KWAME NKRUMAH UNIVERSITY OF SCIENCE AND TECHNOLOGY

KUMASI

DEPARTMENT OF MECHANICAL ENGINEERING

MECHANICAL ENGINEERING LABORATORY

THERMODYNAMICS LABORATORY

THE RELATONTIONSHIP BETWEEN TEMPERATURE AND PRESSURE OF SATURATED STEAM

NAME: RICHARD PUNI

INDEX#: 3756809

GROUP: 7

DATE: 19TH APRIL 2011

The relationship between temperature and pressure of saturated steam.

INTRODUCTION

This experiment is set to investigate the relationship between the

pressure and temperature of saturated steam that was in equilibrium with water. Both

are all at pressures between atmospheric and 150 bar, with a Marcet boiler, which was

developed for this purpose. In this experiment, the liquid(water) is confined in an

enclosure. Thus, neither water nor steam can escape. Hence, as the boiling water is

continuously heated, more steam is produced and this causes the saturated vapour

pressure to increase and also the pressure of the system.

OBJECTIVE

The objective of the experiment is to obtain the relation between the

temperature and pressure of saturated steam and to compare the results with values

obtained from steam tables.

THEORY

From Gibbs’s free function 푑푝 = 푉푑푝 − 푠푑푇 -------- (1)

Since the pressure and temperature are constant during the change of phase, dg is zero, that is, Gibbs’s function is constant. Equating the values of g of the two extreme conditions, we have, gf

= gg,

Where ‘f’ and ‘g’ refer to the saturated liquid and vapour states.

If the pressure is changed by a small amount from P to (P + dp), the saturated temperature will change from Tsat to (T + dt) sat and hence the Gibbs’s function from g to (g + dg).

At this new pressure, we must also have (g + dg) g. It follows that dgf = dgg

From Gibbs’s function, at fixed volume, equation (1) becomes 푑푝 = 푠푑푇---- (2)

The heat added by virtue of which the entropy changes are equal to hfg. However, it is added reversibly at constant temperature.

Hence s= ; where s = entropy

Therefore equation (2) becomes dp = ℎ

And also from Claperyon’s equation: ℎ 푇 푣

ℎ =푇

= =

The significance of the equation is that [dp/dt] sat is the slope of the vapour pressure curve, thus hfg at a given temperature can be determined from the slope of the vapour pressure curve and the specific volumes of saturated vapour at the given temperature.

APPARATUS

Marcet boiler

Thermometer

Blowlamp

Measuring cylinder

Beaker

Tripod stand

DIAGRAM

PRESSURE GAUGE

TAP

CORRECTION CURVE

THERMOMETER

SPHERE CONTAINER

BURSON BURNER

A Marcet Boiler.

PROCEDURE

200 millilitres of water is poured into the spherical container of the Marcet boiler.

The boiler is then heated with the left opened. The water is allowed to boil for a short

time so as to drive off any air in the boiler. The tap is then closed so as to obtain a

closed system. Temperature readings are taken as pressure increases in steps of about

69 kN/m2 (10 lb/sq. in) up to about 1034 kN/m2 (150 lb/sq.in).The source of heat is then

removed and temperature readings taken as the boiler cools at the same pressure as

when the pressure was rising. The mean of the two temperature readings are obtained

and each pressure reading converted into absolute pressure (bar). A graph of absolute

temperature is plotted against absolute pressure (bar).The corresponding values

obtained from steam tables are plotted on the graph in the previous step. A graph of log

T against log P is also plotted.

TABLE OF RESULT

GAUGE PRESSURE

ABSOLUTE PREESURE TEMPERATURE ⁰F

ABSOLUTE TEMPERATURE (K) log T (K) log P (BAR)

STEAM TABLE

(K)

(BAR) (BAR) RISING FALLING RISING FALLING MEAN 0 1.01325 210.92 212.6 372.55 373.75 373.15 2.571883446 0.005716612 373.14

10 × 0.069 1.70325 21632 217.94 375.55 376.45 376 2.575187845 0.231278398 388.34

20 × 0.069 2.39325 241.16 244.94 389.35 391.45 390.4 2.591509809 0.378988068 399.24

30 × 0.069 3.08325 261.96 262.04 400.35 400.95 400.65 2.602765147 0.48900874 407.89

40 × 0.069 3.77325 276.08 276.98 408.75 409.25 409 2.611723308 0.576715581 414.58

50 × 0.069 4.46325 287.06 289.94 414.85 416.45 415.65 2.618727785 0.649651214 420.7

60 × 0.069 5.15325 296.96 300.02 420.35 422.05 421.2 2.624488363 0.712081212 426.05

70 × 0.069 5.84325 305.96 309.02 425.35 427.05 426.2 2.629613445 0.766654468 430.88

80 × 0.069 6.53325 314.96 316.94 430.35 431.45 430.9 2.634376494 0.815129277 435.23

90 × 0.069 7.22325 321.98 323.96 434.25 435.35 434.8 2.638289535 0.858732646 439.33

100 × 0.069

7.91325 327.92 330.98 437.55 439.25 438.4 2.641870545 0.898354886 443.054

110 × 0.069

8.60325 334.04 336.92 440.95 442.55 441.75 2.645176558 0.934662543 446.54

120 × 0.069

9.29325 339.98 343.22 444.25 446.05 445.15 2.648506378 0.96816762 449.845

130 × 0.069

9.98325 345.92 348.8 447.55 448.75 448.15 2.651423401 0.999271947 452.95

140 × 0.069

10.67325 350.06 352.94 449.85 451.45 450.65 2.653839375 1.028296682 455.854

150 × 0.069

11.36325 354.92 357.98 452.55 454.25 453.4 2.656481516 1.055502562 458.644

y = 364.49x0.0888

0

50

100

150

200

250

300

350

400

450

500

0 2 4 6 8 10 12

Absolute temperaturein

kelvin

Absolute pressure in Bar

A graph of Absolute temperature against Absolute pressure

Series1

CALCULATIONS

From the given relation 푇 = 푘푃 taking log of both sides,

We obtain log푇 = 푙표푔푘푃

log푇 = 푛푙표푔푃 + log푘

From the graph, equation of the line is y = 0.0888x + 2.5617, hence,

Log k = 2.5617 ⇒ k = 10 .

푘 = 364.50

Also n is the gradient

푛 = 0.0888

DEDUCTIONS

y = 0.0888x + 2.5617

2.54

2.56

2.58

2.6

2.62

2.64

2.66

2.68

0 0.2 0.4 0.6 0.8 1 1.2

log T

log P

A graph of logT against logP

From graph 1, comparing 푇 = 푘푃 with the equation 푦 = 364.49푥 . the values of n and k (i.e. n = 0.0888 and k = 364.49) are very close to the ones obtained from graph of log T against log P (from calculation above).

PRECAUTIONS

Readings on thermometer and gauge were taken at eye level to avoid errors due to parallax. Tap was opened to ensure moist air and other gases are driven off. Tap was closed tightly to avoid having water leak through. In order to make sure that there was no air in the vessel; the water was heated and allowed to boil for some time before the tap was closed.

CONCLUSION

It can be concluded that the values obtained from the experiment is very close as compared to the ones obtained from the steam table. All the same the values deviate slightly from that obtained from the steam table. At the same absolute pressure, the temperature from the steam table was higher than the ones obtained from the experiment this may be due to inadequate insulation of the pressure vessel.

The absolute temperatures and absolute pressures of saturated steam are directly related even though not perfectly linear over long ranges of readings, there is a corresponding increases in absolute pressure when the absolute temperature increases.

REFERENCE

Applied thermodynamics by T. D. Eastop and A. McConkey

Microsoft Encarta 2009

Richard Puni

BSc. Mechanical Engineering

Kwame Nkrumah University of Science and Technology

19th April 2011