91Ó°ÊÓ

In thermodynamics, the mass flow rate is a crucial concept. It tells us how much mass is passing through a cross-sectional area per unit time. The formula for mass flow rate is simple: \[ \dot{m} = \rho \cdot Q \] \( \dot{m} \) is the mass flow rate in \( kg/s \), \( \rho \) is the density of the fluid in \( kg/m^3 \), and \( Q \) is the volumetric flow rate in \( m^3/s \). For this exercise, we first calculate the density using the ideal gas law and then use it to find the mass flow rate. Remember, density can change with temperature and pressure. For air at the inlet, the density is: \[ \rho_{in} = \frac{100000}{287 \cdot 290} = 1.196 \ kg/m^3 \] With a volumetric flow rate of \( 0.25 \ m^3/s \), the mass flow rate is: \[ \dot{m} = 1.196 \cdot 0.25 = 0.299 \ kg/s \] Understanding the mass flow rate helps in determining other important quantities, such as velocity.

Finding the velocity of the air at different points in the duct is essential. We use the mass flow rate equation \[ \dot{m} = \rho \cdot A \cdot V \] Here, \( A \) is the cross-sectional area and \( V \) is the velocity. Rearranging the formula to solve for velocity: \[ V = \frac{\dot{m}}{\rho \cdot A} \] For the inlet, where \( \rho_{in} = 1.196 \ kg/m^3 \) and \( A = 0.04 \ m^2 \), the velocity is: \[ V_{in} = \frac{0.299}{1.196 \cdot 0.04} = 6.24 \ m/s \] At the outlet, where \( \rho_{out} = 1.012 \ kg/m^3 \), the velocity is: \[ V_{out} = \frac{0.299}{1.012 \cdot 0.04} = 7.39 \ m/s \] These calculations show that the air speeds up as it exits the duct. Always ensure the unit consistency to avoid mistakes.

The rate of heat transfer is directly connected to changes in temperature. Using the first law of thermodynamics for a control volume, the formula is: \[ Q = \dot{m} \cdot c_p \cdot (T_{out} - T_{in}) \] \( Q \) is the heat transfer rate, \( \dot{m} \) is the mass flow rate, \( c_p \) is the specific heat capacity, and \( T_{out} - T_{in} \) represents the temperature difference. For air with \( c_p = 1005 \ J/(kg \cdot K) \), calculate the heat transfer: \[ Q = 0.299 \cdot 1005 \cdot (325 - 290) = 10490 \ W = 10.49 \ kW \] This shows heat is being added to the air, raising its temperature. Understanding heat transfer is vital in designing efficient HVAC systems.

The Ideal Gas Law connects pressure, volume, and temperature of a gas. It's expressed as: \[ PV = nRT \] Here, \( P \) is pressure, \( V \) is volume, \( n \) is the amount of gas in moles, \( R \) is the gas constant, and \( T \) is temperature. For this exercise, we used the specific form: \[ P = \rho RT \] Solving for density \( \rho \): \[ \rho = \frac{P}{RT} \] For air, \( R = 287 \ J/(kg \cdot K) \). At the inlet, \( P_{in} = 100000 \ Pa \), and \( T_{in} = 290 \ K \): \[ \rho_{in} = 1.196 \ kg/m^3 \] At the outlet, \( P_{out} = 95000 \ Pa \), and \( T_{out} = 325 \ K \): \[ \rho_{out} = 1.012 \ kg/m^3 \] This law is fundamental in thermodynamics for solving various gas-related problems.

Thermodynamics is the study of energy and its transformations. It involves understanding how energy transfers between systems, including heat and work. Key principles include:

• First Law: Energy cannot be created or destroyed, only transferred or converted.
• Second Law: Heat flows naturally from hot to cold objects.
• Enthalpy: A measure of total energy in a system.

In this exercise, we used the steady flow energy equation, focusing on heat transfer and fluid flow. By mastering thermodynamics, engineers can design more efficient systems for heating, cooling, and power generation.