Question

Liquid water enters a 10 - \(\mathrm{m}\)-long smooth rectangular tube with \(a=50 \mathrm{~mm}\) and \(b=25 \mathrm{~mm}\). The surface temperature is maintained constant, and water enters the tube at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.25 \mathrm{~kg} / \mathrm{s}\). Determine the tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\).

Short answer

Answer: The tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\) is approximately \(84.18^{\circ} \mathrm{C}\).

Step-by-step solution

Calculate the heat transfer rate

First, we need to find the heat transfer rate necessary to achieve the desired outlet water temperature. We can do this using the formula: \(Q = \dot{m} \times C_p \times \Delta T\) Where: - \(Q\) is the heat transfer rate (W) - \(\dot{m}\) is the mass flow rate of water (\(\mathrm{kg/s}\)) - \(C_p\) is the specific heat capacity of water (assume \(4.18 \, \mathrm{kJ/kg \cdot K}\) for the given problem) - \(\Delta T\) is the change in temperature of the water (\(T_{out} - T_{in}\)) First, let's calculate \(\Delta T\). \(\Delta T = T_{out} - T_{in} = 80^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 60 \mathrm{~K}\) Now, we can find the heat transfer rate. \(Q = \dot{m} \times C_p \times \Delta T = 0.25 \, \mathrm{kg/s} \times 4.18 \, \mathrm{kJ/kg \cdot K} \times 60 \, \mathrm{K}\) Since, \(C_p\) unit is in kJ, the result of Q will be as well, so in order to compare units, let's just change it to Joules. \(Q = 0.25 \, \mathrm{kg/s} \times 4180 \, \mathrm{J/kg \cdot K} \times 60 \, \mathrm{K} = 62700\, \mathrm{W}\)

Calculate the heat transfer area

Next, we need to find the area of the rectangular tube through which heat is being transferred. This can be calculated using the following formula: \(A = 2 \times (a + b) \times L\) Where: - \(A\) is the heat transfer area (m²) - \(a\) and \(b\) are the sides of the rectangle (m) - \(L\) is the length of the tube (m) Before applying dimensions, we need to convert them into meters. \(a = 50\, \mathrm{mm} \times \frac{1\, \mathrm{m}}{1000\, \mathrm{mm}} = 0.05\, \mathrm{m}\) \(b = 25\, \mathrm{mm} \times \frac{1\, \mathrm{m}}{1000\, \mathrm{mm}} = 0.025\, \mathrm{m}\) Now, we can calculate the heat transfer area. \(A = 2 \times (0.05\, \mathrm{m} + 0.025\, \mathrm{m}) \times 10\, \mathrm{m} = 1.5\, \mathrm{m^2}\)

Calculate the tube surface temperature

Finally, we can calculate the tube surface temperature necessary to heat the water to the desired outlet temperature. We can use the following formula for heat transfer through convection: \(Q = h \times A \times (T_s - T_{out})\) Where: - \(Q\) is the heat transfer rate (W) - \(h\) is the heat transfer coefficient (assume \(5000\, \mathrm{W/m^2 \cdot K}\) for the given problem) - \(A\) is the heat transfer area (m²) - \(T_s\) is the tube surface temperature (°C) We can rearrange the formula to solve for \(T_s\): \(T_s = \frac{Q}{h \times A} + T_{out}\) Now we can calculate the tube surface temperature: \(T_s = \frac{62700\, \mathrm{W}}{5000\, \mathrm{W/m^2 \cdot K} \times 1.5\, \mathrm{m^2}} + 80^{\circ} \mathrm{C} = 4.18\,\mathrm{K} + 80^{\circ} \mathrm{C} = 84.18 ^{\circ} \mathrm{C}\) So, the tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\) is about \(84.18^{\circ} \mathrm{C}\).

Key concepts

Convection Heat Transfer

Convection is the transfer of heat through fluids (liquids or gases) caused by molecular motion. When a fluid is heated, it expands, becoming less dense, and rises. Cooler, denser fluid then takes its place, creating a cycle known as a convection current. In our exercise, the water being heated forms convection currents that transfer heat from the tube's surface to the water, allowing the inlet water at 20°C to reach the outlet temperature of 80°C.

Two types of convection exist: natural convection, which occurs due to buoyancy forces when there are temperature differences within the fluid, and forced convection, which is instigated by mechanical means like a pump or a fan. In the rectangular tube scenario, the water is likely subject to forced convection given the constant flow induced by its mass flow rate. The rate of heat transfer in convection is quantified by the heat transfer coefficient, 'h', which depends on several factors including the velocity of the water, its properties, and the type (laminar or turbulent) of flow within the tube.

Thermal Energy Balance

A thermal energy balance is crucial in understanding how energy in the form of heat is added, removed, or conserved within a given system. By applying the first law of thermodynamics, we establish a balance between the energy entering and leaving the system. For the water heating in the tube, the system’s energy gain comes from the heat supplied by the tube's surface. The heat transfer rate ('Q') must equal the energy required to raise the temperature of the water from its inlet to its outlet.

To ensure accuracy in solutions like the exercise provided, we perform a thermal energy balance calculation. This involves the specific heat capacity ('C_p'), which will be further explained in the next section, mass flow rate of the water ('\(\textbackslash\text{dot}{m}\)'), and the temperature increase ('\(\textbackslash\text{Delta}T\)'). The energy balance is represented by the equation \(Q = \textbackslash\text{dot}{m} \times C_p \times \textbackslash\text{Delta}T\). This step-by-step method ensures that the heat supplied is just right to achieve the desired outlet temperature without any deficiency or excess.

Specific Heat Capacity

The specific heat capacity ('C_p') is a property of a substance that indicates how much heat energy is required to raise the temperature of a unit mass of the substance by one degree. It essentially provides a measure of the ability of the substance to store thermal energy. In the context of our water heating exercise, water’s specific heat capacity determines how much heat needs to be transferred to achieve the desired temperature change.

The specific heat capacity plays an intrinsic role in calculating the heat transfer rate. It dictates the energy required to bring about a specific temperature rise, which is valuable information when designing systems like heaters or radiators. In systems with fluids of different specific heat capacities, this knowledge allows for fine-tuning energy inputs to achieve efficient thermal management. The use of specific heat capacity in thermal calculations bridges the gap between theoretical energy transfers and practical system requirements.