Module 1 Property Relationship

MODULE-I --- Property Relationships Relationships for Pure Substances and Mixtures APPLIED THRMODYNAMICS

2014

VTU-NPTEL-NMEICT Project Progress Report

The Project on Development of Remaining Three Quadrants to NPTEL Phase-I under grant in in aid NMEICT, NMEICT, MHRD, MHRD, New Delhi

t c e j o r P T SME Name : C I E M Course Name: N L E web Type of the Course T P N U T Module I V Subject Matter Expert Details

Dr.A.R.ANWAR KHAN Prof & H.O.D Dept of Mechanic Mechanical al Engineering Applied Thermodynamics 

DEPARTMENT OF MECHANIC MECHANICAL AL ENGINEERING, GHOUSIA COLLEGE OF ENGINEERING, RAMANAGARA -562159

Dr.A.R.ANWAR KHAN,Prof & HOD, HOD , GHOUSIA COLLEGE OF ENGINERING, ENGINERING, RAMANAGARA RAMANAGARA Page 1 of 32


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CONTENTS Sl. No.

DISCRETION

1.

Quadrant -2 a. Animations.

t c e j o r Quadrant -3 P T C I E M N L E Quadrant -4 T P N U T V

b. Videos.

c. Illustrations. 2.

a. Wikis.

b. Open Contents

3.

a. Problems.

b. Assignments

c. Self Assigned Q & A. d. Test your Skills.



2014


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CONTENTS Sl. No.

DISCRETION

1.

Quadrant -2 a. Animations.

t c e j o r Quadrant -3 P T C I E M N L E Quadrant -4 T P N U T V

b. Videos.

c. Illustrations. 2.

a. Wikis.

b. Open Contents

3.

a. Problems.

b. Assignments

c. Self Assigned Q & A. d. Test your Skills.



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MODULE-I PROPERTY RELATIONSHI RELATIONSHIP P QUADRANT-2

Animations

t c e j o r P T C I E M N L E T P N U T V

1 )http://www.youtube.com/watch?v=t0sjFKPdvIc 2) http://www.youtube.com/watch?v=0PkOEHMNOLk 3) https://www.google.co.in/search?q=animation+related+to+air+conditioning&tbm=isch &tbo=u&source=univ&sa=X&ei=FtseU6rmOMX9rAeVgIGQAg&ved=0CDMQsAQ& biw=1440&bih=809 4) http://www.dnatube.com/video/7911/Animation-of-how--Air-Conditioning-works 5) http://www.dnatube.com/video/7939/An-animation-of-how--Air-Conditioning-works 6) http://www.dnatube.com/video/9079/Principles-Of-Air-Conditioning 7) http://www.yazaki-airconditioning.com/products/absorption_cooling.html 8) http://www.wisegeek.org/how-does-air-conditioning-work.htm#didyouknowout 9) http://www.youtube.com/watch?v=_lFUlA1PZ8U 10) http://educypedia.karadimov.info/education/mechanicsjavathermo.htm Videos:

1 )http://www.youtube.com/watch?v=t0sjFKPdvIc 2) http://www.youtube.com/watch?v=0PkOEHMNOLk 3) http://www.dnatube.com/video/7911/Animation-of-how--Air-Conditioning-works 4) http://www.dnatube.com/video/7939/An-animation-of-how--Air-Conditioning-works 5) http://www.dnatube.com/video/9079/Principles-Of-Air-Conditioning 6) http://www.youtube.com/watch?v=_lFUlA1PZ8U


MODULE-I --- Property Relationships for Pure Substances and Mixtures APPLIED THRMODYNAMICS

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ILLUSTRATIONS Pure Substance

A pure substance is defined as a substance having a constant and uniform chemical composition. Homogeneous Mixture of Gases

Any mixture of gases in which the constituents do not rennet chemically with one


another and they are in a fixed proportion by weight is referred to as homogeneous mixture of gases and is regarded as a single substance. The properties of such a mixture can be determined experimentally just as for a single substance, and they can be tabulated or related algebraically in the same way. Therefore, the composition of air is assumed invariable for most purposes and air is usually treated as a single substance.

Mixture of Gases

Mixture of gases is generally imagined to be separated into its constituents in such a way that each occupies a volume equal to that of the mixture and each is at the same temperature as the mixture.

EMPIRICAL LAW FOR MIXTURES OF GASES

Consider a closed vessel of volume V at temperature T , which contains a mixture of perfect gases at a known pressure. If some of the mixture were removed, then the pressure would be less than the initial value. If the gas removed was the full amount of one of the

Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA Page 4 of 32


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constituents then the reduction in pressure would be equal to the contribution of that constituent to the initial total pressure. This is not only applicable to pressure but also to internal energy, enthalpy and entropy.

Dalton’s Law

In a mixture of gases, each constituent contributes to the total pressure by an amount which is known as the partial pressure of the constituent. The relationship between the partial pressures of the constituents is expressed by Dalton’s law, as follows: The pressure

of a mixture of gases is equal to the sum of the pressures of the individual constituents when


each occupies a volume equal to that of the mixture at the temperature of the mixture. Dalton’s law is based on experiment and is found to be obeyed more accurately

by gas

mixtures at low pressures. By the conservation of mass: m = m A + m B + mC + ……. or m = ∑mi

(1.1)

By Dalton’s law

P = P A + P B + P C + …..…. or P = ∑ P i

(1.2)

where mi = mass of a constituent; and P i = partial pressure of a constituent.

GIBBS-DALTON LAW

Dalton’s law was reformulated by Gibbs to include a second statement on the

properties of mixtures. The combined statement is known as the Gibbs-Dalton law, and is stated as follows:

The internal energy, enthalpy and entropy of a mixture of gases are respectively equal to the sums of the internal energies, enthalpies and entropies of the individual constituents when each occupies a volume equal to that of the mixture at the temperature of the m ixture. This statement leads to the following equations. mu = m Au A + m Bu B + mC uC + ……. or mu = ∑miui

(1.3)

mh = m Ah A + m Bh B + mC hC + …… or mh = ∑mihi

(1.4)

and ms = m A s A + m B s B + mC sC + ……. or ms = ∑m s i i

(1.5)



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VOLUMETRIC ANALYSIS OF A GAS MIXTURE.

The analysis of a mixture of gases is often quoted by volume as this is the most convenient for practical determinations. Considered a volume V of a gaseous mixture at a temperature T , consisting of three constituents A, B and C. Let each of the constituents be compressed to a pressure P equal to the total pressure of the mixture, and let the temperature remain constant.


Partial pressure of A, P

=

m A R AT

or

A

= P AV

m

A

R AT

V

Also, the total pressure, P =

m A R A T

or

A m A = PV

V

R AT

A

Therefore, In general,

P V = PV A

V i = P i V

P

and ∑V i =

V P ∑ i P

or

A

or

V A

= P A V P

V i = P i V P

(1.6)

since ∑ P i = P ,

∑V i = V

(1.7)

Therefore the volume of a mixture of gases is equal to the sum of the volumes of the individual constituents when each exists alone at the pressure and temperature of the mixture.



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Mixtures of Perfect Gases

Each of the constituents in the mixture is assumed to obey the perfect gas equation of state: for the mixture,

PV = mRT

for the constituent, P V = m R T i

i

or P = or P

i

i

mRT V

=mi R i T V

From dalton’s law, P = ∑ P i = ∑ mi RiT V


mRT

=

V

T

∑

m R i

i

V

that is, mR = ∑m R i i since Ri =

Ro

and R

(1.8)

= Ro

M i

m

=∑

, therefore,

M

mR o

M

=∑

mi R o

and Eq. (1.8) becomes:

M i

mi

(1.9)

M

M

i

The ratio of mass ( m) of a gas to its Molar mass ( M ) is defined as number of moles contain by the volume ot the gas, that is m/ M = n and mi/ M i = ni. Substituting these in Eq. (1.9) we have: n = ∑ni

or

n = n A + n B + nC + ……

(1.10)

Therefore, the number of moles of a mixture is equal to the sum of the moles of the constituents.

THE MOLAR MASS AND SPECIFIC GAS CONSTANT

P iV = m R i iT

or

V ∑ P i = RoT ∑ni

since ∑ P i = P , and therefore, PV = nRoT

P iV = n R i oT

(1.11) (1.12)

∑ ni = n

(1.13)



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This is a characteristic equation for the mixture, which shows that the mixture also acts as a perfect gas. It can be assumed that a mixture of perfect gases obeys all the perfect gas laws. The relationship between the volume fraction and pressure fraction is obtained from ratio of Eq. (1.12) to Eq. (1.13): P iV ni R o T =

PV

or

nRoT

P i

=

P

ni

(1.14)

n

Combination of Eq. (1.14) with Eq. (1.6) gives: P i

P

t c e j o r P T C I E M N L E T P N U T V =

V i = n V

ni

(1.15)

In order to find the specific gas constant for the mixture in terms of the specific gas constants of the constituents, consider the following equations both for the mixture and for a

constituent:

for mixture,

PV = m RT

for a c onstituent

P iV = m R i iT

or ∑ P iV = ∑ m R i iT

then V ∑ P i = T ∑m R i i

since ∑ P i = P , PV = T ∑m R i i

or

mRT = T ∑m R i i

therefore, mR = ∑m R i i R

mi

R = ∑ m

(1.16)

i

where mi/m is the mass fraction of a constituent.

SPECIFIC HEAT CAPACITIES OF A GAS MIXTURE

From the Gibbs-Dalton law, mu = ∑miui at constant volume, u = C VT Therefore, mC VT = ∑miC ViT mC V = ∑mi C Vi

= ∑[(mi /m)C vi ] or CV

(1.17)

Similarly, mh = ∑mihi Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA Page 8 of 32


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at constant pressure, h = C pT Therefore, mC p = ∑ miC pi or

C p = ∑[(mi /m) C pi]

(1.18)

From Eqs. (1.17) and (1.18) C p – C V

= ∑[(mi /m)C pi] – ∑[(mi /m)C vi]

C p – C V

= ∑[(mi /m)(C pi – C vi )]

Also, C p – C V = Ri, therefore,


C p – C V = ∑[( mi / m) Ri R = ∑

Recall Eq. (1.16),

mi

Ri ,

therefore, for the mixture:

m

C p – C V = R

This shows that the following equations can also be applied to a mixture of gases: γ

=

C p

;

R

C v =

C v

γ

; and C p =

−1

γ R

γ

.

−1

UNIVERSAL GAS LAW =

mRoT

pV ~

N

where Ro is the universal constant 8314.4 J/kmol K

Ñ is the relative molecular mass which is 18 for water vapour treated as a gas and 28.96 for dry air treated as a single gas. PARTIAL PRESSURES

The pressure exerted by a gas on the surface of containment is due to the bombardment of the surface by the molecules. The relative distance between molecules is very large so if two or more gases exist in the same space, their behaviour is unaffected by the others and so each gas produces a pressure on the surface according to the gas law above. Each gas occupies the total volume V and Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA Page 9 of 32


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has the same temperature T. If two gases A and B are considered, the pressure due to each is : pa =

ma RoT

p b =

~

Na V m b RoT ~

NV b The total pressure on the surface of containment is

p = p a + p b

This is Daltons Law of partial pressures.


Now let’s see how these laws are applied to mixtures of vapour and air. AIR - VAPOUR MIXTURES

In the following work, water vapour is treated as a gas.

Consider a mixture of dry air and vapour. If the temperature of the mixture is cooled until the vapour starts to condense, the temperature must be the saturation temperature (dew point) and the partial pressure of the vapour ps must be the value of ps in the fluids tables at the mixture temperature.

If the mixture is warmed up at constant pressure so that the temperature rises, the vapour must become superheated. It can be shown that the partial pressure of the vapour and the dry air remains the same as at the saturation temperature.

Let condition (1) be at the saturation condition and condition (2) be at the higher temperature. p is constant so it follows that :

V 1

T

=

V 2

T

1

2

The initial partial pressure of the vapour is: m RoT ps1 = ~s

1

Ns V1 The final pressure of the vapour is : ps2 = ~

s

m RoT 2



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N V s

Since

2

V1 = V2 then p

T

T 1

s1

=p s2

2

By the same process it can be shown that p a1 = pa2 If p is constant then the partial pressures are constant and the partial pressure of the vapour may easily be found by looking up the saturation pressure at the dew point if it is known.


When the air is contact with water, it will evaporate the water and the water will cool down until it is at the saturation temperature or dew point. This idea is used in wet bulb thermometers for example, which measure the dew point. When stable conditions are reached, the air becomes saturated and equal to the temperature of the water and so its temperature is the dew point (t s) in fluids tables.

HUMIDITY

There are two ways to express humidity SPECIFIC AND RELATIVE. SPECIFIC HUMIDITY ω

ω = mass of water vapour/mass of dry air

Starting with the gas law

~

m=

pVN

RoT ~ p VRoTN

ω

~

= s ~ = s ~ s pa VRoTN a pa Na s

ω

p N

=

ps

p a

x

18 28.96

= 0.622

ps

p a

= 0.622 p − ps



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RELATIVE HUMIDITYφ

φ = mass of vapour/maximum possible mass of vapour

The maximum possible mass of water vapour which can be held by air is when the vapour is saturated and the temperature of the mixture is the saturation temperature. mass = Volume/specific volume = V/v When saturated,

v = vg at the mixture temperature.

φ= Alternatively

m V V v s

=

÷ =

a

mg vs va vs

t c e j o p r P T C I E M N L p E p T P N U T V v = V/m vs =

N p s

s

and vg =

RoT φ = s pg

Ng p g

RoT

ps = partial pressure of the actual vapour pg = partial pressure when saturated. ω

= 0.622

φ=

ω

(

ps p and φ = s p − ps pg

−

)

s

0.622pg

Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM Page 12 of 32

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QUADRANT-3

Wikis: 1) http://en.wikipedia.org/wiki/Air_conditioning 2) http://en.wikipedia.org/wiki/HVAC 3) http://simple.wikipedia.org/wiki/Air_conditioner 4) http://z32.wikispaces.com/Air+Conditioning+System 5) http://whirlpool.net.au/wiki/aircon_faq 6) http://commons.wikimedia.org/wiki/File:Air_conditioning_unit-en.svg 7) http://wiki.hometech.com/tiki-index.php?page=HVAC+Control+Tutorial 8) http://climatetechwiki.org/technology/efficient-air-conditioning-systems 9) http://home.howstuffworks.com/ac1.htm 10) https://www.ashrae.org/resources--publications/free-resources/top-ten-things-aboutair-conditioning


Open Contents:

Applied Thermodynamics by R. K. Rajput

Applied Thermodynamics for Engineering Technologists by Eastop Applied Thermodynamics by B. K. Venkanna B. V. S Basic and Applied Thermodynamics by Nag

Applied Thermodynamics by D. S. Kumar

A textbook of applied thermodynamics, steam and thermal ... by S. K. Kulshrestha

Applied thermodynamics by Anthony Edward John Hayes


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QUADRANT-4

Problems

1) The atmospheric conditions are; 20°C and specific humidity of 0.0095 kg/kg of dry air. Calculate the following: (i) Partial pressure of vapour (ii) Relative humidity (iii) Dew point temperature. Solution:

Dry bulb temperature, tdb = 20ºC Specific humidity, W = 0.0095 kg/kg of dry air

t c e j o r P T C I E M N p L p / p E T P N U T V

(i) Partial pressure of vapour, p v:

The specific humidity is given by 0.0095 =0.622 P V/ (1.0132- PV)

W=

0.0095(1.0132 – p v) = 0.622 p v

0.009625 – 0.0095 p v = 0.622 p v

pv = 0.01524 bar.

(ii) Relative humidity φ :

Corresponding to 20ºC, from steam tables, ∴

Relative humidity, φ =

v

vs =

0.0234 bar

vs = 0.01524/0.0234=

0.65 or 65%. (Ans)

(iii) Dew point temperature, tdp :

The dew point temperature is the saturation temperature of water vapour at a pressure of 0.01524 bar,

t dp [from steam tables by interpolation] = = 13.24°C.

2) 0.004 kg of water vapour per kg of atmospheric air is removed and temperature of air after removing the water vapour becomes 20°C. Determine :(i) Relative humidity (ii) Dew point temperature. Assume that condition of atmospheric air is 30°C and 55% R.H. and pressure is 1.0132 bar.


MODULE-I --- Property Relationships for Pure Substances and Mixtures APPLIED THRMODYNAMICS Solution:

Corresponding to 30ºC, from steam tables, pvs = 0.0425 bar ∴

Relative humidity (R.H.), φ =

i.e., 0.55 = P V/0.0425 Pv = 0.02337 bar. ∴

Also the specific humidity, W=

= (0.622*0.02337)/ (1.0132-0.02337)

= 0.01468 kg/kg of dry air. The specific humidity after removing 0.004 kg of water vapour becomes, 0.01468 – 0.004 = 0.01068 kg/kg of dry air and the temperature t db is given as 20ºC.


The partial pressure of water vapour, pv , at this condition can be calculated as follows : W=

0.01068 = 0622Pv/(1.0132-Pv)

or, 0.01068 (1.0132 – p v ) = 0.622 p v or, 0.01082 – 0.01068 p v = 0.622 p v 0.6327 p v = 0.01082 p v = 0.0171 bar

∴

Corresponding to 20ºC, from steam tables, pvs = 0.0234 bar.

(i) Relative humidity, φ = Pv/Pvs = 00171/00234 = 0.73 or 73%. (Ans)

(ii) Dew point temperature, tdp :

Corresponding to 0.0171 bar, from steam tables, t dp = 15°C.

3) The sling psychrometer in a laboratory test recorded the following readings: Dry bulb temperature = 35°C

Wet bulb temperature = 25°C. Calculate the following : (i) Specific humidity (ii) Relative humidity (iii) Vapour density in air (iv) Dew point temperature (v) Enthalpy of mixture per kg of dry air Take atmospheric pressure = 1.0132 bar.


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MODULE-I --- Property Relationships for Pure Substances and Mixtures APPLIED THRMODYNAMICS Solution:

For finding the partial pressure of vapour, using the equation:

Corresponding to 25ºC (from steam tables), (Pvs )wb = 0.0317 bar Substituting the values in the above equation, we get Pv=0.0317 = 0.0317 – 0.0065 = 0.0252 bar. (i) Specific humidity, W=


= (0.622*0.0252)/ (1.0132-0.0252)

= 0.01586 kg/kg of dry air. (Ans)

(ii) Relative humidity, φ = Pv/Pvs

[Pvs = 0.0563 bar corresponding to 35ºC, from steam tables] = 0.447 or 44.7%. (Ans.) (iii) Vapour density :

From characteristic gas equation PvVv = M v Rv T v

Pv = M v R v T v / Vv = ρ v Rv T v

where ρ v =vapour density, (characteristic gas constant Rv= Universal gas constant /Molecularweight of H2O = 8314.3/18 ∴

0.0252 × 10 5 = ρ v × (8314.3/18)× (273 + 35)

∴

ρ v = (0.0252 ×10 5 ×18) /(8314.3× 308) = 0.0177 kg/m 3 (Ans)

(iv) Dew point temperature, tdp : Corresponding to 0.0252 bar, from steam tables (by interpolation), t dp =

= 21.2°C. (Ans)

(v) Enthalpy of mixture per kg of dry air, h : h = Cp tdb + W hvapour = 1.005 × 35 + 0.01586 [h g + 1.88 (t db – tdp )] = 35.175 + 0.01586 [2565.3 + 1.88 (35 – 21.2)] (where hg = 2565.3 kJ/kg corresponding to 35ºC t db ) = 76.27 kJ/kg of dry air.


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4) One kg of air at 35°C DBT and 60% R.H. is mixed with 2 kg of air at 20°C DBT and 13°C dew point temperature. Calculate the specific humidity ofthe mixture. Solution:

For the air at 35°C DBT and 60% R.H. : Corresponding to 35ºC, from steam tables, Pvs = 0.0563 bar Relative humidity, φ = Pv/Pvs p v = φ p vs = 0.6 × 0.0563 = 0.0338 bar

∴

W=

= (0.622*0.0338)/ (1.0132-0.0338) = 0.0214 kg/kg of dry air

t c e j o r P T C I E M N L E T = (0.622*0.015)/(1.0132-0.015) P N U T V

Corresponding to 0.0338 bar, from steam tables, tdp = 26.1ºC

=

Enthalpy, h = c p t db + Wh vapour

= 1.005 t db + W [h g + 1.88 (t db – t dp )]

= 1.005 × 35 + 0.0214 [2565.3 + 1.88 (35 – 26.1)] = 90.43 kJ/kg of dry air.

For the air at 20°C DBT and 13°C dew point temperature : Pv is the vapour pressure corresponding to the saturation pressure of steam at 13ºC. Pv = 0.0150 bar

∴

W=

= 0.00935 kg/kg of dry air

Enthalpy, h = c p t db + Wh vapour

= 1.005 × 20 + 0.00935 [h g + 1.88 (t db – t dp )] = 20.1 + 0.00935 [2538.1 + 1.88 (20 – 13)] = 43.95 kJ/kg of dry air Now enthalpy per kg of moist air

= 58.54 kJ/kg of moist air Mass of vapour/kg of moist air = 0.01316 kg/kg of moist air Specific humidity of mixture = 0.01316/(1-0.01316) = 0.01333 kg/kg of dry air.



5) 90 m 3 of air per minute at 20°C and 75% R.H. is heated until its temperature becomes 30°C. Calculate : (i) R.H. of the heated air. (ii) Heat added to air per minute. Solution:

(i) For air at 20°C and 75% R.H. : Pvs = 0.0234 bar (from steam tables, at 20ºC) Pv = φ × Pvs = 0.75 × 0.0234 = 0.01755 bar tdp = W1 =

= 15.5ºC

= (.622*0.01755)/(1.0132-0.01755) = 0.0109 kg/kg of dry air

Enthalpy, h 1 = c p t db + Wh vapour


= 1.005 × 20 + 0.0109 [h g + 1.88 (t db – t dp)]

= 1.005 × 20 + 0.0109 [2538.1 + 1.88(20 – 15.5)] = 47.85 kJ/kg of dry air (i) Relative humidity of heated air : For air at 30°C DBT :

Since the saturation pressure of water vapour at 30ºC is higher than the saturation pressure of water vapour at 20ºC so it is sensible heating, where p v is same after heating. ∴

Relative humidity, φ = Pv/Pvs =0.01755/0.0425 = 0.412 or 41.2%

Pvs = 0.0425 bar, corresponding to 30ºC)

i.e., Relative humidity of heated air = 41.2% (ii) Heat added to air per minute :

Enthalpy, h 2 = c p t db + Wh vapour

= 1.005 × 30 + 0.0109 × [h g + 1.88 (t db – t dp )]

= 1.005 × 30 + 0.0109 [2556.3 + 1.88 (30 – 15.5)] = 58.31 kJ/kg of dry air

Mass of dry air in 90 m 3 of air supplied

m a = PV /RT = (Pt-Pv)V/RT =((1.0132- 0.01755) × 10 5 ×90) / ( 287×(273+20)) = 106.5 kg/min. Amount of heat added per minute = 106.5 (h2 – h1) = 106.5 (58.31 – 47.85)



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6) 40 m 3 of air at 35°C DBT and 50% R.H. is cooled to 25°C DBT maintaining its specific humidity constant. Determine : (i) Relative humidity (R.H.) of cooled air ; (ii) Heat removed from air. Solution:

For air at 35°C DBT and 50% R.H. : Pvs = 0.0563 bar (At 35ºC, from steam tables) P v = φ × Pvs = 0.5 × 0.0563 = 0.02815 bar

∴

W1 =


h1 = Cp t db 1 + W 1 [ hg 1 + 1.88 ( t db 1 – t dp1 ]


t dp 123ºC (corresponding to 0.02815 bar)

h1 = 1.005 × 35 + 0.0177 [2565.3 + 1.88 (35 – 23)] = 80.98 kJ/kg of dry air

∴

For air at 25°C DBT :

(i) R.H. of cooled air :

Since the specific humidity remains constant the vapour pressure in the air remains constant. φ = Pv/Pvs = 0.02815/0.0317 = 0.888 or 88.8%

i.e., Relative humidity of the cooled air = 88.8%. (Ans.) (ii) Heat removed from air :

h 2 =Cp t db 2 + W 2 [ h g 2 + 1.88 ( tdb 2 – t dp 2 )] = 1.005 × 25 + 0.0177 [2547.2 + 1.88 (25 – 23)] = 70.27 kJ/kg of dry air.

To find mass of dry air (m a ), using the relation :

W1=W2  tdp2=tdp1

PaVa=maRaTa

m a = pa va /R a Ta=(1.0132- 0.02815) ×10 5 ×40 /287×(273+35)= 44.57 kg

∴

Heat removed from 40 m 3 of air

∴

= ma (h 1 – h 2) = 44.57 (80.98 – 70.27) = 477.3 kJ. (Ans) 8) 150 m 3 of air per minute is passed through the adiabatic humidifier. The condition of air at inlet is 35°C DBT and 20 per cent relative humidity and the outlet condition is 20°C DBT and 15°C WBT. Determine the following : (i) Dew point temperature (ii) Relative humidity of the exit air (iii) Amount of water vapour added to the air per minute.



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Solution:

For air at 35°C DBT and 20% relative humidity. Pvs = 0.0563 bar (At 35ºC from steam tables) Pv = φ × P vs = 0.2 × 0.0563 = 0.01126 bar W1 = = 0.00699 kg/kg of dry air (i) The dew point temperature of air which is the saturation temperature of steam corresponding to the pressure 0.01126 bar is 8 + (9 – 8) × ((0.01126- 0.01072)/(0.01150-0.01072))= 8.7ºC


i.e., Dew point temperature = 8.7°C. (Ans.)

(ii) Relative humidity of the exit air : For air at 20ºC DBT and 15ºC WBT.

= 0.0137 bar

W2 =


Relative humidity = φ = Pv/Pvs = 0.585 or 58.5%. (Ans.)

(Pvs = 0.0234 bar, corresponding to 20ºC, from steam tables)

The dew point temperature of air which is the saturation temperature of steam corresponding to 0.0137 bar is 11°C (from steam tables). (Ans.)

The amount of water vapour per kg of dry air = W 2 – W 1 = 0.00852 – 0.00699 = 0.00153 kg

The mass of dry air in 150 m 3 of mixture

m a = pa va /R a Ta=(1.0132- 0.01126) ×10 5 ×150 /287×(273+35)= 170 kg

∴

(iii) The amount of water vapour added to air per minute = m a (W 2 – W1) = 170 × 0.00153 = 0.26 kg/min. 9) An air-water vapour mixture enters an adiabatic saturation chamber at 28°C and leaves at 18°C, which is the adiabatic saturation temperature. The pressure remains constant at 1.0 bar. Determine the relative humidity and humidity ratio of the inlet mixture. Solution:.

The specific humidity at the exit W2S =

=




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The specific humidity at the inlet (equation 10.18) W1 =

=


W 1 = 0.01704 =

∴

Or 0.01704 (1.00 – Pv1) = 0.622 Pv1 Or 0.01704 – 0.01704 Pv1= 0.622 Pv1 Or 0.0639 Pv1= 0.01704 Pv1= 0.02666 bar

∴


Relative humidity = Pv1/Pvs1 =

∴

0.02666/0.03782 = 0.7 or 70%.

10) An air-water vapour mixture enters an air-conditioning unit at a pressure of 1.0 bar. 38°C DBT, and a relative humidity of 75%. The mass of dry air entering is 1 kg/s. The air-vapour mixture leaves the air-conditioning unit at 1.0 bar, 18°C, 85% relative humidity. The moisture condensed leaves at 18°C. Determine the heat transfer rate for the process. Solution:

tdb1 = 38ºC, R.H., φ 1 = 75% t db2 = 18ºC, R.H., φ 2 = 85%

The flow diagram and the process are shown in Figs respectively. At 38°C

From steam tables : P vs = 0.0663 bar, hg1 = 2570.7 kJ/kg Pv = φ × Pvs = 0.75 × 0.0663 = 0.0497 bar

∴

W 1 =


At 18°C

From steam tables : P vs = 0.0206 bar, hg2 = 2534.4 kJ/kg hf2 = 75.6 kJ/kg Pv = 0.85 × 0.0206 = 0.01751 bar W 2 =




Heat transfer rate, q = (W2hg2 – W1hg1) + C p(tdb2 – tdb1)+(W1 – W2)hf2 = (0.01108 × 2534.4 – 0.0325 × 2570.7) + 1.005 (18


– 38) + (0.0325 – 0.01108) × 75.6

= – 55.46 – 20.1 + 1.62 = – 73.94 kJ/kg of dry air.

11) It is required to design an air-conditioning system for an industrial process for the following hot and wet summer conditions : Outdoor conditions ...... 32ºC DBT and 65% R.H.

Required air inlet conditions ...... 25ºC DBT and 60% R.H. Amount of free air circulated ...... 250 m3/min. Coil dew temperature ...... 13ºC.

The required condition is achieved by first cooling and dehumidifying and then by heating. Calculate the following :

(i) The cooling capacity of the cooling coil and its by-pass factor.

(ii) Heating capacity of the heating coil in kW and surface temperature of the heating coil if the by-pass factor is 0.3.

(iii) The mass of water vapour removed per hour.

Solve this problem with the use of psychrometric chart.


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Solution. Refer Fig.



Locate the points ‘1’, ‘5’ and ‘3’ as

 

shown on psychrometric chart. Join the line 15. Draw constant specific humidity line through ‘3’ which cuts the line 1 -5 at point ‘2’.


The point ‘2’ is located in this way.

From psychrometric chart :

h1 = 82.5 kJ/kg, h2 = 47.5 kJ/kg h3 = 55.7 kJ/kg, h5 = 36.6 kJ/kg

W1 = 19.6 gm/kg, W3 = 11.8 gm/kg tdb2 = 17.6ºC, vs1 = 0.892 m3/kg.

The mass of air supplied per minute, ma =250/ 0.892 = 280.26 kg/min.

(i) The capacity of the cooling coil

=( ma (h1 − h2) × 60 )/14000 =(280.26(82.5-47.5) × 60)/14000 = 42.04 TR. (Ans)

The by-pass factor of the cooling coil is given by : BF =

= 0.237 (Ans)

(ii) The heating capacity of the heating coil = ma (h3 – h2) = 280.26 (55.7 – 47.5) = 2298.13 kJ/min = 2298.13/60 kJ/s = 38.3 kW (Ans) The by-pass factor of the heating coil is given by BF =


MODULE-I --- Property Relationships for Pure Substances and Mixtures APPLIED THRMODYNAMICS ∴

tdb6 = 28.2ºC.

Hence surface temperature of heating coil = 28.2 ºC. (Ans)

(iii) The mass of water vapour removed per hour =

= = 131.16 kg/h.


12) It is required to design an air-conditioning plant for a small office room for following winter conditions :

Outdoor conditions ...... 14ºC DBT and 10ºC WBT Required conditions ...... 20ºC DBT and 60% R.H.

Amount of air circulation ...... 0.30 m3/min./person. Seating capacity of office ...... 60.

The required condition is achieved first by heating and then by adiabatic humidifying. Determine the following :

(i) Heating capacity of the coil in kW and the surface temperature required if the by pass factor of coil is 0.4.

(ii) The capacity of the humidifier.

Solve the problem by using psychrometric chart.

Solution. Refer Fig.

_ Locate the points ‘1’ and ‘3’ on the psychrometric chart. _ Draw a constant enthalpy line through ‘3’ and constant specific humidity line through‘1’. _

Locate the point ‘2’ where the above two lines intersect


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From the psychrometric chart : h1 = 29.3 kJ/kg, h2 = h3 = 42.3

kJ/kg t db2 = 24.5ºC, v s1 = 0.817 m3/kg

The mass of air circulated per minute, ma =(0.30*60)/0.817= 22.03 kg/min.

(i) Heating capacity of the heating coil

= ma(h2 – h1) = 22.03 (42.3 – 29.3) = 286.4 kJ/min. = 4.77 kJ/s or 4.77 kW. (Ans)

The by-pass factor ( BF ) of heating coil is given by: BF=

0.4 = ( t db4 – 12) = t db4 – 24.5

∴

i.e., t db4 (coil surface temperature) = 32.8ºC (Ans)

(ii) The capacity of the humidifier

= ma (W 3 −W 1)/1000 × 60 kg/h = 22.03(8.6-6.8)/1000 *60 = 2.379 kg/h. 12) 200 m 3 of air per minute at 15ºC DBT and 75% R.H. is heated until its temperature is 25ºC. Find : (i) R.H. of heated air. (ii) Wet bulb temperature of heated air. Solution:

(iii) Heat added to air per minute. Locate point 1 on the psychrometric chart on intersection of 15ºC DBT and 75% R.H.lines. Through point 1 draw a horizontal line to cut 25ºC DBT line and get point 2.



Read the following values from the psychrometric chart : h 1 = 35.4 kJ/kg h 2 = 45.2 kJ/kg vs1 = 0.8267 m 3 /kg. (i) R.H. of heated air (read from chart) = 41%. (Ans) (ii) WBT of heated air (read from chart) = 16.1ºC. (Ans) (iii) Mass of air circulated per min., m a = 200/0.8267 = 241.9 kg. Heat added to air/min. = m a (h 2 – h 1) = 241.9 (45.2 – 35.4) = 2370.6 kJ.

∴


Frequently asked Questions. 1. What is an ideal gas ?

2. What is the difference between an ideal and a perfect gas ? 3. What are semi-perfect or permanent gases ? 4. Define ‘Equation of state’. 5. Write a short note on Van der Waals’ equation. 6. Define the following terms :

(i) Saturated air ( ii) Dry bulb temperature (iii) Dew point temperature ( iv) Relative humidity (v) Specific humidity. 7. State ‘Dalton’s law of partial pressure’. 8. Describe briefly any two of the following processes :

(i) Sensible heating ( ii) Cooling and dehumidification (iii) Heating and humidification ( iv) Heating and dehumidification.


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9. A vessel of 0.03 m capacity contains gas at 3.5 bar pressure and 35°C temperature.

Determine the mass of the gas in the vessel. If the pressure of this gas is increased to 10.5 bar while the volume remains constant, what will be the temperature of the gas ? For the gas take R = 290 J/kg K. [Ans. 0.118 kg, 650°C] 10. A vessel of spherical shape is 1.5 m in diameter and contains air at 40°C. It is evacuated

till the vacuum inside the vessel is 735 mm of mercury. Determine : ( i) The mass of air pumped out ; (ii) If the tank is then cooled to 10°C what is the pressure in the tank ? The barometer reads 760 mm of mercury. Assume that during evacuation, there is no change in temperature of air.


[Ans. (i) 1.91 kg, ( ii) 3 kPa] 3

11. 100 m of air per minute at 35ºC DBT and 60% relative humidity is cooled to 20ºC DBT

by passing through a cooling coil. Find the following : ( i) Capacity of cooling coil in kJ/h (ii) Amount of water vapour removed per hour, and

(iii) Relative humidity of air coming out and its wet-bulb temperature. [Ans. (i) 1037088 kJ/h, ( ii) 465.36 kg/h, ( iii) 100%, 20ºC]

12. Atmospheric air at 38ºC and 40 per cent relative humidity is to be cooled and

dehumidified to a state of saturated air at 10ºC. The mass rate of flow of atmospheric air entering the dehumidifier is 45.4 kg/h. Neglecting any pressure drop, determine : ( i) The mass of water removed ; ( ii) The quantity of heat removed. [Ans. (i) 0.397 kg/h, (ii) 2287 kJ/h

13. The atmospheric conditions are 30ºC and specific humidity of 0.0215 kg/kg of air.

Determine : (i) Partial pressure of air ( ii) Relative humidity (iii) Dew point temperature. Atmospheric pressure = 756 mm Hg.

[Ans. (i) 14.89 mm of Hg, ( ii) 46.8%, (iii) 17ºC]

14. A mixture of air and water vapour at 1 bar and 25ºC has a dew point temperature of 15ºC.

Determine the relative humidity and specific humidity. [Ans. 53.8%, 0.01078 kg/kg of dry air]



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Assignment:

1) The air supplied to a room of building in winter is to be at 17ºC and have a relative humidity of 60%. If the barometric pressure is 1.01325 bar, calculate the specific humidity. What would be the dew point under these conditions ? [Ans. 0.00723 kg/kg of dry air, 9.18ºC] 2) A mixture of air and water vapour at 1.013 bar and 16ºC has a dew point of 5ºC. Determine the relative and specific humidities. [Ans. 48%, 0.0054 kg/kg of dry air] 3) 100 m3 of air per minute at 35ºC DBT and 60% relative humidity is cooled to 20ºC DBT


by passing through a cooling coil. Find the following : ( i) Capacity of cooling coil in kJ/h (ii) Amount of water vapour removed per hour, and


4) Atmospheric air at 38ºC and 40 per cent relative humidity is to be cooled and dehumidified to a state of saturated air at 10ºC. The mass rate of flow of atmospheric air entering the dehumidifier is 45.4 kg/h. Neglecting any pressure drop, determine : (i) The mass of water removed ; ( ii) The quantity of heat removed. [Ans. (i) 0.397 kg/h, (ii) 2287 kJ/h]

5. 1 kg of air at 24ºC and a relative humidity of 70% is to be mixed adiabatically in a steady

state, steady flow device with 1 kg of air at 16ºC and a relative humidity of 10%. Assuming that the mixing is to be carried out at a constant pressure of 1.0 atm, determine the temperature and relative humidity of the stream leaving the device. [Ans. 19.5ºC, 50%]

6) A balloon of spherical shape is 8 m in diameter and is filled with hydrogen at a pressure of 1 bar abs. and 15°C. At a later time, the pressure of gas is 95 per cent of its original pressure at the same temperature. ( i) What mass of original gas must have escaped if the dimensions of the balloon are not changed ? ( ii) Find the amount of heat to be removed to cause the same drop in pressure at constant volume. [Ans. (i) 5 per cent, ( ii) 3.26 MJ] 7) A constant volume chamber of 0.3 m33 capacity contains 1 kg of air at 20°C. Heat is transferred to the air until its temperature is 200°C. Find : ( i) Heat transferred ; (ii) Change in entropy and enthalpy.



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[Ans. (i) 128.09 kJ, ( ii) 0.339 kJ/kg K, 180.8 kJ] 8) 1 kg of air at 20°C occupying a volume of 0.3 m333 undergoes a reversible constant

pressure process. Heat is transferred to the air until its temperature is 200°C. Determine : (i) The work and heat transferred. (ii) The change in internal energy, enthalpy and entropy. [Ans. (i) 51.5 kJ, 180.8 kJ ; ( ii) 128.09 kJ, 180.8 kJ, 0.479 kJ/kg K] 9) Air expands in a cylinder in a reversible adiabatic process from 13.73 bar to 1.96 bar. If the final temperature is to be 27°C, what would be the initial temperature ? Also calculate the change in specific enthalpy, heat and work transfers per kg of air. [Ans. 524 K, 224.79 kJ/kg, zero, 160.88 kJ/kg]


Self Answered Question & Answer

1) 100 m3 of air per minute at 35ºC DBT and 60% relative humidity is cooled to 20ºC DBT by passing through a cooling coil. Find the following : ( i) Capacity of cooling coil in kJ/h (ii) Amount of water vapour removed per hour, and


2) 1 kg of air at 24ºC and a relative humidity of 70% is to be mixed adiabatically in a steady

state, steady flow device with 1 kg of air at 16ºC and a relative humidity of 10%. Assuming that the mixing is to be carried out at a constant pressure of 1.0 atm, determine the temperature and relative humidity of the stream leaving the device. [Ans. 19.5ºC, 50%]

3) The atmospheric conditions are 30ºC and specific humidity of 0.0215 kg/kg of air.

Determine : (i) Partial pressure of air ( ii) Relative humidity (iii) Dew point temperature. Atmospheric pressure = 756 mm Hg. [Ans. (i) 14.89 mm of Hg, ( ii) 46.8%, (iii) 17ºC] 4) Air expands in a cylinder in a reversible adiabatic process from 13.73 bar to 1.96 bar. If the final temperature is to be 27°C, what would be the initial temperature ? Also calculate the change in specific enthalpy, heat and work transfers per kg of air. [Ans. 524 K, 224.79 kJ/kg, zero, 160.88 kJ/kg]



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5) A vessel of spherical shape is 1.5 m in diameter and contains air at 40°C. It is evacuated

till the vacuum inside the vessel is 735 mm of mercury. Determine : ( i) The mass of air pumped out ; (ii) If the tank is then cooled to 10°C what is the pressure in the tank ? The barometer reads 760 mm of mercury. Assume that during evacuation, there is no change in temperature of air. [Ans. (i) 1.91 kg, ( ii) 3 kPa] 6) 1 kg of air at 27°C is heated reversibly at constant pressure until the volume is doubled and

then heated reversibly at constant volume until the pressure is doubled. For the total path find :(i) The work ; ( ii) Heat transfer ;( iii) Change of entropy. [Ans. (i) 86.14 kJ, ( ii) 728.36 kJ, ( iii) 1.186 kJ/kg K]


7) A mass of air initially at 260°C and a pressure of 6.86 bar has a volume of 0.03 m3. The

air is expanded at constant pressure to 0.09 m3, a polytropic process with n = 1.5 is then carried out, followed by a constant temperature process which completes the cycle. All processes are reversible. Find ( i) The heat received and rejected in the cycle, ( ii) The efficiency of the cycle.Show the cycle on p-v and T-s planes. [Ans. (i) 143.58 kJ, – 20.3 kJ ; ( ii) 38.4%]

Test Your Skills

Choose the Correct Answer :

1. In an unsaturated air the state of a vapour is

(a) wet (b) superheated ( c) saturated ( d ) unsaturated. 2. For saturated air

(a) Wet bulb depression is zero ( b) Wet bulb depression is positive (c) Wet bulb depression is negative ( d ) Wet bulb depression can be either positive or negative. 3. Which one of the following statements is correct ?

(a) Dew point temperature can be measured with the help of thermometer (b) Dew point temperature is the saturation temperature corresponding to the partial pressure of the water vapour in moist air.



(c) Dew point temperature is the same as the thermodynamic wet bulb temperature. (d ) For saturated air, dew point temperature is less than the wet bulb temperature. 4. During sensible heating of moist air, enthalpy

(a) increases ( b) decreases ( c) remains constant ( d ) none of the above. 5. During sensible cooling, wet bulb temperature

(a) decreases ( b) increases ( c) remains constant ( d ) can decrease or increase. 6. Which one of the following statements is correct ?


(a) Evaporative cooling and sensible cooling is the same

(b) Evaporative cooling is a cooling and humidification process

(c) Evaporative cooling is a cooling and dehumidification process (d ) Evaporative cooling is not effective for hot and dry climates.

7. An air washer can work as a

(a) filter only ( b) humidifier only ( c) dehumidifier only ( d ) all of the above.

8. The relative humidity, during sensible heating,

(a) can increase or decrease ( b) increases ( c) decreases ( d ) remains constant. 9. The vapour pressure, during sensible heating of moist air,

(a) increases ( b) decreases ( c) can increase or decrease ( d ) remains constant. 10. The relative humidity, during heating and humidification,

(a) increases ( b) decreases ( c) may increase or decrease ( d ) remains constant. 11. The relative humidity, during cooling and dehumidification of moist air

(a) increases ( b) decreases ( c) can increase or decrease ( d ) remains constant. 12. (a) A perfect gas does not obey the law pv = RT

(b) A perfect gas obeys the law pv = RT and has constant specific heat Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM Page 31 of 32

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Module 1 Property Relationship

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