To solve this problem, we need to apply heat transfer principles and energy balance to determine the tube length and rate of evaporation of water. We'll break the solution into two parts.

1. Energy balance on the hot gases (for the tube length):

We are given the following information:

• Inlet pressure, $P_{in} = 115 \, \text{kPa}$
• Inlet temperature, $T_{in} = 260^\circ \text{C}$
• Exit temperature, $T_{out} = 155^\circ \text{C}$
• Inner surface temperature of the tube, $T_s = 110^\circ \text{C}$
• Mean velocity of gases, $v = 6 \, \text{m/s}$
• Heat transfer coefficient, $h = 125 \, \text{W/m}^2 \cdot ^\circ \text{C}$
• Tube diameter, $D = 4 \, \text{cm} = 0.04 \, \text{m}$
• Specific heat of hot gases, $C_p = 1.025 \, \text{kJ/kg} \cdot ^\circ \text{C} = 1025 \, \text{J/kg} \cdot ^\circ \text{C}$
• Gas constant, $R = 0.287 \, \text{kJ/kg} \cdot ^\circ \text{C} = 287 \, \text{J/kg} \cdot ^\circ \text{C}$

Step 1: Mass flow rate of gases

First, we need to determine the mass flow rate of the gases.

Using the ideal gas law:

$\dot{m} = \frac{P_{in} A v}{R T_{avg}}$

Where:

• $P_{in}$ is the inlet pressure in Pa (since the formula uses SI units),
• $A$ is the cross-sectional area of the tube,
• $v$ is the mean velocity,
• $R$ is the gas constant,
• $T_{avg}$ is the average temperature of the gases in Kelvin.

First, convert the inlet pressure to Pa: $P_{in} = 115 \, \text{kPa} = 115,000 \, \text{Pa}$

Now, calculate the cross-sectional area $A$ of the tube: $A = \frac{\pi D^2}{4} = \frac{\pi (0.04)^2}{4} = 1.2566 \times 10^{-3} \, \text{m}^2$

The average temperature $T_{avg}$ is the mean of the inlet and outlet temperatures in Kelvin: $T_{avg} = \frac{T_{in} + T_{out}}{2} = \frac{260 + 155}{2} = 207.5^\circ \text{C} = 207.5 + 273.15 = 480.65 \, \text{K}$

Now, calculate the mass flow rate $\dot{m}$:

$\dot{m} = \frac{115,000 \times 1.2566 \times 10^{-3} \times 6}{287 \times 480.65}$

Let’s compute this value.

Step 2: Heat lost by gases

The heat lost by the hot gases is given by:

$Q = \dot{m} C_p (T_{in} - T_{out})$

We will calculate this once we know $\dot{m}$.

Step 3: Heat transfer from the tube

The heat transferred from the tube surface to the water is given by:

$Q = h A_s \Delta T_{lm}$

Where:

• $h$ is the heat transfer coefficient,
• $A_s$ is the surface area of the tube,
• $\Delta T_{lm}$ is the logarithmic mean temperature difference between the hot gases and the tube surface.

The surface area $A_s$ of the tube is:

$A_s = \pi D L$

The logarithmic mean temperature difference ( \Delta T_{lm