## Related questions with answers

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 $10,000 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}$. (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 $5 \leq m_{c} \leq 20 \mathrm{kg} / \mathrm{s}$. (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 $R_{f}^{\prime \prime}=0.0002$ and $0.0005 \mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}$.

Solution

VerifiedGiven$\Rightarrow$ Inlet temperature of hot fluid(steam), $T_{h,i}$; Exit temperature of hot fluid(steam), $T_{h,e}$; Mass flow rate of hot fluid(steam), $\dot{m}_{h}$

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

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

From Thermophysical properties table for saturated water(at $1$\ \ atm and assuming mean temperature of $320$K)

$T_{sat}=373.15\ \text{K}$; $\mu_l=577\times 10^{-6} \ \frac{\text{N} \cdot s}{\text{m}^2}$; $k_l=0.640\ \frac{\text{W}}{\text{m} \cdot K}$; $C_{p,c}=4180\ \frac{\text{J}}{\text{kg} \cdot K}$; $Pr=3.77$

$\textbf{(a)}$ 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:

$\text{Re}_D=\dfrac{4 \dot{m}}{\pi D \mu_{l}}$

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

$Re_D=\dfrac{4 \cdot \frac{10}{100}}{\pi (0.02) (577\times 10^{-6})}=11033.27$

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:

$Nu_D=0.023 (Re_D)^{\frac{4}{5}} Pr^{n}$

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

Therefore,

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

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