Question

Hot water \(\left(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows through a 200-m-long PVC \((k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) pipe whose inner diameter is \(2 \mathrm{~cm}\) and outer diameter is \(2.5 \mathrm{~cm}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\), entering at \(40^{\circ} \mathrm{C}\). If the entire interior surface of this pipe is maintained at \(35^{\circ} \mathrm{C}\) and the entire exterior surface at \(20^{\circ} \mathrm{C}\), the outlet temperature of water is (a) \(39^{\circ} \mathrm{C}\) (b) \(38^{\circ} \mathrm{C}\) (c) \(37^{\circ} \mathrm{C}\) (d) \(36^{\circ} \mathrm{C}\) (e) \(35^{\circ} \mathrm{C}\)

Short answer

a) 39°C b) 38°C c) 37°C d) 36°C

Step-by-step solution

Calculate the area of the inner and outer surfaces of the pipe

Compute the inner and outer cross-sectional areas (Ai and Ao) of the PVC pipe using the formula A = πD²/4 where D is the diameter of the pipe. Inner diameter (Di) = 2 cm = 0.02 m Outer diameter (Do) = 2.5 cm = 0.025 m Inner area (Ai) = (π × (0.02)²) / 4 = π × 0.0004 / 4 = 0.000314 m² Outer area (Ao) = (π × (0.025)²) / 4 = π × 0.000625 / 4 = 0.0004909 m²

Calculate the temperature difference across the pipe

Calculate the temperature difference (∆T) between the interior surface temperature and the exterior surface temperature. This is the driving force for heat to flow through the pipe. Interior surface temperature = 35°C Exterior surface temperature = 20°C ∆T = 35 - 20 = 15°C

Calculate the heat loss per unit length of the pipe

Calculate the heat loss (Q) per unit length (L) of the pipe by using the formula Q/L = k * (Ao - Ai) * ∆T / (log(Ao/Ai) * r), where k is the thermal conductivity of the PVC pipe, and r is the radial distance between the inner and outer surfaces. Then, multiply Q/L by the total length of the pipe. k = 0.092 W/m·K L = 200 m r = (Do - Di) / 2 = (0.025 - 0.02) / 2 = 0.0025 m Q/L = (0.092 * (0.0004909 - 0.000314) * 15) / (log(0.0004909 / 0.000314) * 0.0025) = 5.13 W/m Now, calculate the heat loss in the entire pipe. Q = (Q/L) × L = 5.13 × 200 = 1026 W

Calculate the outlet temperature of the water

We will now use the specific heat capacity formula Q = m × cp × ∆T, to calculate the temperature drop in the water flow. Q = 1026 W m = mass flow rate of water = 1 kg/s cp = specific heat capacity of water = 4.179 kJ/kg·K = 4179 J/kg·K Rearrange the formula: ∆T = Q / (m × cp) ∆T = 1026 / (1 × 4179) = 0.2455°C Now, we can calculate the outlet temperature of the water: Outlet temperature = Inlet temperature - ∆T = 40 - 0.2455 = 39.7545°C Considering the given choices, we can approximate the outlet temperature to 39°C. So, the correct answer is (a) 39°C.

Key concepts

Conduction Heat Transfer

Understanding heat transfer is crucial when considering the temperature changes observed in fluids passing through pipes, such as hot water in a PVC pipe. Conduction heat transfer refers to the way heat moves through materials that are in direct contact. Imagine holding one end of a metal rod with a flame at the other end; you'll feel the heat traveling along the rod to your hand. In the context of pipes, heat conduction occurs across the pipe's material from the warmer water inside to the cooler surroundings outside.

The rate at which this heat transfer occurs depends on the material's thermal properties, the temperature difference between the interior and exterior, and the pipe's structure. With a smaller temperature difference or less thermally conductive material, the heat loss would naturally be less, resulting in a smaller difference in outlet water temperature.

Thermal Conductivity

Thermal conductivity, denoted by the symbol 'k', is a measure of how well a material can conduct heat. It represents the amount of heat, in watts, that passes through a meter cube of a substance when there's a temperature difference of one Kelvin between opposite sides. Each material has a unique thermal conductivity value. Higher 'k' values mean better heat conduction.

For example, metals generally have high thermal conductivity, which is why they feel cold to touch; they are efficient at transferring heat away from your warm hand. In our pipe scenario, PVC has a relatively low 'k' value of 0.092 W/m⋅K, limiting how much heat gets lost to the environment as the water travels within. This characteristic helps in maintaining the water's temperature from entry to exit.

Specific Heat Capacity

Specific heat capacity, symbolized as 'cp', is the amount of heat energy required to raise the temperature of one kilogram of a substance by one Kelvin. Liquids like water have a relatively high specific heat capacity, meaning they can absorb a lot of heat before significantly increasing in temperature. For hot water, as in our example, with a 'cp' of 4.179 kJ/kg⋅K, it requires quite a bit of heat loss or gain to alter its temperature.

This property is essential in practical applications, such as the transportation of hot fluids in industrial processes. With a high 'cp', water serves as an effective heat transfer medium, able to carry a significant amount of thermal energy with minimal temperature change, as seen with the mere 0.2455°C drop over 200 meters of the pipe.

Temperature Gradient

The temperature gradient is essentially the change in temperature per unit length and is a driving factor in heat transfer. It is the 'push' that moves thermal energy from a hotter region towards a colder one. This concept is illustrated by the difference in temperature across the wall of the pipe: 35°C inside and 20°C outside, resulting in a gradient that encourages heat to flow out of the pipe.

A steeper temperature gradient means a larger temperature difference over the same distance, which typically results in a faster heat transfer rate. In conduction, the heat flow rate per area is proportional to the temperature gradient. Therefore, if the exterior of the pipe were colder, the gradient would be steeper, increasing the rate at which water cools as it travels through the pipe.