A) Using equation Tm(x)=Tm +((Qdot)(D)(pi)(x))/((Mdot)(Cp)) Water enters a 40-mm

Question

A) Using equation Tm(x)=Tm +((Qdot)(D)(pi)(x))/((Mdot)(Cp))

Water enters a 40-mm-diameter tube at 25 degree C and a nominal pressure of 2 atm. The average velocity is 1.8 m/s. An electrical heating tape wrapped around the tube results in the heating of the water to 30 degree C at a downstream distance of 6 m. Determine the average heat flux at the inside wall of the tube over the 6-m length. Determine the local heat-transfer coefficient at the downstream location if the wall temperature here is 82 degree C.

Explanation / Answer

Let us take the water properties at 30 deg C.

Specific heat Cp= 4180 J/kg-K

Mass flowrate m = [(/4*D2)*V] = 1000*[(/4*0.042)*1.8] = 2.261 kg/s

(a) mCp(Tx-Tin) = Q(DL)

2.261*4180*(30-25) = Q(3.14*0.04*6)

So, Q = 62700 W/m2

(b) hx = Q/(Twallx - Tx)

= 62700/(82-30) = 1205.8 W/m2K

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