KURENAI : Kyoto University Research Information Repository

t= [0.623{ 1 - exp(-1.318 x)}]7/8 (86)
Vx
In Fig.10, a comparison was made between the exact solution of Eq.(85) and
the value calculated by Eq.(86), too. As shown in this figure, Eq.(86)
can approximate the exact solution of Eq.(85) fairly well (within + 5%).
In view of Eqs.(84) and (86), one finally obtains the approximate
solution of Eq.(81),
* *
for 1.318 x < t
for 1.318 x*> t*
* Vx
t=[0.623{ 1 -exp(-1.318xk)}]7/8(87)
S
vx
t[0.623{ 1 - exp(-t*)}17/8(88)
The transient heat transfer coefficient can be obtained by substituting
Eqs.(87) and (88) into Eq.(78). It is given by
* *
for 1.318 x < t
St = 0.0296 Re x °•2[x * ]1/8 (89)
0 .623 { 1 - exp(-1.318x )}
* *
f or 1.318 x > t
St = 0.0296 Re x0•2[x 1l/8(90) 0
.623 { 1 - exp(-t )}
Here, St is Stanton number (St E h/(u WCPp) ). The transient heat transfer
coefficient calculated by Eqs.(89) and (90) is shown in Fig.11 in x vs.
St/(0.0296x*0•125Re 0•2) plot. The steady state heat transfer ciefficinet
x is obtained by putting w +0 in Eq.(89) and given by
•
_0.2
St st = 0.0288 Rex(91)
133
*