Engineering, 12.03.2020 02:24 jaxheisjax532

Hot liquid is flowing in a steel pipe with an inner diameter of D, = 22 mm and an outer diameter of D2 = 27 mm. The inner surface of the pipe is coated with a thin fluorinated ethylene propylene (FEP) lining. The thermal conductivity of the pipe wall is 15 W/m. K. The pipe outer surface is subjected to a uniform flux of 1200 W/m2 for a length of 1 m. The hot liquid flowing inside the pipe has a mean temperature of 180°C and a convection heat transfer coefficient of 50 W/m².K. The interface between the FEP lining and the steel surface has a thermal contact conductance of 1500 W/m2.K. a) Determine the temperatures at the lining and at the pipe outer surface for the pipe length subjected to the uniform heat flux. b) What is the total thermal resistance between the two temperatures? c) The ASME Code for Process Piping (ASME B31.3-2014, A323) recommends a maximum temperature for FEP lining to be 204°C. d) Does the FEP lining comply with the recommendation of the code?

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Answer from: whocares1234

Part a

The temperature of the FEP is 204 C while that of the outer surface is 205.12 C

Part b

The total Thermal resistance is 0.011 C/W

Part c & d

As the temperature of the FEP is 204 which is within the limits of the ASME code thus the FEP lining comply with the recommendation of the code.

Explanation:

As from the given data

Heat flux=q'=1200 W/m^2

Convection heat transfer coefficient hi=50 W/m^2K

T_FEPS=Temperature of the lining which is to be calculated

T_m=Mean temperature which is given as 180C

So the equation of the convection is given as

$\dot{q}=h_i(T_{FEPS}-T_m)\\1200=50(T_{FEPS}-180)\\24=T_{FEPS}-180\\T_{FEPS}=180+24 =204 C$

So the Temperature of the FEP is 204 C

The outer temperature is given as T_s which is equal to

$\dfrac{T_s-T_{FEPS}}{R_eq}=\dot{q}A_{out}$

The total resistance is calculated as

$R_e_q=R_F_E_P_S+R_o_u_t\\R_e_q=\dfrac{1}{hA_i}+\dfrac{ln(D_o/D_i)}{2\pi K L}$

Here

h is given as steel surface contact conductance of 1500 W/m^2K

A_i is given as the area of the internal diameter which is πD_iL

D_o is given as 27 mm=0.027 m

D_i is given as 22 mm=0.022 m

K is given as thermal conductivity of the pipe wall as 15 W/mK

L is given as 1 m so the values are

$R_e_q=\dfrac{1}{hA_i}+\dfrac{ln(D_o/D_i)}{2\pi K L}\\R_e_q=\dfrac{1}{h*\pi*D_i}+\dfrac{ln(D_o/D_i)}{2*\pi* K* L}\\R_e_q=\dfrac{1}{1500*\pi*0.022}+\dfrac{ln(0.027/0.022)}{2*\pi* 15*1}\\R_e_q=0.011 C/W$

$\dfrac{T_s-T_{FEPS}}{R_eq}=\dot{q}A_{out}\\\dfrac{T_s-204}{0.011}=1200*\pi*D_i\\\dfrac{T_s-204}{0.011}=1200*\pi*0.027\\T_s=205.12 C$

So the outer surface temperature is 205. 12 C

Part a

The temperature of the FEP is 204 C while that of the outer surface is 205.12 C

Part b

The total Thermal resistance is 0.011 C/W

Part c & d

As the temperature of the FEP is 204 which is within the limits of the ASME code thus the FEP lining comply with the recommendation of the code.

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