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L. VIJAYAKUMAR AP/MECHANICAL Page 1
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%
Dry bulb temp (td2) = 40°C
Relative humidity (ɸ2) = 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
L. VIJAYAKUMAR AP/MECHANICAL Page 2
Use → td1 = 20°C Relative humidity (ɸ1) = 40%
Enthalpy (h1) = 35KJ/Kg
Humidity (ω1) = 0.0058KJ/Kg of dry air
Step :2
From Psychometric chart
Use → td2 = 40°C Relative humidity (ɸ2) = 40%
Enthalpy (h2) = 90KJ/Kg
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
=
L. VIJAYAKUMAR AP/MECHANICAL Page 3
=
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
L. VIJAYAKUMAR AP/MECHANICAL Page 4
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
From Psychometric chart
Use → td1 = 15°C & ωd1 = 13°C
Enthalpy (h1) = 37KJ/Kg
Step :2
From Psychometric chart
Use → td2 = 25°C & ωd2 = 18°C
Enthalpy (h2) = 51KJ/Kg
L. VIJAYAKUMAR AP/MECHANICAL Page 5
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
L. VIJAYAKUMAR AP/MECHANICAL Page 6
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
Dry bulb temp (td1) = 21°C
Relative humidity (ɸ1) = 55%
Dry bulb temp (td2) = 35°C
Relative humidity (ɸ2) = 60%
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%
From Psychometric chart
V2 = 0.84 m3/Kg
humidity (ω1) = From Psychometric chart
L. VIJAYAKUMAR AP/MECHANICAL Page 7
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
L. VIJAYAKUMAR AP/MECHANICAL Page 8
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
Relative humidity (ɸ1) = 50%
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,
L. VIJAYAKUMAR AP/MECHANICAL Page 9
From Psychometric chart
tω2→ ω2 = 0.0124 & td = 23°C
(h1) = 54.5 KJ/Kg
tω2 = 19°C
Where,
From Psychometric chart
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,
L. VIJAYAKUMAR AP/MECHANICAL Page 10
From Psychometric chart
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
L. VIJAYAKUMAR AP/MECHANICAL Page 11
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
L. VIJAYAKUMAR AP/MECHANICAL Page 12
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
L. VIJAYAKUMAR AP/MECHANICAL Page 13
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
L. VIJAYAKUMAR AP/MECHANICAL Page 14
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
L. VIJAYAKUMAR AP/MECHANICAL Page 15
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
L. VIJAYAKUMAR AP/MECHANICAL Page 16
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 =
=
L. VIJAYAKUMAR AP/MECHANICAL Page 17
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
L. VIJAYAKUMAR AP/MECHANICAL Page 18
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
L. VIJAYAKUMAR AP/MECHANICAL Page 19
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%
L. VIJAYAKUMAR AP/MECHANICAL Page 20
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
L. VIJAYAKUMAR AP/MECHANICAL Page 21
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 ]
L. VIJAYAKUMAR AP/MECHANICAL Page 22
= 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%
L. VIJAYAKUMAR AP/MECHANICAL Page 23
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,
From Psychometric chart use → T2 = 35°C & ɸ = 50%
V2 = 0.89m3/Kg
ma2 = V/V2
ma2 = 30/0.89
ma2 = 33.5Kg/min
Step :3
From Psychometric chart use → T1 = 20°C & ɸ = 50%
h1 = 57.5 KJ/Kg & ω1 = 0.015 Kg/Kg of dry air
From Psychometric chart use → T2 = 35°C & ɸ = 50%
h2 = 81 KJ/Kg & ω2 = ω3 = 0.018 Kg/Kg
Step :4
=
=
L. VIJAYAKUMAR AP/MECHANICAL Page 24
= 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%
L. VIJAYAKUMAR AP/MECHANICAL Page 25
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
L. VIJAYAKUMAR AP/MECHANICAL Page 26
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
The relative humidity (ɸ1) = 30%
The relative humidity (ɸ2) = 65%
The dry bulb temp (td1) = 30°C
The dry bulb temp (td2) = 35°C
To find:
1. ω3
2. Pw
Solution:
Step :1
From Psychometric chart use → td1 = 30°C & ɸ1 = 30%
ω1 = 0.008 Kg/Kg of dry air
From Psychometric chart use → td2 = 35°C & ɸ1 = 65%
ω2 = 0.024 Kg/Kg of dry air
L. VIJAYAKUMAR AP/MECHANICAL Page 27
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
ω3 = 0.0144 Kg/Kg of dry air
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