Theoretical Competition:
12 July 2011
Question 4
Page 1 of 4
4. A Simplified Heat –Engine Model of Tropical Cyclone
troposphere
temperature T2
P
A
Q1,T1
C
B
D
D
C
sea surface
temperature T1
Q2,T2
O
A
V
B
Strom centre
Fig. 1 Showing of the flow of an airmass
Fig. 2 the Carnot’s engine
An airmass  m flows from a high-pressure region A towards the low-pressure region, the storm
centre B, gathering moisture and latent heat of vaporization along the way. Airmass is
supersaturated at B, precipitates rain there releasing the latent heat and rises towards the
troposphere.
4.1 Use the First Law of Thermodynamics to find Q1 (see Fig.2) the term of
L = latent heat of vaporization
 q = mass of moisture picked up by  m between A and B.
PA = pressure at A and PB = pressure at B.
And the universal gas constant R , T1 , molar mass M of the air.
4.2 What is the net work done by this Carnot engine for the airmass  m ?
4.3 Assuming that the work output of the Carnot engine is to increase the kinetic energy of the
airmass, find the expression for vB , the velocity of the eyewall of the strom.
4.4 Taking vA  0 , compute the numerical value for vB using: T1 =313 K, T2 =213 K, R = 8.31
J/mol. K, the relative humidity of moist air = 75%, the saturation vapour pressure = 42 mbar, PA
=1000 mbar, PB = 950 mbar, mean molar mass for air, M = 29 103 kg.mol1 , molar mass for
water = 18 103 kg.mol1 , L = 2.26 106 J.kg 1 , density of dry air = 1.3 kg.m 3 . Note that 1 bar
= 105 pascals.
Theoretical Competition:
12 July 2011
Question 4
Page 2 of 4
4.5 Given that the azimutual velocity v and the radial distance r from the eyewall of the cyclone are
1
such that vr 2  constant, and that the velocity v A at the outer edge of the cyclone is  10 m/s,
calculate the effective radius of the cyclone. [Take rB to be  10 km.]
Theoretical Competition:
12 July 2011
Question 4
Page 3 of 4
SOLUTION
 Q1   U  P V
4.1
[The first law of thermodynamics]
…………….. (1)
Where  U  CP T  L q, T  T1  constant
 T  0, P V   V  P

Hence  Q1  L Q V  P
 Q1  L q 

Q1  Lq 
Q1  Lq 
m
PV 
Note also that
M
 mRT1
M
M
PB
P
1
P
 PdP,
PA
ln
…………….. (3)
RT1
 m RT1
 mRT1
M
…………….. (2)
q 

dq
from A to B
PA
PB
…………….. (4)
4.2 Net work done by the Carnot engine is
W  Q1  Q2
…………….. (5)
And according to the Second Law of Thermodynamics, we have

4.3
Q2
Q
 1
T2
T1
…………….. (6)
 T 
W  1  2  Q1
 T1 
…………….. (7)
 T 
 mRT1 PA 
W  1  2   Lq 
ln 
M
PB 
 T1  
………….. (8)
 T 
 mRT1 PA 
1
ln 
 m  vB2  vA2   1 2   Lq 
2
M
PB 
 T1  
1
vB
  T   q RT1 PA  2 
2

 2  1  2   L

ln   v A 
PB 
  T1    m M



Theoretical Competition:
12 July 2011
Question 4
Page 4 of 4
 T2  303  213
 0.297
1   =
303
 T1 
RT1 PA
ln
 4454 J.kg 1
M
PB
4.4
q  100%  75% of mass at saturation vapour pressure
 0.0075 kg.m3
q
 0.0058 (dimensionless)
m
q
L
 13108 J.kg 1
m
vA  0
 vB  102 m/s = 368 km/h
1
1
1
4.5 From vr 2 = constant, we have vA rA2  vB rB2 , rA  1040 km