Water enters a tube of diameter, D = 60 mm with mass flow rate of 0.01 kg s^-1 as shown in the figure below. The inlet mean temperature is T_m,i = 293 K and the uniform heat flux at the surface of the tube is 2000 W m^-2. For the exit mean temperature of T_m,o = 353 K, the length of the tube, L is ________m (rounded off to 1 decimal place).

Use the specific heat of water as 4181 J kg^-1 K^-1.

The correct answer is 6.8 m (or 6.8).

Step-by-step Explanation:

1. Identify the given parameters from the problem statement and the provided figure:
- Diameter of the tube, $D=60\text{mm}=0.06\text{m}$
- Mass flow rate of water, $\stackrel{·}{m}=0.01{\text{kg s}}^{-1}$
- Inlet mean temperature, $T_{m,i}=293\text{K}$
- Exit mean temperature, $T_{m,o}=353\text{K}$
- Uniform heat flux at the tube surface, $q"=2000{\text{W m}}^{-2}$
- Specific heat of water, $C_p=4181{\text{J kg}}^{-1}{\text{K}}^{-1}$

2. Calculate the total heat transfer rate ($Q$) required to heat the water:
The energy balance equation for internal flow is:
$$Q=\stackrel{·}{m}{C}_p(T_{m,o}-T_{m,i})$$
Substituting the given values into the equation:
$$Q=0.01\times 4181\times (353-293)$$
$$Q=0.01\times 4181\times 60$$
$$Q=2508.6\text{W}$$

3. Relate the heat transfer rate to the surface area and heat flux:
For a tube of diameter $D$ and length $L$, the surface area through which heat is transferred is given by $A_s=\pi DL$. Under uniform surface heat flux $q"$, the total heat rate is:
$$Q=q"A_s=q"\left(\pi DL\right)$$

4. Solve for the tube length ($L$):
Rearranging the formula to solve for $L$:
$$L=\frac{Q}{q"\pi D}$$
Substituting the parameters into the equation:
$$L=\frac{2508.6}{2000\times \pi \times 0.06}$$
$$L\approx 6.8\text{m}$$

Thus, the required length of the tube, rounded off to one decimal place, is 6.8 m.