Name: Julian, John Ralph G. Subject: EC3 - Mechanical Engineering Elective 3 Name of Activity: Assignment 1 – Fundamental Principles of Pumps and Pipings Date Submitted: October 9, 2022 INSTRUCTION PART A: Submit a brief report containing the following: 1. Define, enumerate the assumptions, and based on the assumptions, derive the corresponding Equation of the 1st and 2nd Law of Thermodynamics to following open systems. Provide your own sketch of such system and the nomenclature used in characterizing the variables and properties involved in the system. - - - ο· For Open systems, work is done on/or by mass entering and going out of the system. It is a system in which there is a flow of matter through the boundary and is usually involve with devices that has mass flow. E.g. compressor, turbine, nozzle. 1st law of thermodynamics refers to the conservation of energy, meaning that energy can change from one form to another, but the total amount of energy will always remain constant. Ein=Eout. 2nd law of thermodynamics states that the state of entropy in isolated systems will always increase overtime, and that its increase can never be negative. ds>0. First Law for Open systems ππΈπ π¦π π‘ππ ππ2 ππ2 Μ Μ = π − π + ∑ πΜ π (βπ + + ππ§π ) − ∑ πΜ π (βπ + + ππ§π ) ππ‘ 2 2 Steady State Open System The following can be assumed for steady state open systems a. The state of mass or energy that enters and leaves the system will always be constant and/or will not vary over time. ∑ πΜ π − ∑ πΜ π = 0 = Δππ π¦π π‘ππ - b. The change in mass or energy in the system will always equal to zero. πππ π¦π π‘ππ ππΈπ π¦π π‘ππ = =0 ππ‘ ππ‘ Steady State Energy Equation: substituting the assumption for steady state open system to the first law for open systems, therefore: ππΈπ π¦π π‘ππ ππ2 ππ2 Μ Μ = 0 = π − π + ∑ πΜ π (βπ + + ππ§π ) − ∑ πΜ π (βπ + + ππ§π ) ππ‘ 2 2 π½ππ π½ππ ∴ πΈΜ = πΎΜ − ∑ πΜ π (ππ + + πππ ) + ∑ πΜ π (ππ + + πππ ) π π Note that this equation is also applicable for other steady open systems. Sketch of a steady state open system ο· - m_in m_system m_out E_in E_system E_out Steady Flow Open System The following can be assumed for steady flow open systems. The flow rate may or may not differ at different points or cross section in the system, but it does not change over time. For example: At initial time, ti: Pt.1 Pt.2 πΜ = 1 πΜ = 3 After some time, to: Pt.1 Pt.2 πΜ = 1 πΜ = 3 - For the steady flow energy equation, the steady state energy equation is also applicable to this open system. ο· Steady State Steady Flow (SSSF) Open System The following assumptions can be used for SSSF Open Systems: a. The state and/or flow rate of mass and energy that enters and leaves the system will always be constant and will not vary over time. b. The change in mass or energy in the system will always equal to zero. c. At certain point/cross section in the system, the flow rate may differ in value but will remain constant in respect to time. At initial time, ti: m_in m_system m_out E_in E_system E_out After some time, to: ο· - ο· - m_in m_system m_out E_in E_system E_out Uniform Flow Open System This type of open system is when the magnitude of flow rate in any point of the system is constant. Pt.1 Pt.2 πΜ = 1 πΜ = 1 Steady State Uniform Flow System This type of open system is when the magnitude of flow rate in any point of the system is constant and does not vary over time. At initial time, ti: Pt.1 Pt.2 πΜ = 1 πΜ = 1 After some time, to: Pt.1 Pt.2 πΜ = 1 πΜ = 1 2. Define the Mass Continuity Equation, enumerate the underlying assumptions, and derive the corresponding equation. Provide your own sketch of the system and the nomenclature used. - The mass continuity equation states that the mass is conserved or constant, meaning that the summation of mass that goes in will always be equal to the summation of mass that goes out. ∑ πΜ π = ∑ πΜ π = ππππ π‘πππ‘ Knowing that πΜ = ππΜ ; πππ πΜ = π΄π£ Therefore: ∑(πΜ π = ππ π΄π π£π ) = ∑(πΜ π = ππ π΄π π£π ) = ππππ π‘πππ‘ Μ = ππ¨π ∴π π€βπππππ, πΜ = πππ π ππππ€ πππ‘π π = ππππ ππ‘π¦ πΜ = π£πππ’ππ ππππ€ πππ‘π π΄ = ππππ π π πππ‘πππππ ππππ π£ = π£ππππππ‘π¦ πΜ π = ππ π΄π π£π πΜ π = ππ π΄π π£π PART B: Review Problems. Please attach the solutions. - For the average velocity of the fluid, π£ππ£π ππ π£: Deriving from the parabolic profile y (0, R) 0 v 0 1 (Vmax,0) π π£ππ£π = 2π ∫−π π£ππ¦ – eqn 0 From the equation of sideway parabola : π(π¦ − π)2 = (π£ − β) At vertex (β, π) = (π£πππ₯ , 0) π£ − π£πππ₯ = π(π¦ − 0)2 -- eqn 1 To solve for C, let (π£, π¦) = (0, π ) 1− π£πππ₯ = π(π 2 ) π£ πΆ = − πππ₯ – eqn 2 2 π Substitute eqn 2 to eqn 1 π£πππ₯ ∴ π£ − π£πππ₯ = − 2 π¦ 2 π 1− π£ π£πππ₯ π¦ 2 = (π ) π¦ 2 π£ = π£πππ₯ (1 − (π ) ) -- eqn 3 - Substitute eqn 3 to eqn 0 π πΉ π π ∴ ππππ = ∫ (ππππ (π − ( ) )) π π ππΉ −πΉ πΉ For the volumetric flow rate, πΜ : πΜ = π΄π£ππ£π ∴ π½Μ = - π¨ πΉ π π ∫ (π (π − ( ) )) π π ππΉ −πΉ πππ πΉ For the mass flow rate, πΜ πΜ = ππΜ ππ¨ πΉ π π ∴ πΜ = ∫ (ππππ (π − ( ) )) π π ππΉ −πΉ πΉ References: ο§ ο§ ο§ ο§ ο§ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_M aps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/The _Four_Laws_of_Thermodynamics https://www.quora.com/What-is-the-basic-difference-between-steady-flow-and-uniform-flow https://www.researchgate.net/project/Book-on-Thermodynamics-for-Beginners http://www.csun.edu/~lcaretto/me370/unit05.pdf http://www.mem50212.com/MDME/MEMmods/MEM23006A/thermo/open_systems.html
