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3
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Problem 1
Air enters a heating section at 95 kPa, 12°C, and 30 percent relative humidity at a
rate of 6 m3/min, and it leaves at 25°C. Determine
a) write all balance equations,
b) the rate of heat transfer in the heating section,
c) the relative humidity of the air at the exit
Problem 1
Hava bir kanalın sürekli akışlı ısıtma bölümüne 95 kPa basınç, 12 oC
sıcaklık ve %30 bağıl nemde, 6 m3/min debiyle girmekte ve 25 oC sıcaklıkta
çıkmaktadır.
a) Gerekli denge eşitliklerini yazarak,
b) Isıtma bölümünde havaya olan ısı geçişini,
c) Çıkıştaki havanın bağıl nemini,
d) Girişte havanın hızı 8 m/s ise çıkıştaki hızını hesaplayınız.
Air enters a heating section at a specified state and relative humidity. The rate of
heat transfer in the heating section and the relative humidity of the air at the exit
are to be determined.
Assumptions
- This is a steady-flow process and thus the mass flow rate of dry air remains
constant during the entire process.
- Dry air and water vapor are ideal gases.
- The kinetic and potential energy changes are negligible.
Analysis
The amount of moisture in the air remains constant (𝜔1 = 𝜔2 ) as it flows through
the heating section since the process involves no humidification or
dehumidification. The inlet state of the air is completely specified, and the total
pressure is 95 kPa.
a) write all balance equations
𝑚𝐵𝐸: 𝑚ሶ 1 = 𝑚ሶ 2 = 𝑚ሶ 𝑎
𝐸𝐵𝐸: 𝑚ሶ 𝑎 ℎ1 + 𝑄ሶ 𝑖𝑛 = 𝑚ሶ 𝑎 ℎ2
ℎ1 + 𝑞𝑖𝑛 = ℎ2
ℎ =?
ℎ = ℎ𝑎 + 𝜔ℎ𝑔
𝑐𝑝 𝑇1 + 𝜔1 ℎ𝑔,1 + 𝑞𝑖𝑛 = 𝑐𝑝 𝑇2 + 𝜔2 ℎ𝑔,2
𝑄ሶ 𝑖𝑛
𝐸𝑛𝐵𝐸: 𝑚ሶ 𝑎 𝑠1 + ሶ
+ 𝑆𝑔𝑒𝑛 = 𝑚ሶ 𝑎 𝑠2
𝑇𝑖𝑛
𝑞𝑖𝑛
𝑠1 + + 𝑠𝑔𝑒𝑛 = 𝑠2
𝑇𝑖𝑛
𝑞𝑖𝑛
𝑠𝑎,1 + 𝜔1 𝑠𝑔,1 + = 𝑠𝑎,2 + 𝜔2 𝑠𝑔,2
𝑇𝑖𝑛
𝐸𝑥𝐵𝐸: 𝑚ሶ 𝑎 𝑒𝑥1 + 𝐸𝑥ሶ 𝑄𝑖𝑛 = 𝑚ሶ 𝑎 𝑒𝑥2 + 𝐸𝑥ሶ 𝐷
𝑒𝑥1 + 𝑒𝑥 𝑄𝑖𝑛 = 𝑒𝑥2 + 𝑒𝑥𝐷
𝑒𝑥1 − 𝑒𝑥2 = ℎ1 − ℎ2 − 𝑇0 (𝑠1 − 𝑠2 )
The cp of air can be assumed to be constant at
1.005 kJ/kg·°C in the temperature range −10
to 50°C with an error under 0.2%.
𝑐𝑝 =?
ℎ𝑔 =?
The properties of the air are determined to be
𝑚𝑣 𝑃𝑣 0.622𝑃𝑣
𝜔= = 0.622 =
𝑚𝑎 𝑃𝑎 𝑃−𝑃𝑣
b) Then the rate of heat transfer to the air in the heating section is determined
from an energy balance on air in the heating section to be
c) Noting that the vapor pressure of air remains constant (𝑃𝑣,1 = 𝑃𝑣,2 ) during a
simple heating process, the relative humidity of the air at leaving the heating
section becomes
𝑚𝑣 𝑃𝑣
∅= =
𝑚𝑔 𝑃𝑔
𝑄ሶ
ሶ
𝑚ሶ 𝑎 𝑠1 + 𝑇 𝑖𝑛 + 𝑆𝑔𝑒𝑛 = 𝑚ሶ 𝑎 𝑠2
𝑖𝑛
𝑞
𝑠1 + 𝑇𝑖𝑛 + 𝑠𝑔𝑒𝑛 = 𝑠2
𝑖𝑛
𝑞
𝑠𝑎,1 + 𝜔1 𝑠𝑔,1 + 𝑇𝑖𝑛 +𝑠𝑔𝑒𝑛 = 𝑠𝑎,2 + 𝜔2 𝑠𝑔,2
𝑖𝑛
𝑇 𝑃 298 𝑘𝐽
𝑠𝑎,2 − 𝑠𝑎,1 = 𝐶𝑃𝑎𝑣𝑔 ln 𝑇2 − 𝑅 ln 𝑃2 = 1.005 × ln = 0.045 𝑘𝑔𝐾
1 1 285
𝑘𝑗
𝑠𝑔,1 = 8.865 𝑘𝑔𝐾
𝑘𝑗
𝑠𝑔,2 = 8.557 𝑘𝑔𝐾
𝑘𝐽 13.13 𝑘𝐽/𝑘𝑔
𝑠𝑔𝑒𝑛 = 0.045 𝑘𝑔𝐾 − 0.002768 8.557 − 8.865 − 313 𝐾
𝑠𝑔𝑒𝑛 = 0.0039 𝑘𝐽/𝑘𝑔𝐾
𝑘𝑔 𝑘𝐽
ሶ
𝑆𝑔𝑒𝑛 = 6.938 𝑚𝑖𝑛 × 0.0039 𝑘𝑔𝐾 = 0.027 𝑘𝐽/𝑚𝑖𝑛𝐾
ሶ
𝐸𝑥ሶ 𝐷 = 𝑇0 × 𝑆𝑔𝑒𝑛 = 293 × 0.027 = 7.94 𝑘𝐽/𝑚𝑖𝑛
Problem 4
Air enters a window air conditioner at 1 atm, 32 °C, and 70 percent
relative humidity at a rate of 2 m3/min, and it leaves as saturated air at
15 °C. Part of the moisture in the air that condenses during the process
is also removed at 15 °C. Write mass, energy, entropy and exergy
balance equations and determine the rates of heat and moisture removal
from the air.
a) write all balance equations
𝑚𝐵𝐸: 𝑚ሶ 1 = 𝑚ሶ 2 + 𝑚ሶ 𝑓
𝑚𝐵𝐸 𝑓𝑜𝑟 𝑤𝑎𝑡𝑒𝑟 𝑚ሶ 𝑎 𝜔1 = 𝑚ሶ 𝑎 𝜔2 + 𝑚ሶ 𝑓
𝐸𝐵𝐸: 𝑚ሶ 1 ℎ1 = 𝑚ሶ 2 ℎ2 + 𝑚ሶ 𝑓 ℎ𝑓 + 𝑄ሶ 𝑜𝑢𝑡
𝑚ሶ 1 𝑐𝑝 𝑇1 + 𝜔1 ℎ𝑔,1 = 𝑚ሶ 2 𝑐𝑝 𝑇2 + 𝜔2 ℎ𝑔,2 + 𝑚ሶ 𝑓 ℎ𝑓 + 𝑞𝑜𝑢𝑡
𝑄 ሶ
ሶ
𝐸𝑛𝐵𝐸: 𝑚ሶ 1 𝑠1 + 𝑆𝑔𝑒𝑛 = 𝑚ሶ 2 𝑠2 + 𝑚ሶ 𝑓 𝑠𝑓 + 𝑇𝑜𝑢𝑡
𝑏
𝐸𝑥𝐵𝐸: 𝑚ሶ 1 𝑒𝑥1 = 𝑚ሶ 2 𝑒𝑥2 + 𝑚ሶ 𝑓 𝑒𝑥𝑓 + 𝐸𝑥ሶ 𝑄𝑜𝑢𝑡 + 𝐸𝑥ሶ 𝐷
𝑒𝑥1 − 𝑒𝑥2 = ℎ1 − ℎ2 − 𝑇0 𝑠1 − 𝑠2
𝑒𝑥𝑓 − 𝑒𝑥0 = ℎ𝑓 − ℎ0 − 𝑇0 (𝑠𝑓 − 𝑠0 )
The amount of moisture in the air decreases due to dehumidification (𝜔2 < 𝜔1 ).
The mass flow rate of air is
16
Applying the water mass balance and energy balance equations to the combined cooling
and dehumidification section,
𝑚ሶ 𝑎 𝜔1 = 𝑚ሶ 𝑎 𝜔2 + 𝑚ሶ 𝑓 𝑚ሶ 𝑎 : 𝑑𝑟𝑦 𝑎𝑖𝑟 or
𝑚ሶ 1 𝜔1 = 𝑚ሶ 2 𝜔2 + 𝑚ሶ 𝑓 𝑚ሶ 1 : 𝑑𝑟𝑦 𝑎𝑖𝑟
𝑚ሶ 1 ℎ1 = 𝑚ሶ 2 ℎ2 + 𝑚ሶ 2 𝜔2 + 𝑄ሶ 𝑜𝑢𝑡
