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

2. 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 =  U + W or Q in – Q out =  U + W out - W in

3. 3 Quiz Solution Given: T 1 = 38 K, V 1 = 0.1 m 3, m = 10 kg, T 2 = 320 K, and path is P = P 1 +  (V 2 – V 1 ) Find the work, W, using tables First, note that path is a straight line so that W = (P 1 + P 2 )(V 2 – V 1 )/2 V 1 is given, must find P 1, P 2 and V 2 From tables we see that v f (38 K) < v 1 < v g (38 K) so P 1 = P sat (38 K) = 1.078 MPa

4. 4 Solution Continued The only data for state 2 is T 2 = 320 K We know that P 2 and V 2 must lie on path Have to find P 2 and V 2 that satisfy path equation and v 2 = v(P 2,T 2 = 320 K) in table

5. 5 Iterations for Final State P 2 (MPa)v 2 pathv (P 2,T 2 )v 2 delta 20.0114790.06660.055121 40.0146870.03360.018913 70.0195 0 100.0243130.0138-0.01051

6. 6 Solution Concluded Iterations give P 2 = 7 MPa and V 2 = mv 2 = 0.195 m 3 /kg Can now show W = 0.384 MJ

7. 7 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

8. 8 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 =  U + W = m  u + 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 = Q in – Q out Work done by a system is positive; work done on a system is negative: W = W out – W in

9. 9 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 =  u + 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

10. 10 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

11. 11 Example Calculation Given: 10 kg of H 2 O at 200 o C and 50% quality is expanded to 400 o C at constant pressure Find: Heat Transfer First Law: Q =  U + W = m(u 2 – u 1 ) + W Path: W =  PdV = P 1-2 (V 2 – V 1 ) for constant pressure, P 1-2 = P 1 = P 2 Use property tables for water

12. 12 Find Initial State Properties x 1 = 50% quality implies mixed region at T 1 = 200 o C P 1 = P sat (T 1 = 200 o C) = 1.5538 MPa v 1 = (1 – x 1 ) v f (T 1 = 200 o C) + x 1 v g (T 1 = 200 o C) = 0.06426 m 3 /kg u 1 = (1 – x 1 ) u f (T 1 = 200 o C) + x 1 u g (T 1 = 200 o C) = 1723.0 kJ/kg

13. 13 Find Final State Properties Point 2 has T 2 = 400 o C and P 2 = P 1 = 1.5538 MPa Interpolation in superheat table between 1.40 MPa and 1.60 MPa gives v 2 = 0.19646 m 3 /kg and u 2 = 2950.7 kJ/kg

14. 14 Use Properties to get Answer Q =  U + W =  U + P 1-2 (V 2 – V 1 ) = m(u 2 – u 1 ) + P 1-2 m(v 2 – v 1 ) Q = 14.33 MJ (added to system)

15. 15 Enthalpy Defined property H ≡ U + PV h = H/m = U/m + P(V/m) = u + Pv Example problem has solution that Q = m(u 2 – u 1 ) + P 1-2 m(v 2 – v 1 ) = m[(u 2 + P 2 v 2 ) - (u 1 + P 1 v 1 )] = m(h 2 – h 1 ) For constant pressure process Q =  H H used in open systems and some tables require calculation of u = h - Pv

16. 16 Is Electricity Heat or Work? We know that “resistance heating” is proportional to I 2 R If the system boundary is defined such that electricity crosses boundary then we do I 2 R work on system If resistor is outside system boundary then I 2 R heat is added to system Either approach gives  U = I 2 R