Question

You are given a small pipe with a diameter of 1 mm flowing
cooling fluid (water) at a temperature of 20C. Fluid flow velocity is regulated at a constant speed
of 0.2 m / s and assumed to be uniform. If the pipe surface is given a constant heat flux
of 6000 W / m2
and you assume the analysis is done in a fully develop area, then
based on your design calculations, at what length of the pipe (measured from the beginning of the input)
water temperature will increase to 74C? Note: the value of thermal diffusivity and viscosity
the kinematic is 1,541 ? 10−7 m^2s and 0.556 ? 10−6 m^2s, ? = 0.6367 W / mk

2. Ketika anda kerja praktek di industri oil and gas, kemungkinan besar anda akan menjumpai banyak sekali kasus yang melibatkan proses mixing yang melibatkan beberapa fluida dengan temperatur yang berbeda-beda. Andaikan saat ini anda sedang melakukan kerja praktek di PT Permata Sejahtera dimana anda ditempatkan pada posisi sebagai junior engineer. Anda diberikan tugas oleh pembimbing anda untuk mendesain suatu modul penukar kalor dimana modul ini bekerja berdasarkan prinsip perpindahan panas secara konveksi (dominan). Anda diberikan suatu pipa kecil dengan diameter 1 mm yang dialiri fluida pendingin (air) pada temperatur 20 °C. Kecepatan alir fluida diatur pada kecepatan konstan sebesar 0,2 m/s dan diasumsikan uniform. Jika permukaan pipa tersebut diberikan fluk panas konstan sebesar 6000 W/m² dan anda mengasumsikan analisa dilakukan pada daerah fully develop, maka berdasarkan kalkulasi desain anda, pada panjang pipa berapakah (di ukur dari awal masukan) temperatur air akan meningkat menjadi 74 °C? Note: nilai difusivitas termal dan viskositas kinematiknya adalah 1,541 x 10-7 m´s dan 0,556 x 10-6 m´s, k = 0,6367 W/mk. (10 Poin)

Answer 1

Assumption -

Fully develop region.

Calculate -
Length of the pipe at which water temperature will increase to 74C

Given data -

v (Kinematic viscosity) = 0.556 ? 10−6 m^2s

? = 0.6367 W / mk

Gooow/m2 To Ta 200 14 w

First, calculate whether the flow is laminar or turbulent -

$Re=\frac{vL}{\nu }$

Where

v = velocity of fluid i.e. water

$\nu =$ Kinematic viscosity m2/sec

L = Hydrodynamic or characteristic length i.e. the length at which the flow is in contact.

$Re=\frac{0.2\times 0.001}{0.556 \times 10^{-6} }$

Re= 359.7122 < 2000

For pipe flow, boundary layer growth is almost similar to the flow over a flat plate.

Now using empirical relation of average nusselt number for laminar flow i.e.

Where

Nu = Nusselt number

Re = Reynold number

Pr = Prandtl Number

$\frac{hL}{k}= 0.664(Re)^{0.5}(Pr)^{0.33}$

Where

h = Convection heat transfer coefficient

L = D(For cylinder) =Hydrodynamic or characteristic length

k = Thermal conductivity in W/m-K

$Pr = \frac{\mu C_p}{k}$

$Pr = \frac{\nu }{\alpha }=\frac{\mu C_p}{k}$

Where

Kinematic viscosity

$0.556\times 10^{-6}= \frac{\mu}{1000}$

Therefore,

Dynamic viscosity

$\mu=0.556\times 10^{-3}Pa-sec$

Specific heat of water = 4.18 KJ/Kg-K

? = 0.6367 W / mk

$Pr = \frac{0.556\times 10^{-3}\times 4.18\times 1000}{0.6367}$

or

$Pr = \frac{\nu }{\alpha }=3.6119$ (Use this one because both values are given in question directly)

Now assume 74 C temperature occurs at distance x from the leading edge.

Therefore the Reynold number will be -

$Re=\frac{0.2\times x}{0.556 \times 10^{-6} }$

Now assume 74 C temperature occurs at distance x from the leading edge.

Therefore,

Put the above values in empirical relation -

$h_{local}= 0.332\times \frac{0.6367}{x}(359712.2x)^{0.5}(3.6119)^{0.33}$

$h_{x}= \frac{193.6883}{\sqrt{x}}$

See at the surface of the pipe, velocity is zero or approximately zero. So heat flux at the surface is given as -

$\frac{Q}{A}=h_{x}(T_s-T_w)$

$6000=\frac{193.6883}{\sqrt{x}}(74-20)$

$\mathbf{x= 3.038 m}$

If you have any doubt or answer does not match please write in comment section.

Thank you.