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A feedwater heater that supplies a boiler consists of a shell-and-tube heat exchanger with one shell pass and two tube passes. One hundred thin-walled tubes each have a diameter of 20 mm and a length (per pass) of 2 m. Under normal operating conditions water enters the tubes at 10 kg/s and 290 K and is heated by condensing saturated steam at 1 atm on the outer surface of the tubes. The convection coefficient of the saturated steam is 10,000W/m2K10,000 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}. (a) Determine the water outlet temperature. (b) With all other conditions remaining the same, but accounting for changes in the overall heat transfer coefficient, plot the water outlet temperature as a function of the water flow rate for 5mc20kg/s5 \leq m_{c} \leq 20 \mathrm{kg} / \mathrm{s}. (c) On the plot of part (b), generate two additional curves for the water outlet temperature as a function of flow rate for fouling factors of Rf=0.0002R_{f}^{\prime \prime}=0.0002 and 0.0005m2K/W0.0005 \mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}.

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Given\Rightarrow Inlet temperature of hot fluid(steam), Th,iT_{h,i}; Exit temperature of hot fluid(steam), Th,eT_{h,e}; Mass flow rate of hot fluid(steam), m˙h\dot{m}_{h}

Inlet temperature of cold fluid(water), Tc,i=290 KT_{c,i}=290\ \text{K}; Exit temperature of cold fluid(water) is Tc,eT_{c,e}; Mass flow rate of cold fluid(water), m˙c=10 kgs\dot{m}_{c}=10\ \frac{\text{kg}}{\text{s}}

Number of tubes, N=100N=100; Diameter, D=0.02 mD=0.02\ \text{m}; Length per pass, L=2L=2m; Convectiive heat transfer coefficient, ho=10000 Wm2Kh_o=10000\ \frac{\text{W}}{\text{m}^2 \cdot K}

From Thermophysical properties table for saturated water(at 11\ \ atm and assuming mean temperature of 320320K)

Tsat=373.15 KT_{sat}=373.15\ \text{K}; μl=577×106 Nsm2\mu_l=577\times 10^{-6} \ \frac{\text{N} \cdot s}{\text{m}^2}; kl=0.640 WmKk_l=0.640\ \frac{\text{W}}{\text{m} \cdot K}; Cp,c=4180 JkgKC_{p,c}=4180\ \frac{\text{J}}{\text{kg} \cdot K}; Pr=3.77Pr=3.77

(a)\textbf{(a)} In order to determine the type of flow, finding Reynolds number. Reynolds number for a circular tube in case of internal flow is given by:

ReD=4m˙πDμl\text{Re}_D=\dfrac{4 \dot{m}}{\pi D \mu_{l}}

where μl\mu_l is the viscosity of the liquid flowing and m˙\dot{m} is mass flow rate per tube.

ReD=410100π(0.02)(577×106)=11033.27Re_D=\dfrac{4 \cdot \frac{10}{100}}{\pi (0.02) (577\times 10^{-6})}=11033.27

If $Re_D \geq 2300 $ in case of internal flow through a circular tube, then flow is Turbulent.

Hence, given flow is Turbulent.

For fully developed (hydrodynamically and thermally) turbulent flow in a smooth circular tube, the Nusselt number may be obtained from the Dittus-Boelter equation:

NuD=0.023(ReD)45PrnNu_D=0.023 (Re_D)^{\frac{4}{5}} Pr^{n}

where n=0.4n=0.4 for heating and n=0.3n=0.3 for cooling. Since, it is a case of heating. Hence value of nn is 0.40.4

Therefore,

NuD=0.023(11033.27)45(3.77)0.4=67.054Nu_D=0.023 (11033.27)^{\frac{4}{5}} (3.77)^{0.4}=67.054

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