Question

Hot water with the mass flow rate of 3 kg/s and temperature of 90 C enters a 2 inch copper pipe with the length of 3 m. The pipe is exposed to cold air at the constant temperature of 1C and wind velocity is 5 m/s. Find the outlet temperature of water and the rate of heat loss from the pipe.

Answer 1

Given data:

• Air temp = 1 C
• Inlet water temp = 90 C
• Dia of pipe = 2 in
• Mass flow rate of water = 3 kg/s
• Wind Velocity = 5 m/s

We will first find the external convection heat transfer coefficient,

hoy nut =?

Reynold's Number =

VD Re V

where

• V = Velocity of external flow = 5 m/s
• D = Outer dia of pipe = 2 in = 0.0508 m
• v = Kinematic Viscosity of air at 1 C = 13.37e-6 m2/s

VD Re V

$Re=\frac{5\times 0.0508}{13.37e-6 }=18997.756$

The appropriate Nusselt Number from Heat Transfer Data book is:

$Nu=C\times (Re^{m})\times (Pr^{0.333})$

where

• C = 0.193
• m = 0.618
• Re = 18997.756
• Pr = Prandtl Number for air at 1 C = 0.736

$Nu=C\times (Re^{m})\times (Pr^{0.333})$

$Nu=0.193\times (18997.756^{0.618})\times (0.736^{0.333})=76.82$

Also, we know, $Nu=\frac{h_{out}D}{k}$

where

• h(out) = ?
• D = Outer dia of pipe = 2 in = 0.0508 m
• k = Thermal conductivity of air at 1 C = 0.02371 W/mK

$Nu=\frac{h_{out}D}{k}$

$76.82=\frac{h_{out}\times 0.0508}{0.02371}=>h_{out}=35.85\:W/m^2K$

Now, we shall find the internal convective heat transfer coeff =

hin =?

Reynold's Number =

VD Re V

where

• V = Velocity of internal flow = $\frac{m}{\rho A}=\frac{3}{965.3\times 0.002027}=1.533\:m/s$
  • m = mass flow rate of water = 3 kg/s
  • $\rho$ = density of water at 90 C = 965.3 kg/m3
  • A = Internal cross-sectional area of pipe =
    11 D2 II x 0.05082 = 0.002027 m2 4 4
    
    • D = 2 in = 0.0508 m
• D = Inner Dia of pipe = 2 in = 0.0508 m [Considering negligible thickness]
• v = Kinematic Viscosity of water at 90 C = 3.258e-7 m2/s

VD Re V

$Re=\frac{1.533\times 0.0508}{3.258e-7 }=239057.602$

The appropriate Nusselt Number from Heat Transfer Data book is:

$Nu=0.023\times (Re^{0.8})\times (Pr^{0.3})$

where

• Re = 239057.602
• Pr = Prandtl Number for water at 90 C = 2

$Nu=0.023\times (Re^{0.8})\times (Pr^{0.3})$

$Nu=0.023\times (239057.602^{0.8})\times (2^{0.3})=568.6$

Also, we know, $Nu=\frac{h_{in}D}{k}$

where

• h(in) = ?
• D = Inner dia of pipe = 2 in = 0.0508 m
• k = Thermal conductivity of water at 90 C = 0.6613 W/mK

$Nu=\frac{h_{in}D}{k}$

$568.6=\frac{h_{in}\times 0.0508}{0.6613}=>h_{in}=7402.21\:W/m^2K$

-----

There are 3 thermal resistances present:

• Internal Heat Convection (R1)
• Conduction through walls - Can be ignored as the thickness is not given to us in the question.
• External Heat Convection (R2)

-----

Internal Heat Convection (R1):

Thermal Resistance = $R1=\frac{1}{h_{out}A}$

where

• h(out) = 35.85 W/m2K
• A = Area of heat trasnfer = $\prod DL=\prod \times 0.0508\times 3=0.4788\:m^2$
  • D = 0.0508 m
  • L = Length of pipe = 3 m

Thermal Resistance = $R1=\frac{1}{35.85\times 0.4788}=0.05825\:K/W$

-----

External Heat Convection (R2)

Thermal Resistance = $R2=\frac{1}{h_{in}A}$

where

• h(out) = 7402.21 W/m2K
• A = Area of heat trasnfer = $\prod DL=\prod \times 0.0508\times 3=0.4788\:m^2$ ​​​​​​​​​​​​​​
  • D = 0.0508 m
  • L = Length of pipe = 3 m

Thermal Resistance = $R2=\frac{1}{7402.21\times 0.4788}=0.0002822\:K/W$

------

The total resistance = $R_{total}=R1+R2=0.05825+0.0002822=0.05854\:K/W$

Thus, the total heat transferred in steady state:

$m\times c\times (T_{in}-T_{out})=\frac{T_{mean}-T_{air}}{R_{total}}$

where

• T(mean) = mean temp of water = $\frac{T_{in}+T_{out}}{2}$ ​​​​​​​
  • T(in) = inlet temp of water = 90 C
  • T(out) = outlet temp of water = ?
• c = specific heat of water = 4187 J/kgK
• m = mass flow rate of water = 3 kg/s
• T(air) = outside temp of air = 1 C (given)

$m\times c\times (T_{in}-T_{out})=\frac{T_{mean}-T_{air}}{R_{total}}$

$3\times 4187\times (90-T_{out})=\frac{\frac{90+T_{out}}{2}-1}{0.05854}=>\mathbf{T_{out}=89.41\:^{\circ}C\:(Ans.)}$

The amount of heat transfer can be written as:

$Q=\frac{T_{mean}-T_{air}}{R_{total}}=\frac{\frac{90+89.41}{2}-1}{0.05854}=>\mathbf{Q=1515.44\:W\:(Ans.)}$​​​​​​​

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