Question

Combustion gases passing through a 3-cm-internaldiameter circular tube are used to vaporize waste water at atmospheric pressure. Hot gases enter the tube at $115\mathrm{kPa}$ and ${250}^{\circ }\mathrm{C}$ at a mean velocity of $5\mathrm{m}/\mathrm{s}$, and leave at ${150}^{\circ }\mathrm{C}$. If the average heat transfer coefficient is $120\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$ and the inner surface temperature of the tube is ${110}^{\circ }\mathrm{C}$, determine $(a)$ the tube length and (b) the rate of evaporation of water.

Short answer

Answer: The tube length is 43.7 cm, and the rate of evaporation of water is 1.11 x 10^-5 kg/s.

Step-by-step solution

Calculate the Tube's Cross-Sectional Area

We are given the internal diameter of the tube, which is $3cm$. Therefore, the radius of the tube is $r=\frac{3}{2}cm$ or $0.015m$. Now, let's calculate the cross-sectional area $(A)$ of the tube using the formula for the area of a circle: $$A=\pi r^2$$ Substitute the radius: $$A=\pi (0.015m{)}^2=7.07\times {10}^{-4}{m}^2$$

Calculate the Mass Flow Rate of the Gases

We are given the mean velocity of the gases $V=5m/s$. We can use the equation for mass flow rate $(\dot{m})$ to find the rate at which the gases are passing through the tube: $$\dot{m}={\rho }_{g}AV$$ However, we don't have the density $({\rho }_g)$ of the gases provided. We can find the density using the Ideal Gas Law: $${\rho }_g=\frac{P}{RT}$$ where $P$ is pressure, $R$ is the specific gas constant for air $(287J/kg\cdot K)$, and $T$ is the temperature in Kelvin. First, convert the inlet temperature to Kelvin: $$T_1={250}^{\circ }C+273.15=523.15K$$ Now, find the density of the gases: $${\rho }_g=\frac{115\times {10}^{3}Pa}{(287J/kg\cdot K)(523.15K)}=0.768kg/m^3$$ Finally, calculate the mass flow rate of the gases: $$\dot{m}=(0.768kg/m^3)(7.07\times {10}^{-4}{m}^2)(5m/s)=2.71\times {10}^{-3}kg/s$$

Calculate the Heat Transfer Rate

We are given the average heat transfer coefficient $h=120W/m^2\cdot K$. We also know the temperature difference between the inner surface of the tube and the inlet gases, $\mathrm{\Delta }T=T_1-T_s$, where $T_1$ is the inlet gas temperature and $T_s$ is the tube surface temperature. Calculate the temperature difference: $$\mathrm{\Delta }T=523.15K-383.15K=140K$$ Now, to find the heat transfer rate $(Q)$, we use the equation: $$Q=hA\mathrm{\Delta }T$$ Substitute the given values: $$Q=(120W/m^2\cdot K)(7.07\times {10}^{-4}{m}^2)(140K)=11.88W$$

Calculate the Tube Length

To calculate the tube length $(L)$, we need to find the energy required to heat the gases from their inlet temperature to their exit temperature. We can use the heat capacity of the gases and their mass flow rate to find this energy, as follows: $$Q=\dot{m}{c}_p\mathrm{\Delta }{T}_{2}L$$ where $\mathrm{\Delta }{T}_2=T_2-T_1$ is the temperature difference between the inlet and outlet of the tube, and $c_p$ is the specific heat capacity of the gases at constant pressure. Rearrange the equation to find the tube length: $$L=\frac{Q}{\dot{m}{c}_p\mathrm{\Delta }{T}_2}$$ The specific heat capacity of the combustion gases is typically around $1000J/kg\cdot K$. Now, find the temperature difference between the inlet and outlet of the tube: $$\mathrm{\Delta }{T}_2=523.15K-423.15K=100K$$ Calculate the tube length: $$L=\frac{11.88W}{(2.71\times {10}^{-3}kg{s}^{-1})(1000J/kg\cdot K)(100K)}=0.0437m=43.7cm$$ Now, we have found the tube length $(a)43.7cm$.

Calculate the Evaporation Rate of Water

We can use the conservation of energy principle to find the rate at which water evaporates from the tube. All the heat transferred to the water is used to raise its temperature from ambient temperature to the boiling point and then to evaporate it. Assuming the water enters at ambient temperature, the energy required to evaporate the water can be calculated as follows: $$Q={\dot{m}}_w(c_{p_w}\mathrm{\Delta }{T}_w+L_v)$$ where ${\dot{m}}_w$ is the mass flow rate of the evaporating water, $c_{p_w}$ is the specific heat capacity of water, $\mathrm{\Delta }{T}_w$ is the difference between the inlet water temperature and boiling point, and $L_v$ is the latent heat of vaporization. Rearrange to find the evaporation rate of water: $${\dot{m}}_w=\frac{Q}{c_{p_w}\mathrm{\Delta }{T}_w+L_v}$$ The specific heat capacity of water is $4180J/kg\cdot K$, the latent heat of vaporization is $2.26\times {10}^{6}J/kg$, and the difference between the inlet water temperature and the boiling point is ${100}^{\circ }C-{25}^{\circ }C={75}^{\circ }C$. Calculate the evaporation rate of water: $${\dot{m}}_w=\frac{11.88W}{(4180J/kg\cdot K)(75K)+(2.26\times {10}^{6}J/kg)}=1.11\times {10}^{-5}kg/s$$ We found the rate of evaporation of water $(b)1.11\times {10}^{-5}kg/s$. In conclusion: $(a)$ The tube length is $43.7cm$. $(b)$ The rate of evaporation of water is $1.11\times {10}^{-5}kg/s$.

Key concepts

Mass Flow Rate

Understanding the mass flow rate is crucial when examining processes involving fluids or gases in motion. In thermodynamics, the mass flow rate, denoted by $\dot{m}$, refers to the mass of a substance passing through a given surface per unit time. It's mathematically expressed as $\dot{m}=\rho AV$, where $\rho$ is the density of the fluid or gas, $A$ is the cross-sectional area through which the substance is moving, and $V$ is the velocity of the substance.

For gases, the density can be dynamically calculated using the ideal gas law, as shown in the exercise's solution. This is particularly essential in engineering applications, such as heating or cooling systems, where the characteristics of the fluid or gas change according to the conditions of pressure and temperature, affecting the mass flow rate and consequently the overall system performance. When calculating the mass flow rate for gases, the temperature must be in Kelvin, and the pressure in Pascals to use the ideal gas law correctly.

Relevance in Heat Transfer

Within the context of heat transfer, knowing the mass flow rate is vital for determining the amount of heat a fluid can carry away or deposit over time. The mass flow rate is directly proportional to the heat transfer rate; thus, for a higher $\dot{m}$, the system can either absorb or release more heat. This concept plays a pivotal role in designing equipment like heat exchangers, where controlling the temperature of different fluids is necessary for efficient operation.

Ideal Gas Law Application

The ideal gas law is a fundamental equation in thermodynamics and physical chemistry, providing a simple relation between the pressure ($P$), volume ($V$), temperature ($T$), and amount of an ideal gas. The law is typically stated as $PV=nRT$ or $PV=mRT$, where $n$ is the number of moles, $R$ is the universal gas constant, and $m$ represents the mass of the gas with $R$ being the specific gas constant for a particular gas.

In the particular scenario of the exercise, the ideal gas law is reformulated to find the density ${\rho }_g=\frac{P}{RT}$, facilitating the calculation of the mass flow rate when the volume flow rate and the gas constant are known. The application of the ideal gas law assumes that the gas being considered behaves ideally, which is a reasonable approximation under normal conditions for many gases, including air and combustion gases in the discussed problem.

This law becomes a powerful tool when determining properties of gases required for further calculations in various engineering problems. For instance, the ideal gas law enables you to connect the macroscopic properties of a gas sample under different conditions — a critical aspect when dealing with processes like combustion, compression or expansion of gases, and even atmospheric studies. In heat transfer exercises or real-world applications, the ideal gas law allows engineers and scientists to estimate the behaviour of gases in response to changes in temperature or pressure.

Energy Conservation in Evaporation

The energy conservation principle is a foundational concept in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of evaporation, energy conservation becomes a key factor in understanding and quantifying the process.

Evaporation involves converting liquid water into water vapor, which requires energy in the form of heat. This energy, known as the latent heat of vaporization ($L_v$), is absorbed by the water as it changes state, without a rise in temperature. The total amount of heat $Q$ needed to evaporate a mass $m_w$ of water is given by the equation $Q=m_w(c_{p_w}\mathrm{\Delta }{T}_w+L_v)$, where $c_{p_w}$ is the specific heat capacity of water, and $\mathrm{\Delta }{T}_w$ is the temperature change that the water undergoes before reaching the boiling point.

In the textbook problem, after calculating the heat transferred from the gases to the water, the energy conservation principle is employed to compute the rate of water evaporation. The heat lost by the gas as it cools is equal to the heat absorbed by the water in raising its temperature and causing evaporation. These calculations are fundamental in engineering applications such as boiler design, refrigeration, and even climate modeling where evaporation plays a significant role in energy transfer within the Earth's hydrological cycle.