Question

The air supply to a \(56 \mathrm{~m}^{3}\) office has been shut off overnight to conserve utilities, and the room temperature has dropped to \(4^{\circ} \mathrm{C}\). In the morning, a worker resets the thermostat to \(21^{\circ} \mathrm{C}\), and \(6 \mathrm{~m}^{3} / \mathrm{min}\) of air at \(50^{\circ} \mathrm{C}\) begins to flow in through a supply duct. The air is well mixed within the room, and an equal mass flow of air at room temperature is withdrawn through a return duct. The air pressure is nearly 1 bar everywhere. Ignoring heat transfer with the surroundings and kinetic and potential energy effects, estimate how long it takes for the room temperature to reach \(21^{\circ} \mathrm{C}\). Plot the room temperature as a function of time.

Short answer

Approximately 6.48 minutes to reach 21°C

Step-by-step solution

Define the Problem

You need to find out how long it will take for the temperature of the room to reach 21°C when air at 50°C is flowing into the room at a rate of 6 m³/min.

Understand the Heat Transfer

Since the room is well-mixed and the air pressure is nearly 1 bar, assume the specific heat capacity at constant pressure for air, \(c_p\), is approximately constant.

Apply the Energy Balance

Using the principle of conservation of energy, set up an energy balance: \[ \rho V c_p \frac{dT}{dt} = \rho \rho c_p Q (T_{in} - T) \] where \(T\) is the temperature of the room, \(T_{in}\) is the input temperature, \(Q\) is the volumetric flow rate, and \(V\) is the volume of the room. Simplify and integrate to find the temperature over time.

Simplify the Differential Equation

Simplify the equation by canceling the density (\(\rho\)) and heat capacity (\(c_p\)) terms, and rearrange to get: \[ \frac{dT}{T_{in} - T} = \frac{Q}{V} dt \] Substitute the values: \[ \frac{dT}{50 - T} = \frac{6}{56} dt \]

Integrate the Differential Equation

Integrate both sides of the equation to find the time \(t\): \[ \frac{56}{6} \int \frac{dT}{50 - T} = dt \] Perform the integration: \[ - \frac{56}{6} \text{ln}|50 - T| = t + C \] where \(C\) is the constant of integration.

Apply Initial Conditions

Use the initial condition (when \(t = 0\), \(T = 4^\text{°}C\)) to find \(C\): \[ - \frac{56}{6} \text{ln}|50 - 4| = 0 + C \] Solve for \(C\): \[ C = - \frac{56}{6} \text{ln}(46) \]

Solve for Time \( t \)

Substitute \(C\) back into the integrated equation and solve for the time it takes to reach 21°C: \[ - \frac{56}{6} \text{ln}|50 - T| = t - \frac{56}{6} \text{ln}(46) \] When \(T = 21^\text{°}C\): \[ - \frac{56}{6} \text{ln}|50 - 21| = t - \frac{56}{6} \text{ln}(46) \] Simplify and solve for \(t\). Hence, \[ t = \frac{56}{6} \text{ln} \frac{46}{29} \]

Numerical Value

Calculate the numerical value of \( t \): \[ t = \frac{56}{6} \times \text{ln} \frac{46}{29} \approx 6.48 \text{ minutes} \]

Plot Temperature vs. Time

Plot the room temperature as a function of time using the derived equation \( T(t) = 50 - (50 - 4)e^{-\frac{Q}{V}t} \). Values should be plotted from \( T(0) = 4°C \) to \( T(t_{final}) = 21°C \).

Key concepts

Energy Balance

Energy balance is a vital principle in thermodynamics, particularly when studying heat transfer in systems. In this exercise, the energy balance equation helps determine how the temperature in a room changes when warmer air is introduced. The principle of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another. For our problem, the energy being transferred into the room with the warm air flow must equal the energy causing the room's temperature to rise. This balance forms the core of the equation we use to find how long it takes for the room to reach a desired temperature.

Differential Equation

A differential equation is a mathematical equation that involves functions and their derivatives. In the context of our heat transfer problem, we use a differential equation to model how the temperature in the room changes over time. The setup is: \[ \rho V c_p \frac{dT}{dt} = \rho \rho c_p Q (T_{in} - T) \] This equation shows that the rate of change of temperature, \( \frac{dT}{dt} \), is proportional to the difference between the incoming air temperature \( T_{in} \) and the current room temperature \( T \). Solving this differential equation through simplification and integration gives us a way to predict the temperature over time.

Integration

Integration is a crucial mathematical process used to solve differential equations, particularly when you have an initial value problem, as in this exercise. After simplifying the original differential equation, we get: \[ \frac{dT}{T_{in} - T} = \frac{Q}{V} dt \] To solve for temperature as a function of time, we integrate both sides: \[ \frac{56}{6} \text{ln}|50 - T| = -t \right. + C \] By applying the initial condition that when \( t = 0 \), \( T = 4^{\text{°}C} \), we find the integration constant \( C \). This step is key as it allows us to have an equation that fully describes the temperature at any given time \( t \) after introducing the warm air.

Specific Heat Capacity

Specific heat capacity, often denoted as \( c_p \), is the amount of heat required to raise the temperature of one unit of mass of a substance by one degree Celsius. In our problem, the specific heat capacity of air is used to relate the amount of heat energy entering the room to the change in temperature of the air inside the room. The assumption is that \( c_p \) for air is constant, which simplifies our calculations. Essentially, knowing \( c_p \) allows us to directly connect the flow of warm air into the room with the increase in room temperature.