Answered: The liquid food is flowed through an…

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter5: Analysis Of Convection Heat Transfer

Section: Chapter Questions

Problem 5.12P

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Question

The liquid food is flowed through an uninsulated pipe at 90 ° C. The product flow rate is 0.3 kg / s and has a density of 1000 kg / m³, specific heat 4 kJ / (kg K), a viscosity of 8 x 10-6 Pa s, and a thermal conductivity of 0.55 W / (m) K). Assume that the change in viscosity is negligible. The internal diameter of the pipe is 20 mm with a thickness of 3 mm made of stainless steel (k = 15 W / [m ° C]). The outside temperature is 15 ° C. If the outer convective heat transfer coefficient is 18 W / (m² K), calculate the heat loss at steady state per meter pipe length. a. Find the convection coefficient in the pipe = Answer W / m² ° C. b. Calculate heat loss per meter pipe length = Answer watt.

Expert Solution

Step 1

Given data:

The temperature inside the pipe is $T_{i} = 90 ° C$

The mass flow rate of the liquid food is $\dot{m} = 0.3 \frac{k g}{s}$

The density of the liquid food is $ρ = 1000 \frac{k g}{m^{3}}$

The specific heat of the liquid food is $C = 4 \frac{k J}{k g \cdot K}$

The viscosity of the liquid food is $ν = 8 \times 10^{- 6} P a \cdot s$

The thermal conductivity of the liquid food is $K_{l} = 0.55 \frac{W}{m \cdot K}$

The internal diameter of the pipe is $D_{i} = 20 m m = 0.02 m$

The internal radius of the pipe is $R_{i} = \frac{D_{i}}{2} = \frac{0.02}{2} = 0.01 m$

The thickness of the stainless steel is $t = 3 m m = 0.003 m$

The outer radius of the pipe is $R_{o} = R_{i} + t = 0.01 + 0.003 = 0.013 m$

The thermal conductivity of stainless steel is $K_{s} = 15 \frac{W}{m ° C}$

The outside temperature of the pipe is $T_{o} = 15 ° C$

The outer convective heat transfer coefficient is $h_{o} = 18 \frac{W}{m^{2} ° C}$

Step 2

a. Calculate the Reynolds number of liquid flood as follows:

$R e = \frac{V D}{ν} R e = \frac{\dot{4 m}}{π ν ρ D} R e = \frac{4 \times 0.3}{π \times 8 \times 10^{- 6} \times 1000 \times 0.02} R e = 2387.3 T h e f l o w i s c l o s e t o l a m i n a r f l o w$

The flow is laminar, and flow is inside the pipe.

So, assume the constant wall temperature condition.

Step 3

Now, Calculate the convection coefficient inside the pipe as follows:

$\begin{array}{rcl} N u & = & 3.66 \\ \frac{h_{i} D_{i}}{K_{l}} & = & 3.66 \\ \frac{h_{i} \times 0.02}{0.55} & = & 3.66 \\ h_{i} & = & 100.65 \frac{W}{m^{2} \cdot K} \end{array}$

b. Calculate the heat loss from the pipe per meter length as follows:

$Q_{l o s s} = Q_{c o n d} + Q_{c o n v} Q_{l o s s} = \frac{T_{i} - T_{o}}{\frac{1}{2 π h_{i} R_{i} L} + \frac{\ln (\frac{R_{o}}{R_{i}})}{2 {πK}_{s} L} + \frac{1}{2 {πh}_{o} R_{o} L}}$

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