final lab report for thermos 2(mech)
TRANSCRIPT
Cape Peninsula University of TechnologyBellville Campus
Department of Mechanical Engineering
THERMODYNAMICS LABORATORY
THE DRYNESS FRACTION OF THE STEAM
By L.Nyandu
Subject: APT200SLecturer: S Makhomo
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:
LAB 1: THE DRYNESS FRACTION OF THE STEAM
The aim of this experiment is to determine the dryness fraction of wet steam.
1. Apparatus• Steam boiler plant.
• Separating and throttling calorimeter.
• Measuring beaker
2. Theoretical background
According to Philip (1998) the separating calorimeter is a vessel used initially to separate some of the moisture from the steam, to ensure superheat conditions after throttling. The steam is made to change direction suddenly; the moisture droplets, being heavier than the vapor, drop out of suspension and are collected at the bottom of the vessel.The throttling calorimeter is a vessel with a needle valve fitted on the inlet side. The steam is throttled through the needle valve and exhausted to the condenser. Suppose M kg of wet steam with a dryness fraction of x enters the separating calorimeter. The vapor part will have a mass of x M kg and the liquid part will have a mass of M kg. In the separating calorimeter part of the liquid, say1 kg will be separated from the wet steam. Hence the dryness fraction of the wet steam will increase to x1 which will pass through the throttling process valve. After the throttling process the steam in the throttling calorimeter will be in superheated state.
3. Procedure1. Start the boiler and supply steam to the separating and throttling
calorimeter unit.
2. Start the cooling water flow through the condenser.
3. Open steam valve and allow the steam to flow through the calorimeters to warm through the steam.
4. Open the throttle valve and adjust to give a pressure at exhaust of about 5cm Hg measured on the manometer.
5. Drain the separating calorimeter.
6. Start the test and take readings at 2-3 minutes intervals.
7. When a reasonable quantity of condensate is collected measure the quantity of separated water and the quantity of condensate.
Figure2. T-S diagram of the separating and throttling calorimeter.From the steady flow energy equation: Q – W = hC - hB
Since throttling takes place over a very small distance, the heat transfer is negligible Q = 0. Then the steady flow energy equation for the throttling process becomes,
hC = hB
Hence, enthalpy after throttling = enthalpy before throttlinghC=hf + xhfg
If the pressure of the steam before throttling, the pressure and temperature of the steam after throttling, are known the value of x can be calculated using steam tables.
Dryness Fraction =
Therefore, ws
so mm
mx
+=
Where, Ms the mass of dry steam and Mw is the mass of suspended water separated from the calorimeter.
Results
●C
A B ● ●
P2
P1
T2
T1
T
S
Absolute Pressure = Gauge Pressure + Atmospheric pressure
=
=
= 841.33 KPa and = 30 KPa
= 112.5 and = 69.06
The above written values were calculated using interpolation method and values from the steam tables.
Separating calorimeter only (xo):
Ms: 1830mL = 1.83L = 1.83
Mw: 94mL = 0.94L= 0.94
=
=
= 0.95
Throttling calorimeter only x1
Specific enthalpy before throttling at P1 = specific enthalpy after throttling at P2
(Using interpolation)
= +
=
=
This means it lies outside the dome
=
= 0,951
Table 1: Lab readingsReading Units
Volume of Steam (Vs) 0.00183 m3
Volume of Water (Vw) 0.000094 m3
Pressure before throttling (P1) 840 kPaPressure After throttling (P2) 30 kPaTemperature of steam before throttling (t1) 112,5 oCTemperature of steam After throttling (t2) 69,06 oC
4. Discussion Using the readings that were recorded in the lab the dryness fraction of the steam could be found, using the theoretical equations. From the results obtained the dryness fraction is 0.957 at x0 and 1.0064 at x1. The combined separating and throttling calorimeter was found by using equation 7 where both x0 and x1 were multiplied to get 0.951.
5. Conclusion It can be concluded that the experiment was successful. The dryness fraction of the steam was found using the readings found in the lab.
6. Recommendations The equipment in the lab should be either replaced or maintained in order to get more accurate readings. The students should be given the opportunity to record more than one value per reading so a mean can be obtained or so that there can be more certainty for each reading.
Cape Peninsula University of TechnologyBellville Campus
Department of Mechanical Engineering
THERMODYNAMICS LABORATORY
THE ENTHALPIES AND ENTROPIES
By Student name Student no Signature
L. Nyandu
Subject: APT200SLecturer: S Makhomo
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:
LAB 2. THE ENTHALPIES AND ENTROPIES
(a) = 740 + 101
= 840 KPa = 0.841 MPa
=
6.835688 = + x.Sfg
= 0.2821 + x (8.40686) X = 0.78
@ = .
4.187 x 18
= 75.366 KJ/Kg
= 75,36 + 0.78 (2458,26)
= 1987,89 KJ/Kg
=
= 2382,89 KJ/Kg
= 6.835688
@ =210
(b) @
=
= 0.2677 + 0.78(8,334)
= 6.835 KJ/Kg
@
(C) = -
= 1987.89 -75.366
= 1912.526 KJ/Kg
= -
= 2382.89 – 75.366
= 2307.53 KJ/Kg
= -
= 2860.29 – 75.366
= 2785.03 KJ/Kg
(d) = -
= 6.835 – 0.2677
= 6.557 KJ/Kg
= 8.1763 – 0.2677
= 7.9086 KJ/Kg
= 6.8345 – 0.2677
= 6.567KJ/Kg
(e)= -
= 2382.89 – 1987.89
= 395 KJ/Kg
= -
= 2860.296 – 1987.89
= 872.405 KJ/Kg
(f)
= 8.1263 – 6.83
= 1.341 KJ/Kg
= 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
Cape Peninsula University of TechnologyBellville Campus
Department of Mechanical Engineering
THERMODYNAMICS LABORATORY
BOILER EFFICIENCY & CONDENSERS
By Student name Student no
Signature
L.Nyandu
Subject: APT200SLecturer: S Makhomo
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:
LAB 3. THE EFFICIENCIES
Boiler efficiency
Equivalent Evaporation
Rackine efficiency
Lab 3 readingsReading
Units
Volume of fuel 5.9 x10-4 m3
Time 2.22 minMass flow rate of steam 318.18 kg/hCV of fuel 45.5 MJ/kgDensity of fuel 0.84 kg/L
Cape Peninsula University of TechnologyBellville Campus
Department of Mechanical Engineering
THERMODYNAMICS LABORATORY
BOILER EFFICIENCY & CONDENSERS
By Student name Student no
Signature
L.Nyandu
Subject: APT200SLecturer: S Makhomo
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:
LAB 4. THE CONDENSERS
Pressure inside the condenser
Dryness fraction of turbine exhaust steam entering condenser
(a) Condenser efficiency(ηc)
Vacuum efficiency (ηvac)
Pa = ρmghm
= 13.6 x 9.81x hm(m) / 1000
Note: Relative density of mercury is 13.6PA = Pg + 0.133h (kPa) or PA = Pg + hm / 7.5 (kPa)
Lab 4 readingsReading
Units
Mass flow rate of cooling water (Mc/w) 5.09 kg/minInlet temperature to condenser of cooling water (tw1)
21 oC
Outlet temperature to condenser of cooling water (tw2)
27 oC
Condenser Pressure (Pc) 75 kPaCondensate Temperature (tk) 46 oC