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Unit Two Quiz Solutions and Unit Three Goals

Unit Two Quiz Solutions and Unit Three Goals. Mechanical Engineering 370 Thermodynamics Larry Caretto February 18, 2003. Outline. Quiz Solution Finding work as integral over path Combining work and state questions Unit three – heat, internal energy and the first law of thermodynamics

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Unit Two Quiz Solutions and Unit Three Goals

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  1. Unit Two Quiz Solutions and Unit Three Goals Mechanical Engineering 370 Thermodynamics Larry Caretto February 18, 2003

  2. Outline • Quiz Solution • Finding work as integral over path • Combining work and state questions • Unit three – heat, internal energy and the first law of thermodynamics • Heat, Q, is energy in transit due only to a temperature gradient • Internal energy is a property • Q = DU + W or Qin – Qout = DU + Wout - Win

  3. Solution Continued • The only data for state 2 is T2 = 320 K • We know that P2 and V2 must lie on path • Have to find P2 and V2 that satisfy path equation and v2 = v(P2,T2 = 320 K) in table

  4. Iterations for Final State

  5. Solution Concluded • Iterations give P2 = 7 MPa and V2 = mv2 = 0.195 m3/kg • Can now show W = 0.384 MJ

  6. Unit Three Goals • As a result of studying this unit you should be able to • find properties more easily than you were able to do after units one and two • describe the path for a process and determine the work with more confidence • understand that heat is energy in transit due only to a temperature difference

  7. Unit Three Goals Continued • understand the meaning of the internal energy as a property giving the among of energy stored in a body. • use the first law as Q = DU + W = mDu + W in problem solving • use the sign convention for heat and work • Heat added to a system is positive; heat removed from a system is negative: Q = Qin – Qout • Work done by a system is positive; work done on a system is negative: W = Wout–Win

  8. Unit Three Goals Continued • we do not have to worry about the sign convention when we use W = PdV; we automatically get the correct sign • use the first law as q = Du + w where q = Q/m and w = W/m, in problem solving • find the internal energy in tables from any definition of the state in either the one-phase or mixed region

  9. Unit Three Goals Concluded • find the internal energy from the enthalpy as u = h – Pv • use the enthalpy in place of the internal energy when enthalpy is given in a table but internal energy is not • work problems, possible involving trial-and-error solutions, using the first law, property tables and a path equation

  10. Example Calculation • Given: 10 kg of H2O at 200oC and 50% quality is expanded to 400oC at constant pressure • Find: Heat Transfer • First Law: Q = DU + W = m(u2 – u1) + W • Path: W = PdV = P1-2 (V2 – V1) for constant pressure, P1-2 = P1 = P2 • Use property tables for water

  11. Find Initial State Properties • x1 = 50% quality implies mixed region at T1 = 200oC • P1 = Psat(T1 = 200oC) = 1.5538 MPa • v1 = (1 – x1) vf(T1 = 200oC) + x1 vg(T1 = 200oC) = 0.06426 m3/kg • u1 = (1 – x1) uf(T1 = 200oC) + x1 ug(T1 = 200oC) = 1723.0 kJ/kg

  12. Find Final State Properties • Point 2 has T2 = 400oC and P2 = P1 = 1.5538 MPa • Interpolation in superheat table between 1.40 MPa and 1.60 MPa gives v2 = 0.19646 m3/kg and u2 = 2950.7 kJ/kg

  13. Use Properties to get Answer • Q = DU + W = DU + P1-2 (V2 – V1) = m(u2 – u1) + P1-2 m(v2 – v1) • Q = 14.33 MJ (added to system)

  14. Enthalpy • Defined property H ≡ U + PV • h = H/m = U/m + P(V/m) = u + Pv • Example problem has solution that Q = m(u2 – u1) + P1-2 m(v2 – v1) = m[(u2 + P2v2) - (u1 + P1v1)] = m(h2 – h1) • For constant pressure process Q = DH • H used in open systems and some tables require calculation of u = h - Pv

  15. Is Electricity Heat or Work? • We know that “resistance heating” is proportional to I2R • If the system boundary is defined such that electricity crosses boundary then we do I2R work on system • If resistor is outside system boundary then I2R heat is added to system • Either approach gives DU = I2R

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