- #1

Tiberious

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## Homework Statement

I've completed this problem and received the below feedback but have gone to a state of complete loss on what to do to amend. Can anyone provide a prompt ?

Heat transfer coefficient is correct. But heat transfer rate is wrong - the question says Heat transfer per UNIT AREA, so why would you multiply your heat flux by AREA??

Liquid ammonia is heated as it flows at a mean velocity of 2 m s-1 through a circular pipe. The pipe, which has an internal diameter of 75 mm, is at a uniform temperature of 27°C, and the ammonia at a section 1.2 m from the inlet to the pipe has a temperature of -23°C.

Use the following information to estimate the local heat transfer flux at l = 1.2 m. Note, the properties of ammonia liquid have been taken at -23°C, except where stated.

Liquid ammonia properties:

ρ Density=600 kg m^(-3)

C_p Specific heat capacity=4.86 kJ kg^(-1) K^(-1)

μ Dynamic Viscosity (at 27°C)=1.19∙10^(-4 ) kg m^(-1) s^(-1)

μ Dynamic Viscosity= 2.05∙10^(-4 ) kg m^(-1) s^(-1)

k Thermal conductivity= 5.11∙10^(-4 ) kW m^(-1) s^(-1)

## The Attempt at a Solution

Heat transfer correlations:

N_u=1.86 〖R_e〗^(1/3) 〖P_r〗^(1/3) 〖(d/l)^(1/3) (μ/μ_w )〗^0.14for laminar flow

N_u=0.023 〖R_e〗^0.8 〖P_r〗^0.33 for turbulent flow.

Determining the Reynolds number,

Re=(ρ∙u∙L)/μ=(ρ∙u∙D)/μ=(600∙2∙0.075)/(2.05∙〖10〗^(-4) )=439024.4

Determining the Prandtl number,

Pr=(C_p∙μ)/k=(4.86∙〖10〗^3∙2.05∙〖10〗^(-4))/(5.11∙〖10〗^(-4)∙〖10〗^3 )=1.95

Re > 2600, so flow is turbulentNu=0.023∙〖Re〗^0.8∙〖Pr〗^0.33=0.023∙〖439024.4〗^0.8∙〖1.95〗^0.33=936.3

h=Nu∙k/d

=936.3∙(5.11∙〖10〗^(-4)∙〖10〗^3)/0.075

=6379.5 W/(m^2∙K)

Φ=h∙A∙(T_s-T_f )

=h∙π∙D∙L∙(T_s-T_f )

=6379.5∙π∙0.075∙1.2∙(27-(-23))

Answer:

=90188.W

=90.2 kW