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ENTC 370: Announcements. Homework assignments No.4: Assigned Problems: 4.13, 4.23, 4.32, 4.37, 4.44, 4.46, 4.52, 4.81 4.85. Due Thursday , October 9 th before 10:50 am For more information, go to: http://etidweb.tamu.edu/classes/entc370. ENTC 370: Announcements. Exam I:

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## ENTC 370: Announcements

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**ENTC 370: Announcements**• Homework assignments No.4: • Assigned Problems: • 4.13, 4.23, 4.32, 4.37, 4.44, 4.46, 4.52, 4.81 4.85. • Due Thursday, October 9th before 10:50 am • For more information, go to: • http://etidweb.tamu.edu/classes/entc370**ENTC 370: Announcements**• Exam I: • Thursday, Oct 23rd • Chapters 1 – 5 • Homeworks1 – 5 • Closed book/closed notes • Students will be allowed to bring own equation sheet • Double-sided is ok • Size: 8½ x 11**Specific Heat**• Indicator of energy storage capabilities of various substances (physical property) • Defined as the energy required to raise the temperature of a unit mass of a substance by one degree • Two types: • Specific Heat at constant volume, CV • Specific Heat at constant pressure, CP**These are property relations or independent of the type of**process Valid for any substance Units: kJ/kg-K or kJ/kg-° C Subscripts v and P stand for constant specific volume and constant pressure, respectively. Formal Definitions of Specific Heat**Internal Energy (u), Enthalpy (h) and Specific Heats (cp &**cv) of Ideal Gases • For gases, u and h only depends on Temperature: u = f(T) In the case of air: Use Table A-17**Example**For 5 min • Determine the final temperature, T2 • Identify type of fluid, type of device, energy flows • What remains constant? • Write down 1st Law and solve for unknown (T2)**Example**Mass of piston is such that a 350 kPa pressure is required to move it. http://brod.sfsb.hr/test/testhome/vtAnimations/animations/chapter05/closed/processes/uniform/index.html Heat is added until the volume doubles. Find final temperature, work done by the air and total heat transferred to air.**Internal Energy, Enthalpy and Specific Heats of Solids and**Liquids • Solids and liquids are incompressible substances, r ≠ f(Pressure); density remains rather constant regardless of pressure • CP = CV = C ← Good approximation for small changes in temperature**Internal Energy, Enthalpy and Specific Heats of Solids and**Liquids • For solids • For liquids • Constant Pressure Process • Constant Temperature Process**Example**• A 50 kg iron block at 80 °C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25 °C. Determine the temperature when thermal equilibrium is reached. • Identify material type (solid, liquid, etc.) • Write down 1st law for the described system • Solve for unknown**Chapter 5: Mass and Energy Analysis of Control Volumes**(Open Systems) Conservation of Mass Principle • Continuity for any fluid (gas or liquid) • Mass flow rate In = Mass Flow Rate out • r1*A1*v1 = r2*A2*v2 • Continuity for liquids (incompressible flow) • A1*v1 = A2*v2**Mass Flow Rates**Mass flow rate for any open system: Mass flow rate: The average velocity Vavg is defined as the average speed through a cross section.**Volume Flow Rates**Definition of average velocity: Volume flow rate: The volume flow rate is the volume of fluid flowing through a cross section per unit time. The average velocity Vavg is defined as the average speed through a cross section.**Conservation of Mass Principle**Mass In – Mass Out = Net Change in Mass within system**Conservation of Mass Principle**The conservation of mass principle for a control volume: The net mass transfer to or from a control volume during a time interval t is equal to the net change (increase or decrease) in the total mass within the control volume during t.**Conservation of Mass for General Control Volume**The conservation of mass principle for the open system or control volume is expressed as**Conservation of Mass Principle**General conservation of mass: General conservation of mass in rate form: or**Control surface**Control Volume Mass in system, m(t) Mass flow into control surface, kg/s An unsteady situation: dmC.V./dt >0**A steady situation: dmC.V./dt =0**Control surface Control Volume Mass outflow Mass inflow**Steady Flow Process**During a steady-flow process, the total amount of mass contained within a control volume does not change with time (mCV = constant). Then the conservation of mass principle requires thatthe total amount of mass entering a control volume equal the total amount of mass leaving it.**Steady-state, Steady-Flow Conservation of Mass**Since the mass of the control volume is constant with time during the steady-state, steady-flow process, the conservation of mass principle becomes**Flow Work**• P=F/A (definition of pressure) • F = P*A (finding force using pressure) • Work = Force*Distance • Work = F*L = P*A*L = P*Volume • Work/mass = P*v**Total Energy of Simple Compressible System**Consists of three parts without flow work: Internal Energy Kinetic Energy Potential Energy

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