MAHALAKSHMImahalakshmiengineeringcollege.com/pdf/mec/IIIsem/ME2202/UNIT 5 - 16 Mark.pdf · QUESTION...

L. VIJAYAKUMAR AP/MECHANICAL

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI- 621213.

QUESTION BANK

DEPARTMENT: MECH SEMESTER – III

SUBJECT CODE: EE2201 SUBJECT NAME: ENGINEERING THERMODYNAMICS

UNIT-V PSHYCOMETRIC

1. Air at 20°C , 40% relative humidity is mixed adiabaticlly with air at 40°C , 40% RH in the ratio

of 1Kg former with 2Kg of latter (on dry basis). Find the final condition (humidity and enthalpy)

of air [AU MAY 2008]

Given:

Dry bulb temp (td1) = 20°C

Relative humidity (ɸ1) = 40%



Ratio of 1Kg of former with 2Kg of latter =

To find:

The final condition of air

Enthalpy (h1 , h2 , h3)

Humidity (ω1 , ω2 , ω3)

Solution:

Step :1

From Psychometric chart


Use → td1 = 20°C Relative humidity (ɸ1) = 40%

Enthalpy (h1) = 35KJ/Kg

Humidity (ω1) = 0.0058KJ/Kg of dry air

Step :2


Use → td2 = 40°C Relative humidity (ɸ2) = 40%


Humidity (ω2) = 0.019KJ/Kg of dry air

Step :3

=

=

2(90 – h3) = 1(h3 - 35)

180 – 2 h3 = h3 – 35

180 – 2 h3 - h3 + 35 = 0

-3 h3 + 215 = 0

3 h3 = 215

h3 = 215/3

h3 = 71.6666KJ/Kg

Step :4

=


=

2(0.019 – ω3) = 1(ω3 – 0.0058)

0.038 – 2 ω3 = ω3 – 0.0058

0.038 – 2 ω3 - ω3 + 0.0058 = 0

-3 ω3 + 0.0438 = 0

3 ω3 = 0.0438

ω3 = 0.0438/3

ω3 = 0.0146Kg of vap / Kg of dry air


2. 30m3/min of moist air at 15°C DBT and 13°C WBT are mixed with 12m3/min of moist air at

25°C DBT and 18°C WBT. Determine DBT and WBT of the mixture. Assuming the barometric

pressure is 1 atm [AU DEC 2008]

Given:

Volume of air [Va1] = 30m3/min

Volume of air [Va2] = 12m3/min

Dry bulb temp [td1] = 15°C

Dry bulb temp [td2] = 25°C

Wet bulb temp [tω1] = 13°C

Wet bulb temp [tω2] = 18°C

To find:

1. Dry bulb temp of mixture[td]mixing

2. Wet bulb temp of mixture[ωd]mixing

Solution:

Step :1


Use → td1 = 15°C & ωd1 = 13°C


Step :2


Use → td2 = 25°C & ωd2 = 18°C



Step :3

=

=

2.5 =

2.5(37 – h3) = 1(h3 - 51)

92.5 – 2.5 h3 = h3 – 51

92.5 – 2.5 h3 - h3 + 51 = 0

-3.5 h3 + 143.5 = 0

3.5 h3 = 143.5

h3 = 143.5/3.5

h3 = 41KJ/Kg

Step :4

From Psychometric chart Corresponding to, h3 = 41KJ/Kg

The Dry bulb temp [td3] = 17.5°C

The Wet bulb temp [tω1] = 14.6°C


3. Air conditioning plant is required to supply 60m3 of air per minute at a DBT of 21°C and 55%

R.H. The outside air is at DBT of 35°C and 60% R.H. Determine the mass of water drained and

capacity of the cooling coil. Assume the air conditioning plant is first to dehumidity and then to

cool the air. [AU MAY 2005]

Given: Problem based on Dehumidification and the cooling.

Volume (V) = 60/60 m3/min = 1m3/sec





To find:

1. Mass of water drained (mw)

2. Capacity if cooling coil (Q)

Solution:

Step :1

Mass of water drained in Kg/sec (mw)

mw = ma [ω1 - ω2]

Where,

Ma = V/V2 = 1/0.84 = 1.1904 Kg/sec

Where,

V2 = Use → td1 = 21°C & ɸ = 55%


V2 = 0.84 m3/Kg

humidity (ω1) = From Psychometric chart


Use → td2 = 35°C & ɸ = 60%

(ω1) = 0.022 Kg/Kg of dry air

humidity (ω2) = From Psychometric chart

Use → td1 = 21°C & ɸ = 55%

(ω2) = 0.0086 Kg/Kg of dry air

Enthalpy (h1) = From Psychometric chart

Use → td2 = 35°C & ɸ = 60%

(h1) = 89.5 KJ/Kg

Enthalpy (h2) = From Psychometric chart

Use → td2 = 25°C & ɸ = 55%

(h2) = 43.5 KJ/Kg

mw = ma [ω1 - ω2]

mw = 1.1904 [0.022 – 0.0086]

mw = 0.01595 Kg of water/sec

mw = 0.01595 × 3600

mw = 57.4285 Kg of water/hr

Step :2

Capacity of cooling coil (Q)

Q = ma [h1 - h2]

Q = 1.1904 [89.5 – 43.5]

Q = 54.7584KJ/sec


4. Air is initially at a temp of 23°C and a Relative humidity of 50% . It’s moisture content is

increased from 0.0086 KJ/Kg of air to 0.0124KJ/Kg of air.

1. Find the increase in wet bulb temp , dry bulb temp and specific humidity.

2. Determine Change in Relative Humidity.

3. If 200m3 of air is passed through the system every minute and the density of air is

1.3Kg/m3 , calculate amount of moisture added per min. [AU NOV 2006]

Given:

Volume (V) = 200m3

Initial dry bulb temp (td1) = 23°C

humidity (ω1) = 0.0086 KJ/Kg of air

humidity (ω2) = 0.0124KJ/Kg of air


Density of air (ρair) = 1.3Kg/m3

To find:

1. Increase in wet bulb temp.

2. Increase in dry bulb temp.

3. Change in Relative Humidity.

4. Amount of moisture added per min.

Solution:

Step :1

Increase in wet bulb temp =tω2 - tω1

Where,



tω2→ ω2 = 0.0124 & td = 23°C

(h1) = 54.5 KJ/Kg

tω2 = 19°C

Where,


tω1→ ω1 = 0.0086 & td = 23°C

(h2) = 45.5 KJ/Kg

tω2 = 16.2°C

Increase in wet bulb temp =tω2 - tω1

=tω2 - tω1

= 19 – 16.2

= 2.8°C

Increase in dry bulb temp =td2 - td1

= 0

Note,

td = Constant

Step :2

Increase in Specific enthalpy = h2 - h1

=54.5 – 45.5

= 9KJ/Kg

Step :3

Increase in Relative Humidity = ɸ2 - ɸ1

Where,



ɸ2 → h2 = 54.5KJ/Kg & ω2 = 0.0124

(tω2) = 19°C use

ɸ2 = 68%

Increase in Relative Humidity = ɸ2 - ɸ1

= 68 – 50

= 18%

Step :4 Amount of moisture added per min

mw = ma [ω2 – ω1]

Ma = V/V1

Where,

V1 = 1/ ρ1 = 1/1.3 = 0.7692 m3/Kg

Ma = V/V1 = 200/0.7692 = 260 Kg of dry air/min

So ,

mw = ma [ω2 – ω1]

mw = 260 [0.0124 – 0.0086]

mw = 0.988 Kg of moisture/min


5. For the atmospheric air at room temp of 30°C and relative humidity of 60%. Determine

partial pressure of air , humidity ratio , dew point temp , density and enthalpy of air.[AU MAY

2011]

Given:

Temproom = 30°C + 273 = 303K

Relative humidity (ɸ) = 60%/100 = 0.6

To find:

1. Partial pressure of air (Pair)

2. Humidity ratio (ω)

3. Dew point temp (tdew)

4. Vapour density (ρv)

5. Enthalpy of air (hair)

Solution:

Step :1

Partial pressure of air (Pair) = Patm - Pvapour

Where ,

Vapour pressure (Pvapour) = ɸ - Psaturation

Where ,

From steam table at (Troom)

Psaturation = 0.04242bar

Pvapour = ɸ × Psat


Pvapour = 0.6 × 0.04242

Pvapour = 0.025452bar × 100

Pvapour = 2.5452 KN/m2

Partial pressure of air (Pair) = Patm - Pvapour

Pair = 101.325 – 2.5452

Pair = 98.43798 KN/m2

Step :2

Humidity ratio (ω) = 0.622

= 0.622

ω = 0.01602 KJ/Kg of dry air

Step :3 To find dew point temp (tdp)

From steam table at Pv = 0.025452

(tdp) = 21.10°C

Step :4 To find vapour density (ρv)

ρv =

Where,

PvVv = mRvTv

Vv = =

Vv = 55m3/Kg


ρv = = 1/55

Vapour density (ρv) = 0.01818Kg/m3

Step :5 To find enthalpy of air

h= 1.005 × Troom + ω[2500 + 1.88×30]

= 1.005 × 30 + 0.01602[2500 + 1.88×30]

= 30.15 + 40.9535

h= 71.1035KJ/Kg


6. Atmospheric air at 1.0132 bar has a DBT of 32°C and a WBT of 26°C. Compute

1. The partial pressure of water vapour ,

2. The specific humidity ,

3.The dew point temp ,

4. The relative humidity ,

5. Degree of saturation ,

6. The density of air in the mixture ,

7. The density of vapour in the mixture ,

8. The enthalpy of the mixture. Use thermodynamic table only. [AU DEC 2008].

Given:

Patm = 1.0132bar × 100 = 101.32KN/m2

Dry bulb temp (DBT) = 32°C + 273 = 305K

Wet bulb temp (WBT) = 26°C + 273 = 299K

To find:

1. Pv

2. The specific humidity ,

3. tdp

4. ɸ

5. μ

6. ρa

7. ρv

8. h


Solution:

Step :1 To find partial pressure of water vapour (Pv)

Pv = Ps -

Where,

Saturation pressure (Ps) corresponding to tWBT

From steam table corresponding to tW = 26°C

Ps = 0.03360bar

Pv = 0.03360 -

Pv = 0.03360 – 0.00393519

Pv = 0.02966 bar × 100

Pv = 2.9664 KN/m2

Step :2 To find specific humidity (ω)

ω = 0.622

= 0.622

ω = 0.01875 Kg/Kg of dry air

Step :3 To find dew point temp (tdp)

From steam table , corresponding to

Pv = 0.02966 bar

tdp = 22.6°C


Step :4 To find relative humidity (φ)

φ =

Where,

Pvs = Corresponding to tDBT = 32°C

Pvs = 0.04753 bar

φ =

φ =

φ = 0.6240 ×100

φ = 62.40%

Step :5 To find degree of saturation (μ)

μ = =

μ =

μ =

μ = 0.6126 × 100

μ = 61.26%

Step :6 To density of air in mixture(ρair)

ρair =

=


ρair = 1.1235 Kg of air/m3

Step :7 Density of vapour in the mixture (ρv)

ρv =

Where,

Rv = = = 0.4618 KJ/ Kg. K

ρv =

ρv = 0.02105 Kg of vapour/m3

Step :8 To find enthalpy of mixture (hmix)

hmix = 1.005 td + ω[hg + 1.88 (td – tdp)]

Where,

Hg = corresponding to tDBT = 32°C

Hg = 2560 KJ/Kg.K

hmix = 1.005 ×32 + 0.01875[2560 + 1.88 (32 – 22.6)]

hmix = 32.16 - 48.3313

hmix = 80.4913 KJ/Kg


7. A sample of moist air at 1atm and 25°C has a moisture content of 0.01 by volume.

Determine the humidity ratio, the partial pressure of water vapour , the degree of saturation,

the relative humidity and dew point temp. [AU MAY 2010]

Given:

Pb = 1atm

Td = 25°C + 273 = 298K

Volume (V) = 0.01

To find:

1. ω

2. Pv

3. μ

4. ɸ

5. td

Solution:

Step :1

Specific humidity (ω) = ρwater×

=

ω = 0.01 Kg/Kg of dry air

Step :2


From Psychometric chart corresponding to specific humidity (ω) = 0.01 and dry bulb temp

25°C, mark point 1.

Pv = 11.5 mm Hg

φ = 50%

tdp = 14°C

Pv =

Pv = 0.015342 bar ×100 = 1.5342KN/m2

Step :3 To find degree of saturation

μ =

Where,

Pvs = = = 0.0306bar

μ =

μ =

μ = 0.4936 × 100

μ = 49.36%


8. It is required to design an air conditioning plant for a small office room for following winter

conditions.

Out door conditions = 14°C DBT and 10°C WBT

Required conditions = 20°C DBT and 60% R.H

Amount of air circulation 0.30 m3/min/person.

Seating capacity of office = 60

(1) Required condition is achieved first by heating and then by Adiabatic humidifying Determine the following (2) heating capacity of the coil in KW and the surface temperature required if the by pass factor of coil is 0.4. [AU DEC 2011]

Solution:

Step :1

Enthalpy (h1) → From Psychometric chart use WBT = 10°C

h1 = 29.5 KJ/Kg

V1 = 0.825 m3/Kg

Step :2

Enthalpy (h2) → From Psychometric chart use DBT = 20°C and 60% R.H

H2 = 42 KJ/Kg

Step :3

Amount of dry air supplied (ma1)

Ma1 =

Ma1 =

Ma1 = 0.363 Kg/sec


Step :4

Capacity of heating coil = Ma1 [h2 - h1 ]

= 0.363[42 – 29.5 ]

= 4.5375 KW

Step :5

By pass factor =

0.4 =

0.4 [ts – 14] = ts – 20

0.4ts – 5.6 = ts – 20

0.4ts – 14 - ts + 20 = 0

-0.6ts +14.4 = 0

0.6ts = 14.4

ts = 14.4/0.6

ts = 24°C

ts = 24°C + 273

ts = 297K

Step :6

Capacity of humidifier = Ma1 [ω2 - ω1 ]

From Psychometric chart use → td2 = 20°C & 60% RH → ω3 = 0.009

WBT 10°C & 60% RH → ω2 = 0.006

Capacity of humidifier = Ma1 [ω2 - ω1 ]

= 0.363 [0.009 – 0.006 ]


= 1.089 ×10-3Kg/sec

9. Saturated air at 20°C at a rate of 70m3/min is mixed adiabatically with the outside air at

35°C and 50% relative humidity at a rate 30m3/min . Assuming that the mixing process occurs

at a pressure of 1 atm. Determine the specific humidity , the relative humidity , the dry bulb

temp and the volume of flow rate of mixture. [AU DEC 2011]

Given:

Temp (T1) = 20°C

Volume (V1) = 70m3/min

Temp (T2) = 35°C

Volume (V2) = 30m3/min

Relative humidity (ɸ) = 50%

Pressure (P1) = 1 atm.

To find:

1. The specific humidity (ω) ,

2. The relative humidity (ɸ3) ,

3. The dry bulb temp (td3)

4. The volume of flow rate of mixture (Vm).

Solution:

Step :1

Amount of dry air supplied (ma1) = V/V1

Where,

From Psychometric chart use → T1 = 20°C & ɸ = 50%


V1 = 0.85m3/Kg

ma1 = V/V1

ma1 = 70/0.85

ma1 = 82.35Kg/min

Step :2

Amount of dry air supplied (ma2) = V/V2

Where,


V2 = 0.89m3/Kg

ma2 = V/V2

ma2 = 30/0.89

ma2 = 33.5Kg/min

Step :3


h1 = 57.5 KJ/Kg & ω1 = 0.015 Kg/Kg of dry air


h2 = 81 KJ/Kg & ω2 = ω3 = 0.018 Kg/Kg

Step :4

=

=


= 2.459

0.018 - ω3 = 2.459[ω3 - 0.015]

0.018 - ω3 = 2.459ω3 - 0.036885

0.018 – 0.036885 = ω3 + 2.459ω3

0.054885 = 3.245ω3

0.054885/3.245 = ω3

ω3 = 0.0168 Kg/Kg of dry air

Step :5

=

=

= 2.3333

81 - h3 = 2.3333[h3 - 57.5]

81 - h3 = 2.3333h3 - 134.16666

81 – 134.16666 = h3 + 2.3333h3

215.16666 = 3.3333h3

215.16666/3.3333 = h3

h3 = 64.5506 Kg/Kg of dry air

Step :6

From Psychometric chart use → h3 = 64.5506 & ω3 = 0.0168

The relative humidity (ɸ3) = 80%


The dry bulb temp (td3) = 24°C

V3 = 0.865m3/Kg

Step :7

mass of air mixture (ma3) = ma1 + ma2

= 82.5 + 33.5

ma3 = 115.85 Kg/min

Step :8

The volume of flow rate of mixture (Vm) = ma3 × V3

= 115.85 × 0.865

The volume of flow rate of mixture (Vm) = 100 m3/min


10. two steam of moist air , one having flow rate of 3Kg/sec at 30°C and 30% Relative

humidity, other having flow rate of 2Kg/s at 35°C and 65% Relative humidity, get mixed

adiabatically. Determine specific humidity and partial pressure of water vapour after mixing .

Take Cp steam = 1.86 KJ/Kg.K [AU MAY 2011]

Given:

Mass (m1) = 3Kg/s

Mass (m2) = 2Kg/s





To find:

1. ω3

2. Pw

Solution:

Step :1

From Psychometric chart use → td1 = 30°C & ɸ1 = 30%


From Psychometric chart use → td2 = 35°C & ɸ1 = 65%



Step :2

=

=

1.5 =

1.5[0.008 – ω3] = ω3 - 0.024

0.012 – 1.5ω3 = ω3 - 0.024

0.012 + 0.024 = ω3 + 1.5 ω3

0.036 = 2.5ω3

ω3 = 0.036/2.5


Step :3

From Psychometric chart use ω3 = 0.0144 Kg/Kg of dry air

Partial pressure of water vapour (Pw) = 16.5mm of Hg

Note,

760mm of Hg = 1.013 bar

16.5mm of Hg =( 1.013/760 ) × 16.5

Partial pressure of water vapour (Pw) = 0.02199 bar

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