Estimate the constant specific heats for R-134a from Table

Problem 129HP Chapter 3

Fundamentals of Thermodynamcs | 8th Edition

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Problem 129HP

Estimate the constant specific heats for R-134a from Table B.5.2 at 100 kPa and 125°C. Compare this to the specific heats in Table A.5 and explain the difference.

TABLE B.5.2 Superheated R-134a

 Temp. v u h s v u h s (°C) (m3/kg) (kJ/kg) (kJ/kg-K) (kJ/kg-K) (m3/kg) (kJ/kg) (kJ/kg) (kJ/kg-K) 50 kPa (−40.67°C) 100 kPa (−26.54°C) Sat. 0.36889 354.61 373.06 1.7629 0.19257 362.73 381.98 1.7456 -20 0.40507 368.57 388.82 1.8279 0.19860 367.36 387.22 1.7665 -10 0.42222 375.53 396.64 1.8582 0.20765 374.51 395.27 1.7978 0 0.43921 382.63 404.59 1.8878 0.21652 381.76 403.41 1.8281 10 0.45608 389.90 412.70 1.9170 0.22527 389.14 411.67 1.8578 20 0.47287 397.32 420.96 1.9456 0.23392 396.66 420.05 1.8869 30 0.48958 404.90 429.38 1.9739 0.24250 404.31 428.56 1.9155 40 0.50623 412.64
Step-by-Step Solution:

Solution 129HP

Step 1 of 3:

We are going to compare the constant specific heats for R-134a from the tables B.5.2 and A.5 for the given temperature and pressure values.

The given pressure P = 100 kPa

The given temperature T = 125°C

The table B.5.2 for the corresponding temperatures is listed below.

 Temp. v u h s v u h s (°C) (m3/kg) (kJ/kg) (kJ/kg-K) (kJ/kg-K) (m3/kg) (kJ/kg) (kJ/kg) (kJ/kg-K) 50 kPa (−40.67°C) 100 kPa (−26.54°C) 120 0.63835 480.44 512.36 2.2128 0.31797 480.16 511.95 2.1555 130 0.65479 489.63 522.37 2.2379 0.32626 489.36 521.98 2.1807

The specific enthalpy at T120 (h1) = 511.95 kJ/kg

The specific enthalpy at T130 (h2) = 521.98 kJ/kg

The specific internal energy at T120 (u1) = 480.16 kJ/kg

The specific internal energy at T130 (u2) = 489.36 kJ/kg

The approximate specific heats at the temperature T are given by

The specific heat at constant pressure is expressed as

Putting the values

The specific heat at constant volume is expressed as

Putting the values

Step 2 of 2

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