Question

a) What is the physical interpretation of the wet-bulb temperature? b) During winter time a room is maintained at a steady temperature and humidity by an air-conditioning system shown. 200 m/min of room return air at 30°C, 50% relative humidity is first mixed with 40 m/min of outdoor at 10°C, 20% relative humidity. The mixed air is then introduced to a heater where the temperature of the air is raised to 45°C and supplied to the room. (Note that air and the steam in the heater do not mix. Steam simply supplies the heat.) Do the following using the ASHRAE Psychrometric chart only: 1) Plot the processes on the ASHRAE Psychrometric chart and submit. Show all the states clearly on the chart. Determine: ii) the required capacity of the steam heater, kW III) the relative humidity, , of the air supplied to the room,% IV) Is the calculated relative humidity of the supply air acceptable in common engineering practices? Explain why. v) the required mass flow rate of steam through the heater if saturated vapor is supplied at 150 kPa and exits the heater as saturated liquid at the same pressure, kg/min Stap 150 Pa 200 Mixed air 49°C Room 2 10.30 Steam heater Sat 150 PS

Answer 1

SUBPART A

When a thermometer is used to take the reading of the atmospheric air, but its bulb is covered in a wet cloth with air flowing over it, then the temperature recorded is the WBT of the room air. Now the important thing to note is that air can only absorb a certain amount of moisture in itself. When the moisture from the wet-bulb/cloth is converting into vapour, then it takes place by absorbing some heat from the bulb, thus lowering its temperature. Hence the WBT is always lesser than the DBT or normal temperature. Now as the relative humidity increases, the ability to absorb of the room air to absorb the moisture decreases. When it gets completely saturated, the moisture on the bulb does not vaporize. Hence it does not absorb any heat from the bulb, and thus the WBT becomes same as the DBT. Hence the significance of WBT is that it gives us an idea about the moisture content in the atmosphere. Closer the WBT to the actual temperature, higher is the moisture content of the atmosphere

SUBPART B

Important assumptions are

• Mixing is adiabatic in nature
• The heater is 100% efficient.

part 1

The process as depicted on a psychrometric chart is as follows:- (Please note that due to copyright issues I am not allowed to use ASHRAE psychrometric chart)

Specific volume (m®/kg) Relative humidity (%) 100 90 80 180/ 70 Absolute humidity (kg/kg) 60 50 40 -0.0550 170) 30 -0.0500 160/40 -0.0450 150 140) -0.0400 1220V 130 354 -0.0350 120 20- 110 -0.0300 100 30 000 90 -0.0250 80 -0.0200 70/25 60 to 0986 -0.0150 return 50 mix exit 40 -0.0100 30 20 -0.0050 1001 05096 outside 0.0209 10 15 04840 10.8810.900 0.920 0.940 -0.0000 65 20 25 30 35 40 45 50 55 60

part 2

From the chart, we aquire the following data which we will need in our calculations

• Volume flow rate at 'mix' V = 200+40 = 240 m3/min = 4 m3/s
• density at 'mix' rho = 1.17 kg/m3
• specific enthalpy at 'mix' h1 = 55.2 kJ/kg
• specific enthalpy at 'exit' h2 = 74.2 kJ/kg

Thus the capacity of the steam heater needs to be Q = V*rho*(h2-h1) = 4 * 1.17 * (74.2-55.2) = 88.92 kW

part 3

From the chart, we see that the exit relative humidity is at 18.7%

part 4

Frankly speaking, only a vague answer can be presented to the asked question. This is because "common engineering practices" can vary over a wide range, whose requirements vary accordingly. Our output air has high temperature and low relative humidity. This means that evaporation of water is rapid, and there are little to no chances of problems caused by high moisture content occurring. Hence this is perfect for electronic devices, for which moisture is bad. On the other hand, high evaporation rates can cause heat to be extacted from materials fast enough to cause shrinkage. This can be bad for materials like metals and concretes as they may give rise to surface defects (similar to our lips cracking during winter).

part 5

At 150 kPa, the enthalpy of saturated vapour is hw1 = 2693.11 kJ/kg, and the enthalpy of saturated liquid is hw2 = 467.08 kJ/kg

Thus the required mass flow rate of steam is Q/(hw1-hw2) = 88.92/(2693.11 - 467.08) = 0.0399 kg/s

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