WK 7 TUTE AnswerSheet
description
Transcript of WK 7 TUTE AnswerSheet
TransportPhenomenaCHEN3001Week 7 tutorial solutions
1. Calculate average h value based on the rate of heat generated by the microchip,
TThAVq ss .
Then, assuming that the flow is laminar, we can use the equation for laminar flow over a flat
plate, 315.0 PrRe664.0Nu .
Use the film temperature (Tf = 42.5 deg. C) to evaluate the properties of air: Kinematic
viscosity = 17x10^‐6 m2/s, Pr = 0.705, thermal conductivity = 0.0273 W/m.K.
Bulk fluid velocity = 3.95 m/s.
Check Reynolds number to see if the flow is indeed laminar!
The solution to the boundary layer equations shows that for maximum heat transfer, we
want to have the thinnest boundary layer over the surface of the chip. To minimise the
boundary layer thickness and maximum the heat transfer coefficient, we want to orient the
short dimension of the chip parallel to the flow.
2. Q = Heat transfer rate = Heat flux x surface area =5.7x10^3 x (pi D L) = 2.04 x 10^3 kJ/hr
Q = (mass flow rate) cp (Texit – Tinlet)
Texit = 16.0713 deg. C
Using Colburn analogy at T(average) of approximately 16 deg. C,
Average velocity = 3.75 m/s
Re = 95242 (Turbulent)
fF = 0.0046
Cf/2 = fF/2 = St Pr^2/3 ‐‐> St = (0.0046/2) / (7^(2/3)) = 0.001059
St = h/(rho V Cp) ‐‐> h = 9851 W/m2.K
q = h(Ts – Tfluid) ‐‐> Ts = 16.23 deg. C
3. Use the following properties of oil at 20 deg. C (this is the inlet temperature; so, this is an
approximation only. The calculations may have to be re‐done if the exit temperature is <<
inlet temperature):
Density = 888 kg/m3, thermal conductivity = 0.145 W/m.oC, kinematic viscosity =
901 x 10^‐6 m2/s, heat capacity = 1880 j/kg.oC, Pr = 10400.
Check Reynolds number to determine length of entrance region ‐‐> Re = 666, so:
DLt PrRe0575.0 =119,480 m.
So, the entrance length region of the pipe is much longer than the pipe itself! This is typical
of fluids with very high values of Pr number.
Use equation for thermally developing region: 32PrRe04.01
PrRe065.066.3
LD
LDNu
So, h = 18.0 W/m2.oC. For the case of constant surface temperature,
a.
Lp
s
is
es hCm
A
TT
TT
exp
So, Texit = 19.71 deg. C. As the mean temperature of the oil doesn’t decrease very much
from 20 deg. C, we don’t need to re‐evaluate the properties of the oil at the average
temperature.
b. Determine log mean temperature difference, ie
ie
TT
TTT
lnln , then calculate heat
transfer rate, lnThAQ Ls = ‐67.4 kW
c. As flow is laminar, f = 64/Re. Then, D
LvfP
2
2 =1.14 x 10^5 Pa
Pumping power required = P x volume flow rate
= 16.1 kW
4. This is a conjugate heat transfer problem, i.e.
oo
io
ii
oir
hDk
DD
hD
TTQ
1
2
)/ln(1,,
Calculate h value inside the pipe:
As flow is fully turbulent, we can use Dittus‐Boelter equation. Properties of water at
the mean temperature: thermal conductivity = 0.67 W/m.K, Pr = 2.2.
k
DhNu i
D 3154 PrRe023.0 ‐‐> hi = 2765 W/m2.K
Calculate h value outside of pipe, using the Zukauskas correlation:
k
DhCNu i
s
nmDD
4
1
Pr
PrPrRe
where:
Re C m
1‐40 0.75 0.4
40‐1000 0.51 0.5
1000‐2x105 0.26 0.6
2x105 – 106 0.076 0.7
If Pr ≤ 10, n = 0.37, otherwise n = 0.36.
Note: I realised I haven’t talked about this correlation in class. The benefit of this correlation
is that the fluid properties are all evaluated at the bulk fluid temperature ( T ), except for
Prs, which is evaluated at the surface temperature. Now, as we don’t know the surface
temperature of the pipe at this stage, let’s assume that the change in Prandtl number
between sT and T is negligible, so sPrPr is approximately 1.
Properties of air at the bulk fluid temperature (25 deg. C):
Kinematic viscosity = 15.89 x 10^‐6 m2/s, thermal conductivity = 0.0263 W/m.K, Pr =
0.707.
Re = 31,466
ho = 120 W/m2.K
The rate of heat transfer can then be calculated: Q = 489 W/m.