Answer:
heat rate= 1281W
length = 15.8m
Explanation:
we have this data to answer this question with
Tmi = 85 degrees
Tmo = 35 degrees
Ts = 25 dgrees
flow rate = 25 degrees
using engine oil property from table a-5
Tm = Tmo - TMi/2 = 333k
u =0.522x10⁻²
k = 0.26
pr = 51.3
cp = 2562 J/kg.k
mcp(Tmo-Tmi) =
0.01 x 2562(35-85)
= 1281 W
we find the change in Tim
= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]
= -50/ln0.167
= -50/-1.78976
= 27.9°c
we finf the required reynold number
4x0.01/πx0.003x0.522x10⁻²
= 0.04/0.00004921
= 812.8
= 813
we find approximate correlation
NuD = hd/k
NuD = 3.66
3.66 = 0.003D/0.26
cross multiply
0.003D = 3.66x0.26
D = 3.66x0.26/0.003
= 317.2
As = 1281/317x27.9
= 0.145
As = πDL
L = As/πD
= 0.145/π0.003
= 0.145/0.009429
L = 15.378
