i will includea study guide

SOLUTIONS
1. Answer
PART -1
Tube side: Sea wate
Water temperature at inlet TC1= 20 0C
Water temperature at outlet TC2= 30 0C
Length of heat exchanger L= 3 mete
No. of tubes n = 3
Outer Diameter do= 1.315” = 0.033401 mete
Inner Diameter di = 1.049”= 0.0266446 mete
Annulus - engine oil
Temperature at inlet Th1 = 60 0C
Temperature at outlet Th2= 40 0C
Mass flow rate mh= 5 kg/s
Outer Diameter Do= 4.5” = 0.1143 mete
Inner Diameter Di = 4.206”= 0.1068324 mete
Properties of sea water at temperature (30+20)/2= 250C
Density ρ c= 997.1 kg/m3, μ= 10.07 x 10-4 N.s/m2
Cp= 4181 J/Kg K, K= 0.599 W/m. k
Pr= 7.05
Properties of engine oil at temperature (40+60)/2= 500C
Density ρ h= 870 kg/m3, μ= 0.123 N.s/m2
Cp= 2006 J/Kg K, K= 0.141 W/m. k
Pr= 1749, μ w= 0.3 N.s/m2
Annulus- Heat transfer coefficient:-
De= 4 Ac/ (3Πdo)
Where,
De= equivalent dia. for transfer of heat
Ac = (Π/4) [Di2-3do2] = (Π/4) [(0.1068324)2-3(0.033401)2]
Ac = 0.00633524 m2
De= 4 Ac/ (3Πdo) = De= 4(0.00633524) / (3Π x0.033401) = 0.080 m
De= 0.080 m
Similarly we can get the value of hydraulic dia. DH:-
DH= 4 Ac/ (ΠDi+3Πdo) = 4(0.00633524) / (0.1068324 Π + 3Π x0.033401)
DH=0.0389 m
Reynolds number Re:-
Re= (ρ h um DH)/ μ
Where,
u m = m/( ρ h x Ac) = 5/(870 x 0.00633524) = 0.9072 m/s
Re= (870 x 0.9072 x 0.0389)/ 0.123 = 249.61
Hence flow will be lamina
According co-relation of Sieder & Tate:-
Nu = 1.86 [Re Pr DH /L] 1/3 [ μ / μ w] 0.14
Nu = 1.86 [249.61 x 1749 x 0.0389 /3] 1/3 [0.123/0.3] 0.14
Nu = 1.86 x 15.73= 29.2578
h o= Nu. K/ De = 29.2578 x 0.141 / 0.080 = 51.56W/m2. k
h o= 51.56 W/m2. K
Tubes- Heat transfer coefficient:-
Q= m c Cp c (TC2-TC1) = m h Cp h (Th1-Th2)
m c = m h Cp h (Th1-Th2) / [Cp c (TC2-TC1)]
m c = 5x 2006 x20 / [4181x 10] = 4.798 kg/s
m c =4.798 kg/s
u=mc/ (ρ c x Ai x n), where Ai= (Π/4) di2 = (Π/4) (0.0266446)2 = 0.00055758 m2
u= 4.798 / (997.1 x 0.00055758 x 3) = 2.8766 m/s
Reynolds number Re:-
Re= (ρ c u di)/ μ = (997.1 x 2.8766 x 0.0266446) / (10.07 x 10-4) = 75892.33
Re= 75892.33
Form the co-relation of Petuknov & Krillov:-
f= [3.64 log Re -3.28]-2 = [3.64 log (75892.33) -3.28]-2
f= 0.004766
Nu= (f/2) Re Pr / [1.07 +12.7(f/2)0.5 (Pr2/3-1)]
Nu= [(0.002383) x 75892.33 x 7.05]/ [1.07 +12.7(0.002383)0.5 (7.052/3-1)]
Nu= 1275 / 2.7293 = 467.15
Nu= 467.15
h i= Nu. K/di
h i= 467.15 x 0.599 /0.0266446 = 10502 W/m2. K
Overall heat transfer co-efficient
As per standard
Fouling resistances are as following
R f o = 0.000176 m2. K/w, and R f i = 0.000088 m2. K/w
K=43 W/m K
U o= 1/ [(ro / ri hi) + (ro R f i
i) + (ro /k) log (ro
i) + R f o + (1/ ho)]
U o= 1/ [(0.00011936) + (0.00011031458) + (0.00003811) + 0.000176 + (0.01939)]
U o= 1/ 0.01983378458 = 50.419 W/m2. K
U o= 50.419 W/m2. K
Total pressure drops in Annulus :-
∆P=4f (2LN hp/Dh) ρ u2/2
Where,
f= 16/Re = 16/249.61=0.064
Before going further we have to find heat transfer area
Q= m h Cp h (Th1-Th2) = 5x 2006 x20 = 200600 = 200.6 KW
Q= U o x A o x ∆T lm
∆T lm = [∆T1 -∆T2] / [log (∆T1 /∆T2)] = 30 / [log4] = 49.828
200600 = U o x A o x ∆T lm = 50.419 x 49.828 x A o
A o = 79.85 m2
Surface area, A s = 2Πd0L = 0.6295 m2
N hp = A o / (n x A s) = 42.28 = 42 approx
DH=0.0389 m (as found above)
∆P=4f (2LN hp/Dh) ρ u2/2 = 4 x 0.064 x (2 x 3 x 42 /0.0389) x 870 x (0.9072)2/2
∆P=4 x 0.064 x (2 x 3 x 42 /0.0389) x 870 x (0.9072)2/2 = 593726.2 Pa
∆P= 593.73 K Pa
PART-2
As per question we have to use the same dimensions and value as in part 1
If we are using finned tube double pipe heat exchanger, the value of heat transfer co-efficient hi will be same but the value of ho will be affected due to consideration of fins
So now we have to calculate the value of heat transfer co-efficient ho
Reynolds number Re:-
Re= (ρ h um DH)/ μ
Where,
u m = m/( ρ h x Ac)
Ac = (Π/4) [Di2-do2 Nt] - (δ Hf Nf Nt)
Where,
No. of tubes, Nt = = 3 (given)
Fin thickness, δ= 0.9 mm = 0.0009 m (assumed)
Fin height, Hf = 0.0127 m (assumed)
No. of fins per tubes, Nf = 30 (assumed)
MOC = ca
on steel
Thermal conductivity K = 52 W/mk
Ac = (Π/4) [Di2-do2 Nt] - (δ Hf Nf Nt)
Ac = (Π/4) [(0.1068324)2-3(0.033401)2] – (0.0009 x 0.0127 x 30 x 3)
Ac = 0.00530654 m2
Hence,
u m = m/( ρ h x Ac) = 5/(870 x 0.00530654) = 1.083 m/s
Hydraulic dia.:-
DH= 4 Ac/ [Π (Di+ Nt do) + 2 Hf Nf Nt] =
DH= 4 Ac/ [Π (0.1068324 + 3 x 0.033401) + (2 x 0.0127 x 30 x 3)]
DH= (4 x 0.00530654) / 2.936 = 0.00723 m
Re= (ρ h um DH)/ μ = (870 x 1.083 x 0.00723)/ 0.123 = 55.38
Hence flow will be lamina
According...