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Tutorial 1
Mathematical Modeling and
Engineering Problem Solving
Lecture Notes
Dr. Rakhmad Arief Siregar
Universiti Malaysia Perlis
Applied Numerical Method
for Engineers
1
P. 1.1



Water accounts for roughly 60% of total body weight.
Assuming it can be categorized into six regions, the
percentages go as follows. Plasma claims 4.5% of the body
weight and is 7.5% of total body water. Dense connective
tissue and cartilage occupies 4.5% of the total body weight
and 7.5% of the total body water. Interstitial lymph is 12%
of the body weight, which is 20% of the total body water
and 4.5% total body weight. If intracellular water is 33% of
the total body weight and transcellular water is 2.5% of the
body water,
What percent of total body weight must the transcellular
water be?
What percent must the transcellular water be?
2
Solution 1.1
Components Total Body Weight Total Body Water
1 Plasma
4.5%
7.5%
2 Dense
4.5%
7.5%
3 Interstitial
12%
20%
4 Inaccessible
4.5%
7.5%
5 Intracellular
33%
55?%
6 Trancellular
1.5? %
2.5%
60%
100%
Total
3
P. 1.2


A group of 30 students attend a class in a
room that measures 10 m by 8 m by 3 m.
Each student takes up about 0.075 m3 and
gives out about 80W of heat (1W = 1J/s).
Calculate the air temperature rise during the
first 15 minutes of the class if the room is
completely sealed and insulated.
4
Notes



Assume the heat capacity, Cv , for air is 0.718 kJ/(kg K)
Assume air is an ideal gas at 20 ̊C and 101.325 kPa.
The heat absorbed by the air Q is related to the mass of the
air, m, the heat capacity, and the change in temperature by
the following relationship:
T2
Q  m  Cv dT  mCv (T2  T1 )
T1

The mass of air can be obtained from the ideal gas law:
m
PV 
RT
Mwt
Mwt: 28.97 kg/kmol (air)
R: 8.314 kPa m3/(kmol K)
5
Solution 1.2

Compute total heat absorbed Q of students
80
s
kJ
Qstudents  30 st  J / s 15 min  60

 2160 kJ
st
min 1000 J

Computer the mass
m
PV 
RT
Mwt
m
PVMwt
RT
(101.325 kPa )(10m  8m  3m  30  0.075 m3 )( 28.97 kg/kmol )
m
 286.3424 kg
3
(8.314 kPa m /( kmol K)(( 20  273.15)K)

Compute change of temperature
Qstudents
2160 kJ
T 

 10.50615 K
mC v
(286.3424 kg )(0.718 kJ/(kg K))
6
Solution 1.2
The temperature rise after 15 minutes:
 T = (20 ̊C +273.15) + 10.50615
T = 303.65615 K
or
 T = 30.50615 ̊C

7