New Section 9 Page 1
New Section 9 Page 2
New Section 9 Page 3
Problem 2 (30 points)
A concentric tube heat exchanger in the parallel flow arrangement is used to cool hot engine oil
from Th,i to Th,o using cold water. The hot fluid (engine oil), with thermal conductivity k and
specific heat cp,h = cp, enters the inner tube at a mass flow rate m h = m . The flow can be assumed
as laminar and fully developed along the entire tube length. The cold fluid (water) with specific
heat cp,c = 2cp enters the outer tube at Tc,i and mass flow rate m c = 0.5m . The inner tube diameter
is D, and the wall thickness can be neglected.
cp,c, , ṁc, 𝑇𝑇𝑐𝑐,𝑖𝑖
𝑇𝑇ℎ,𝑜𝑜
𝐷𝐷
cp,h, , ṁh, 𝑇𝑇ℎ,𝑖𝑖
(a) Assuming uniform inner tube wall temperature, evaluate the heat transfer coefficient hi of
the hot fluid, in terms of the given symbols.
For fully developed laminar flow with constant wall temperature, according to Eq. 8.55,
hi D
= 3.66
k
k
hi = 3.66
D
Nu
=
D
(b) Express the outlet temperature of the cold fluid Tc,o and the effectiveness of the heat
exchanger ε, in terms of the given symbols.
m h c p ,h (Th ,i − =
Th ,o ) m c c p ,c (Tc ,o − Tc ,i )
⇒ Tc ,o =Tc ,i +
m c p
(Th ,i − Th ,o )
1/ 2m * 2 c p
⇒ Tc ,o = Tc ,i + Th ,i − Th ,o
=
ε
Ch (Th ,i − Th ,o ) m h c p ,h (Th ,i − Th ,o ) Th ,i − Th ,o
q
=
=
=
qmax Cmin (Th ,i − Tc ,i ) m h c p ,h (Th ,i − Tc ,i ) Th ,i − Tc ,i
ε
=
Cc (Tc ,o − Tc ,i ) m c c p ,c (Tc ,o − Tc ,i ) Tc ,o − Tc ,i
q
=
=
=
qmax Cmin (Th ,i − Tc ,i ) m c c p ,c (Th ,i − Tc ,i ) Th ,i − Tc ,i
Or
1
Division (circle one):
1
2
3
4
5
Last _______________ , First _______________
(c) The convective heat transfer coefficient ho of the cold fluid is found to be ho = hi. Express
the tube length L needed for the heat exchanger, in terms of the given symbols.
Solution approach 1: LMTD method:
=
q m h c p ,h (Th ,i − T=
UA
h ,o )
1 1 1
=
+ ;
U hi ho
∆T1 − ∆T2
ln(∆T1 / ∆T2 )
(1)
π DL
A=
∆T1 = Th ,i − Tc ,i ;
∆T2 = Th ,o − Tc ,o = 2Th ,o − Th ,i − Tc ,i
All plug in (1)


mc
Th ,i − Tc ,i
⇒ L = p ln 
hiπ D  2Th ,o − Th ,i − Tc ,i



Solution approach 2: ε-NTU method. From Table 11.4
NTU = −
=
NTU
ln [1 − ε (1 + Cr ) ]
(1)
1 + Cr
Th ,i − Th ,o
UA
ε
Cr 1;=
; =
Cmin
Th ,i − Tc ,i
1
 p
=
=
hi ; A π DL
; Cmin mc
2
All plug in (1)
=
U


mc
Th ,i − Tc ,i
⇒ L = p ln 

hiπ D  2Th ,o − Th ,i − Tc ,i



Solution approach 3: Internal flow with constant wall temperature. If part d is solved to
get Ts
Ts =
Th ,i + Tc ,i
2
 PL 
Ts − Th ,o
= exp  −
hi 
 mc


Ts − Th ,i
p


P =πD
(1)


mc
Th ,i − Tc ,i
⇒ L = p ln 
hiπ D  2Th ,o − Th ,i − Tc ,i



2
Division (circle one):
1
2
3
4
5
Last _______________ , First _______________
(d) Again the convective heat transfer coefficient ho of the cold fluid is found to be ho = hi.
Express the inner tube wall temperature at the inlet (Ts,i) and outlet (Ts,o) respectively, in
terms of the given symbols, and show that they are equal. (Therefore, the assumption of
uniform inner tube wall temperature made in part (a) is reasonable).
Tc,i
Tc,o
Ro”
Ro”
Ts,i
Ts,o
Ri”
Ri”
Th,i
Th,o
"
R=
i
qi"
=
1 1
= = Ro"
hi ho
Th ,i − Ts ,i Ts ,i − Tc ,i
=
Ri"
Ro"
Th ,i + Tc ,i
⇒ Ts =
2
Similarly,
qo"
=
Th ,o − Ts ,o Ts ,o − Tc ,o
=
Ri"
Ro"
Ts ,o
⇒=
⇒ Ts ,o
Th ,o + Tc ,o Th ,o + (Th ,i + Tc ,i − Th ,o ) Th ,i + Tc ,i
=
=
2
2
2
Ts ,i
=
3
Problem 3 (30 points)
A small flying animal (a Porg) leaves the water and flies into the air. Please consider the heat and
mass transfer on its wet wings as it takes flight. Assume its wing can be approximated as a flat
plate. The velocity of air is 10 m/s and the length of the wing is 0.1m. The width into the page is
0.3m. Don’t forget units in your answers.
ρair= 1.2 kg/m3, Tair=7°C, Ts=27°C, RH = 75%, 𝐷𝐻2 𝑂 𝑖𝑛 𝑎𝑖𝑟 = 2.3x10-5 m2/s, hfg = 2262 kJ/kg, νair
= 1.5e-5m2/s, αair= 2.0e-5 m2/s, kair = 0.024 W/mK, cp= 1.0 kJ/kgK
RH = 75%
Tair = 7°C
v = 10 m/s
Ts = 27°C
Evaporation from
wet surface
L = 0.1
m
Nusselt number correlations:
(a) (7 pts) Choose the appropriate correlation for the average convection over the wing and
calculate the relevant Nusselt number.
1
Division (circle one): 1 2 3
4
5
Last _______________ , First _______________
(b) (5 pts) Find heat (q) lost from convection (per unit length) on one side of the wing.
2
Division (circle one): 1 2 3
4
5
Last _______________ , First _______________
(c) (6 pts) Using the heat mass analogy, find the relevant dimensionless numbers for mass
transfer of water vapor evaporating from the wing surface.
(d) (8 pts) Solve for hm (mass transfer coefficient) and the evaporation rate of water away from
the wing, assuming it is thoroughly soaked. Hint: What must be the Relative Humidity (RH) at
the surface be for a phase change like evaporation to occur?
3
Division (circle one): 1 2 3
4
5
Last _______________ , First _______________
e) (4 pts) For the top surface of one wing, find the heat transfer from evaporation.
4