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Lana71 [14]
2 years ago
6

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 expec

ted 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%)
Engineering
1 answer:
Scrat [10]2 years ago
8 0

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%

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<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

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