𝑠ℎ𝑎𝑓𝑡 = −2 + 1.8 = −0.2 𝑘𝑊
• Formula 𝑑𝐸/𝑑𝑡 = −0.2[1 − 𝑒(−0.05𝑡)] − (−0.2) = 0.2𝑒−0.05𝑡
• To obtain the change in energy, we perform an integration with the formula ∆𝐸 = ∫ 0.2𝑒−0.05𝑡 𝑑𝑡 = 4[1 − 𝑒(−0.05𝑡)]
• Use a spreadsheet to tabulate 𝑄 and 𝑊 in kW and the change in energy ∆𝐸 in kJ as functions of time from t = 0 s to t = 120 s.

Next Lecture

• Energy balance for power, refrigeration, and heat pump cycles.