1 / 21

Unit One Quiz Solutions and Unit Two Goals

Unit One Quiz Solutions and Unit Two Goals. Mechanical Engineering 370 Thermodynamics Larry Caretto February 11, 2003. Outline. Solution to quiz Look at individual problems Solutions available on line Unit two – work and paths Unit goals available on line Group exercise on Thursday

jory
Télécharger la présentation

Unit One Quiz Solutions and Unit Two Goals

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Unit One Quiz Solutions and Unit Two Goals Mechanical Engineering 370 Thermodynamics Larry Caretto February 11, 2003

  2. Outline • Solution to quiz • Look at individual problems • Solutions available on line • Unit two – work and paths • Unit goals available on line • Group exercise on Thursday • Quiz on Tuesday, February 18

  3. Problem One Solution • Given: P = 2 MPa, T = 200 K, V = 2 m3 • Find the mass, m, using tables • First, find the specific volume from superheat tables v(2 MPa,200 K) = 0.04164 m3/kg • m = 48.03 kg

  4. T, K P,MPa 200 v 2 h s 0.0416 269.92 4.2172 Searching the Table

  5. Problem Two Solution • Given: P = 2 MPa, T = 200 K, V = 2 m3 • Find the mass, m, using ideal gas equation • Still have m = V/v with v = RT/P • m = 48.55 kg, a 1.3% error

  6. Problem Three Solution • Given: Neon at 200 K and 2 MPa cooled at constant volume to 30 K • Find: Final state • For constant volume process initial specific volume = final specific volume = 0. 04164 m3/kg from problem one • Final state is v = 0. 04164 m3/kg and T = 30 K

  7. Problem Three Continued • Check vf and vg at T = 30 K • Find vf = 0.000869 m3/kg < v = 0.4164 m3/kg < vg = 0.05016 m3/kg • Mixed, so P = Psat(30 K) = 0.2238 MPa

  8. Unit Two Goals • As a result of studying this unit you should be able to • find properties more easily than you were able to do after completing unit one • describe the path for a process • find the work as the integral of PdV • find the work as the area under the path on a P-V diagram

  9. Unit Two Goals Continued • understand the difference between the applied force (or pressure) and the system force (or pressure). • recognize that work is always the integral of the applied force (or pressure) over distance (or volume) • recognize when the applied pressure and the system pressure are the same

  10. More Unit Two Goals • use the description of the path as an equation • recognize the difference between the path equation and the equation of state • use the path equation and the equation of state in a trial-and-error procedure to find the final state • use the path equation and the equation of state in a trial-and-error procedure to find the final state when the "equation of state" is a set of tables

  11. Analysis of Work • Start with dW = Fdx • dW = (PA)d(V/A) = PdV • Note that dimensions are (F/L2) times L3 • E. g., pascal-m3 gives N-m or joules • psia-ft3 times 144 in2/ft2 gives ft-lbf • W = PdV over path • Work depends on path

  12. Simple Path • Here we have an initial point (1), a final point (2) • Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2 • Work is positive 1 P 2 V

  13. Integrating a Line • Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2 • Work is positive because if V2 > V1 (and pressure is always positive) • Work is positive when system expands (system does work on surroundings) 1

  14. Reverse Path • Here the initial point (1) and final point (2) are reversed • Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2 • Work is negative 2 P 1 V

  15. Integrating a Reverse Line • In the previous chart the initial point (1) and final point (2) are reversed • Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2 • Work is negative because V2 < V1 • Work is negative when system is compressed (work done on system) 1

  16. More Complex Path • Here we have an initial point (1), a final point (4) and two intermediate points (2 and 3) • Work = area under path = trapezoid area plus rectangle area 2 3 P 4 1 V

  17. How Do We Specify State • To integrate PdV, we need path equation, P(V) • Path equation is integrated along path in terms of P and V • Can specify states in terms of T • Need to use equation-of-state or tables to get v from (P,T) and V = m v

  18. Example Calculation • Given: 10 kg of H2O at 200oC and 50% quality is expanded to 400oC at constant pressure • Find: Work • Equation: W = PdV = P(V2 – V1) for constant pressure • How do we find P, V1 and V2 from data given?

  19. Example Continued • Use tables for H2O to find P and v • Get total volume as V = m v • Initial state is in mixed region so the constant pressure, P = Psat(200oC) = 1.5538 MPa and v = vf + x (vg – vf) • At 200oC, vf and vg, respectively, = 0.001157 and 0.12736 m3/kg • For x1 = 0.5, v1 = 0.06426 m3/kg

  20. Example Concluded • V1 = m v1 = 0.6426 m3 for m = 10 kg • Point 2 has T = 400oC and P = initial P = 1.5538 MPa • From superheat table interpolation this final state has v = 0.19646 m3/kg • Since m = 10 kg, V2 = 1.9646 m3 • W = (1.5538 MPa)(1.9646 – 0.6426) m3

  21. Units for Work • Basic idea is that Pa•m3 gives J, kPa•m3 gives kJ, MPa•m3 gives MJ, • For other pressure units need to convert into kJ; here W = 2,054 kJ • Without unit conversion we could get the work as 2.054 MJ

More Related