Water heating in a tube, Dittus-Boelter

(a) Reynolds number

$$Re = \frac{\rho V D}{\mu} = \frac{992.2\times1\times0.025}{653\times10^{-6}} = 3.80\times10^4$$

The flow is turbulent ($Re > 10\,000$) and the tube is long, so the Dittus-Boelter equation applies.

(b) Heat transfer coefficient

For heating of the fluid the exponent is $n = 0.4$:

$$\begin{aligned} Nu &= 0.023\,Re^{0.8}Pr^{0.4}\\ &= 0.023\,(37\,986)^{0.8}(4.32)^{0.4} = 190.4 \end{aligned}$$ $$h = \frac{Nu\,k}{D} = \frac{190.4\times0.631}{0.025} = 4805\ \text{W/m}^2\text{K}$$

(c) Heat gained

Mass flow rate:

$$\dot m = \rho V\frac{\pi D^2}{4} = 992.2\times1\times\frac{\pi(0.025)^2}{4} = 0.487\ \text{kg/s}$$ $$Q = \dot m c_p (T_{out}-T_{in}) = 0.487\times4179\times20 = 40\,700\ \text{W}$$

(d) Tube length

For constant wall temperature the mean temperature difference is the logarithmic mean:

$$\Delta T_{lm} = \frac{(90-30)-(90-50)}{\ln\left(\dfrac{90-30}{90-50}\right)} = \frac{60-40}{\ln1.5} = 49.3\ ^\circ\text{C}$$ $$Q = h\,(\pi D L)\,\Delta T_{lm}\;\Rightarrow\; L = \frac{40\,707}{4805\times\pi\times0.025\times49.3} = 2.19\ \text{m}$$

Check: $L/D = 87.5 > 60$, so the entrance effect is small and the correlation is acceptable.

Answer: $Re = 3.8\times10^4$; $h = 4805\ \text{W/m}^2$K; $Q = 40.7$ kW; $L \approx 2.2$ m.