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Q. 14.3

Water at 40 °C at a mass flow rate of 0.5 kg/s enters a 2.5 cm ID tube whose wall is maintained at a uniform temperature of 90 °C. Calculate the tube length required for heating the water to 60 °C. Also determine the pressure drop across the length of pipe.

Verified Solution

Given data is written down

$\dot{m}=0.5 kg / s$

D = 25 mm

$T_{B, 0}=40^{\circ} C$ $T_{B, L}=60^{\circ} C$ $T_{w}=90^{\circ} C$

Water properties are evaluated at $T_{m}=\frac{T_{B, 0}+T_{B, L}}{2}=\frac{40+60}{2}=50^{\circ} C$. From table of properties of water, we have

$\nu=0.568 \times 10^{-6} m ^{2} / s$, $k=0.064 W / m ^{\circ} C$ , $P r=3.68$ and $\rho=990 kg / m ^{3}$

Hence, the dynamic viscosity of water is $\mu=\rho \nu=990 \times 0.568 \times 10^{-6}=5.623 \times 10^{-4}kg/ms$

The velocity of water in the tube may be calculated as

$U=\frac{4 \dot{m}}{\pi D^{2} \rho}=\frac{4 \times 0.5}{\pi \times 0.025 \times 990}=1.029 m / s$

The Reynolds number is then given by

$R e_{D}=\frac{U D}{\nu}=\frac{1.029 \times 0.025}{0.568 \times 10^{-6}}=45285$

The flow is turbulent and hence we may use the Sieder and Tate correlation, taking into account the variation of properties, to calculate the Nusselt number. From tables, we read off the kinematic viscosity and density at wall temperature of 90 °C as

$\nu_{w}=0.329 \times 10^{-6} m ^{2} / s ^{\prime} \rho_{w}=967.4 kg / m ^{3}$

Hence, the dynamic viscosity of water at the wall temperature is

$\mu_{w}=\rho_{w} \nu_{w}=967.4 \times 0.329 \times 10^{-6}=3.183 \times 10^{-4} kg / m s$

Hence, we have

$\frac{\mu}{\mu_{w}}=\frac{5.623 \times 10^{-4}}{3.183 \times 10^{-4}}=1.767$

Clearly the constant property assumption would not be valid. The Sieder and Tate correlation is appropriate. Hence, using Eq. 14.28, we have

$N u_{D}=0.027 \operatorname{Re}_{D}^{0.8} \operatorname{Pr}^{\frac{1}{3}}\left(\frac{\mu}{\mu_{w}}\right)^{0.14}$  (14.28)

$N u_{D}=0.027 \times 45285^{0.8} \times 3.68^{\frac{1}{3}} \times 1.767^{0.14}=239.5$

The mean value of the heat transfer coefficient is then given by

$\bar{h}=\frac{N u_{D} k}{D}=\frac{239.5 \times 0.640}{0.025}=6131.2 W / m ^{2}{ }^{\circ} C$

With the various temperatures that are specified, we may calculate the temperature differences ratio as

$\frac{T_{w}-T_{B, L}}{T_{w}-T_{B, 0}}=\frac{90-60}{90-40}=\frac{30}{50}=0.6$

The Stanton number may be calculated as

$S t=\frac{N u_{D}}{\operatorname{Re}_{D} P r}=\frac{239.5}{45285 \times 3.68}=0.001437$

Hence, using Eq. 14.32,

$\frac{T_{w}-T_{B, L}}{T_{w}-T_{B, 0}}= e ^{-\left(4 \times S t \times \frac{L}{D}\right)}$  (14.32)

$e^{-4 \cdot S t \cdot \frac{L}{D}}=0.6$

or solving for pipe length

$L=-\frac{0.025 \times \ln (0.6)}{4 \times 0.001437}=2.22 m$

Pressure drop is calculated assuming that the tube is smooth. Equation 14.24(b) is appropriate and hence

$f=\frac{0.184}{R e_{D}^{0.2}} \text { for } R e_{D}>2 \times 10^{4}$  (14.24)(b)

$f=\frac{0.184}{45285^{0.2}}=0.0216$

The pressure drop over the length of the tube is then given by

$-\Delta p=f \cdot \frac{L}{D} \cdot \frac{\rho U^{2}}{2}=0.0216 \times \frac{2.2}{0.025} \times \frac{990 \times 1.029^{2}}{2}=1005.3 Pa$