Water flows through a tube of 3 cm internal diameter and length 20 m, The outside surface of the tube is heated electrically so that it is subjected to uniform heat flux circumferentially and axially. The mean inlet and exit temperatures of the water are 10°C and 70°C, respectively. The mass flow rate of the water is 720 kg/h. Disregard the thermal resistance of the tube wall. The internal heat transfer coefficient is 1697 W/m²K. Take specific heat Cp of water as 4.179 kJ/kgK. The inner surface temperature at the exit section of the tube is __________ °C (round off to one decimal place).

The correct answer is 85.6.

Step-by-Step Explanation:

First, we identify the given data from the problem statement and the accompanying schematic diagram:
- Internal diameter of the tube, $D=3\text{cm}=0.03\text{m}$
- Length of the tube, $L=20\text{m}$
- Mass flow rate of water, $\stackrel{·}{m}=720\text{kg/h}=\frac{720}{3600}\text{kg/s}=0.2\text{kg/s}$
- Mean inlet temperature of water, $T_{w,\text{inlet}}=10°\text{C}$
- Mean exit temperature of water, $T_{w,\text{exit}}=70°\text{C}$
- Heat transfer coefficient, $h=1697{\text{W/m}}^2\text{K}$
- Specific heat capacity of water, $C_p=4.179\text{kJ/kgK}=4179\text{J/kgK}$

1. Determine the rate of heat transfer (Q):
Using the global energy balance for the water flowing through the heated tube:

$Q=\stackrel{·}{m}\times C_p\times (T_{w,\text{exit}}-T_{w,\text{inlet}})$

Substituting the given parameters:

$Q=0.2\times 4179\times (70-10)$

$Q=0.2\times 4179\times 60=50148\text{W}$

2. Calculate the surface area of the tube (A):
The internal surface area through which heat is transferred to the water is given by:

$A=\pi \times D��L$

$A=\pi \times 0.03\times 20=0.6\pi \approx 1.88496{\text{m}}^2$

3. Calculate the uniform heat flux (q"):
Since the outside surface is heated to provide a uniform heat flux axially and circumferentially:

$q"=\frac{Q}{A}$

$q"=\frac{50148}{1.88496}\approx 26604.34{\text{W/m}}^2$

4. Apply Newton's Law of Cooling at the tube exit:
At the exit section of the tube, the heat flux is related to the tube wall inner surface temperature and the mean exit temperature of the water by:

$q"=h\times (T_{\text{tube},\text{exit}}-T_{w,\text{exit}})$

Substituting the known values:

$26604.34=1697\times (T_{\text{tube},\text{exit}}-70)$

$T_{\text{tube},\text{exit}}-70=\frac{26604.34}{1697}\approx 15.68$

$T_{\text{tube},\text{exit}}=70+15.68\approx 85.6°\text{C}$