Cape Peninsula University of TechnologyBellville Campus

Department of Mechanical Engineering

THERMODYNAMICS LABORATORY

THE ENTHALPIES AND ENTROPIES

By L.Nyandu

Subject: APT200SLecturer: Mr Maoda

Evaluation Criteria

Introduction: (Aim for each lab, Background, List of the apparatus, Procedure etc)

10%

Result: (Calculations, Correct method, etc) 55%

Explanations (did you explain what you are doing rather than put formulas.)

10%

Discussion: (Discussion of the results, do they make sense? Any possible errors, etc)

10%

Conclusion and Recommendations: (Did we achieve our aims? What do we need to do to improve our results)

5%

Presentation, layout and neatness: (Cover page, Typed/ print neat, report format, etc)

10%

Total 100%

Date of submission: 28/10/2014

LAB 2: THE ENTHALPIES AND ENTROPIES

The aim of this experiment is to determine the enthalpies and entropies of the steam in different points.

Results obtain from experiment

(a) P|¿|¿= 740 + 101 = 840 kPa = 0.841 MPa

S¿ = Swet 6.835688 = SF + x.Sfg = 0.2821 + x (8.40686) X = 0.78

h0 @ t 0= CP. T 0¿4.187 x 18 = 75.366 KJ/Kg

hwet=¿ h0+xhfg = 75,36 + 0.78 (2458,26)

= 1987,89 KJ/Kg

hg0= h fg−¿ h0 = 2382,89 KJ/Kg

S¿ = 6.835688 h¿=2860.314KJ /Kg @t f=210℃

(b) S0 @ t o=s f@180 c=0.2677KJ /Kg

S fgo=8,44 kJ /kg

Swet= S0+xSfg 0 = 0.2677 + 0.78(8,334) = 6.835 KJ/Kg

Sg0=Sfg 0−S0 ¿6.644626KJ /Kg

S¿ @ t f=6.835KJ /Kg

(C) ∆ hwet= hwet - h0 = 1987.89 -75.366 = 1912.526 KJ/Kg

∆ hg= hg - ho

= 2382.89 – 75.366 = 2307.53 KJ/Kg

∆ h¿= h¿ - h0 = 2860.29 – 75.366 = 2785.03 KJ/Kg

(d) ∆ Swet= Swet - S0 = 6.835 – 0.2677 = 6.557 KJ/Kg

∆ Sg=Sg−So = 8.1763 – 0.2677 = 7.9086 KJ/Kg

∆ S¿=S¿−So = 6.8345 – 0.2677 = 6.567KJ/Kg

(e) ∆ hg= hg - hwet = 2382.89 – 1987.89 = 395 KJ/Kg

∆ h¿= h¿ - hwet = 2860.296 – 1987.89

= 872.405 KJ/Kg

(f) ∆ Sg=Sg−So = 8.1263 – 6.83 = 1.341 KJ/Kg

∆ S¿=S¿−So = 6.835688 – 6.835688 = 0KJ/Kg

Lab 2 readingsReadings Units

Boiler Pressure 841 kPaTemperature of feed water (to) 18 oCTemperature of superheated steam (tf) 210 oC