Question

A 2"" Sch 40 stainless steel (k = 14.9 W/m-K) pipe is to be used as the interior pipe of a double pipe heat exchanger. The expected heat transfer coefficients of the hot and cold fluids are 1800 W/m2-K and 1600 W/m2-K respectively. Determine the total resistance to heat flow per unit length (i.e. R  L) with and without the resistance of the pipe wall. What is the percentage difference? (18.2%)

Answer 1

Answer:

the percentage difference is E = = 55.7% respect to the real value ( with resistance of the wall)

Explanation:

the resistance to heat flow R is

R=  ( 1/h₁ + ln (re/ri)/(2π*k*L) + 1/h₂)

where

h₁= heat transfer coefficient for the hot fluid = 1600 W/m²K

h₂= heat transfer coefficient for the cold fluid = 1800 W/m²K

re/ri = ratio between external radius and internal radius of the pipe = 2.375 /2.067  for 2"" Sch 40 stainless steel

k= thermal conductivity = k = 14.9 W/m-K

L = length of the pipe = 1 ( unit length)

therefore replacing values

R₁=  1/h₁ + ln (re/ri)/(2π*k*L) + 1/h₂ = 1/1600 W/m²K + ln(2.375 /2.067)/(2π*14.9 W/m-K*1) + 1/1800 W/m²K = 2.664 *10⁻³m²K/W

when the resistance of the pipe wall is neglected then R would be

R₂=  1/h₁ + 1/h₂ =  1/1600 W/m²K 1+ 1/1800 W/m²K =1.180 *10⁻³ m²K/W

the percentage difference between the total resistance with and without the pipe wall resistance per unit length  is

E = 1-(R₂/L) /( R₁/L) = 1- R₂/R₁=1- (1.180 *10⁻³/2.664 *10⁻³) = 0.557= 55.7%

E = = 55.7%