semester test 1 MTX311 2016 worked out solutions. The memorandum has solutions are 100% correct including alternative working methods. The solutions are done in a step by step matter
1A (15) 1.2 m³/s of a gas/vapour mixture at 40°C, 1 bar, consisting of air and water vapour, is
fed to a cooling tower, see fig. 1. The mixture has a volume fraction of water vapour of 4.4%. In the
cooling tower, the mixture is cooled down to 20°C and a relative humidity of 90%. How much heat
has to be removed from the cooling tower and how much condensate is formed?
Strategy:
Note first both these conditions and the quantifies you have to calculate sound perfect for using the
psychrometric chart, which is by far the easiest way to solve this problem. Therefore, the strategy is
to use the pshychrometric chart; amount of condensate is mΔω, amount of heat extracted =mΔh.
First calculate the amount of (dry) air and relative humidity (is easier to calculate than the humidity
ratio) state 1
Use psychrometric chart to find humidity ratio and enthalpy of state 1 and state 2 (end state, 20°C,
φ=90%. The difference between state 2 and 1 gives Δω and Δh
Calculate condensate mass flow rate using mc= -mdry air Δω and the amount of heat that has to be
extracted using Q=ΔH= mdry air Δh
Amount of dry air:
1.2m³m/s, 4.4% H2O and 95.6% dry air;
Dry air volumetric flow rate is v=0.956*1.2=1.147m³/s
Specific volume of the dry air: PV=RT V=RT/P=0.287*313.15/100=0.899m³/kg
Dry air mass flow rate m=v/V=1.147/0.899=1.275kg/s
Relative humidity:
Pressure fraction Pi/Ptot = volume fraction for ideal gas mixtures: Partial pressure of water vapour:
Pi=4.5%*100=4.5kPa
Saturation pressure: Psat,water(40°C)=7.384 (table B1.1)
Φ=PH2O/Psat=4.4/7.384=0.609=59.6%
Now use psychrometric chart: (top right corner):
h=130.2kJ/kg dry air
ω=0.0284 kg H2O/kg dry air
more accurate even, and therefore better is to calculate ω using:
=0.0286 kg H2O/kg dry air
But both methods will give you full mark
Now determine enthalpy and humidity ratio of state 2:
Graph:
h=73. 2kJ/kg dry air
ω=0.0136 kg H2O/kg dry air
or calculate: water vapour partial pressure =φpsat=0.9*2.339=2.105kPa
=0.0134 kg H2O/kg dry air
Δh=h2-h1=73.2-130.2=-57 kJ/kg dry air
Δω=ω2-ω1=0.0134-0.0286=-0.0152 kg H2O/kg dry air
, which means that we have to extract 57kJ and 15.2g of water from every kg of dry air:
heat extracted: Q=ΔH= mdry air Δh=1.275*-57=-72.68 kJ (- for out)
condensate formed: mc= mdry air Δω =1.275*-0.0152=0.01938kg/s=19.38 g/s
1B
Strategy.
The water in the cooling tower mantle has to be cooled down to 20°C. this is only possible when the
temperature inside the cooling tower, after evaporation of the water, is less than 20°C. The air in the
cooling tower is cooled by evaporation of water; For a given dry air temperature, the lowest
temperature of the wet air is achieved when an excess liquid water is present; the water evaporates
forming a gas mixture of 100% relative humidity. We are therefore looking for the dry bulb
temperature for a given wet bulb temperature of 20°C at 40% relative humidity. Use the
psychrometric chart;
TWB=20°C, φ=40% →Tdry bulb=30.2°C.
1C
First solve combustion with stoichiometric oxygen (make table)
Add nitrogen to get combustion with stoichiometric air, then add excess air to get full reaction
equation
Reaction equation Table:
in out
C H O N C H O N
1 C4H12N2 4 12 2
7 O2 14
4 CO2 4 8
6 H2O 12 6
1 N2 2
total 4 12 14 2 4 12 14
4 carbon gives 4 CO2 (needing 8 oxygen atoms)
12 hydrogen forms 6 water (needing 6 oxygen atoms)
2 nitrogens forms one N2
6+8=14 oxygen atoms needed for combustion; non are present in the fuel,
therefore 14/2=7 oxygen molecules have to be supplied for stoichiometric combustion
Reaction equation with stoichiometric oxygen:
C4H8(NH2)2 +7O2 4CO2 +6H2O +N2
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