Question

A 10 ft long double pipe heat exchanger consisting of a 1 in sch 40 (1.315 in OD, 1.029 in ID) inner pipe within a 4 in sch 40 (3.998 in ID) outer pipe uses chilled water to cool hot glycerin (the water flows in the annulus and glycerin in the inner pipe). The pipes are constructed of AISI 302 stainless steel. The mean velocities of the water and glycerin are 4 ft/s and 1.3 ft/s, respectively. If the average bulk temperatures of the water and glycerin are 45 ◦F and 180 ◦F, respectively, compute the overall heat transfer coefficient based on the outer area of the 1 in diameter pipe. Neglect fouling.

Answer 1

given that :

velocity of glycerine = 1.3 ft/s =39.624 cm/s

velocity of water = 4 ft/sec =121.92 cm/s

mean temperature of water t1 = 45 F =7.22 C

  mean temperature of glycerine t2 = 180 F =82.22 C

from table of properties ,

for water at 7.22 C ,  ρ =999.8 kg/m³ , k=0.575 w/m-k , Pr =10.42 , v (kinematic viscosity) =1.425*10^-6 m²/s

for glycerine at 82.22 C ,  ρ =1226.67 kg/m³ , k=0.3066 w/m-k , Pr =2100, v (kinematic viscosity) =1.63*10^-4 m²/s

length of heat exchanger (L) = 10 ft = 304.8 cm

inner diameter of outer pipe(D ) = 3.998 inch = 10.15 cm

outer diameter of inner pipe(D0) = 1.315 inch = 3.34 cm

inner diameter of inner pipe (Di)= 1.029 inch = 2.6137 cm

inner surface area of inner pipe (Ai)= pi*Di*L =2502.77 cm²

outer surface area of inner pipe (A0)= pi*D0*L =3198.24 cm²

inner surface area of outer pipe (A)= pi*D*L = 9718.98 cm²

now , reynolds no for glycerine is given by ,

Re(gly) = 39.624 *0.01*2.6137*0.01/(1.63*10^-4 ) =63.536 , hence laminar flow

so nusselt no will be Nu =4.36 ( in case of laminar pipe flow assuming constant heat flux )

so, hi * 2.6137*0.01/0.3066 =4.36

so hi = 51.15 w/m²-k  

similarly for water, Re(water) = 121.92*0.01*(10.15-3.34)*0.01/1.425*10^-6 =14566.23 , turbulant flow

so, nusselt no is given by (Nu) =0.023Re^0.8Pr^0.4 = 125.76

so, h0 *(10.15-3.34)*0.01/0.575 = 125.76

h0 = 1061.85 w/m²-k

where hi and h0 are convective heat transfer coefficients on glycerine and water side respectively.

now , equivalent thermal resistance neglecting thermal resistance of wall is given by,

Req = 1/hiAi +1/h0A0

= {1/(51.15*2502.77) + 1/( 1061.85*3198.24)}*10⁴

= 0.08106 k/w

hence , overall heat transfer coefficient based on outer area of inner pipe (glycerine pipe)is given by ,

U0 *A0 = 1/Req

U0 =1/( Req*A0) = 1/(0.08106*3198.24)*10⁴

therefore, U0 =38.573 w/m²-k