91Ó°ÊÓ

A steam radiator is used to heat a \(60-\mathrm{m}^{3}\) room. Saturated steam at 3.0 bar condenses in the radiator and emerges as a liquid at the saturation temperature. Heat is lost from the room to the outside at a rate $$\dot{Q}(\mathrm{kJ} / \mathrm{h})=30.0\left(T-T_{0}\right)$$ where \(T\left(^{\circ} \mathrm{C}\right)\) is the room temperature and \(T_{0}=0^{\circ} \mathrm{C}\) is the outside temperature. At the moment the radiator is turned on, the temperature in the room is \(10^{\circ} \mathrm{C}\). (a) Let \(\dot{m}_{\mathrm{s}}(\mathrm{kg} / \mathrm{h})\) denote the rate at which steam condenses in the radiator and \(n(\mathrm{kmol})\) the quantity of air in the room. Write a differential energy balance on the room air, assuming that \(n\) remains constant at its initial value, and evaluate all numerical coefficients. Take the heat capacity of air \(\left(C_{v}\right)\) to be constant at \(20.8 \mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\) (b) Write the steady-state energy balance on the room air and use it to calculate the steam condensation rate required to maintain a constant room temperature of \(24^{\circ} \mathrm{C}\). Without integrating the transient balance, sketch a plot of \(T\) versus \(t,\) labeling both the initial and maximum values of \(T\) (c) Integrate the transient balance to calculate the time required for the room temperature to rise by \(99 \%\) of the interval from its initial value to its steady-state value, assuming that the steam condensation rate is that calculated in Part (b).

Step by step solution

01

Write a differential energy balance

Due to the conservation of energy principle, the rate of energy increase within the room comes from heat added through condensation minus the heat loss towards the outside. You can express this mathematically as:\[ \frac{d\left(nCvT\right)}{dt} = \dot{m}_{\text{s}}h_{\text{fg}} - \dot{Q} \]where \(d\left(nCvT\right)/dt\) is the rate of energy increase in the room, \( \dot{m}_{\text{s}}h_{\text{fg}}\) represents the heat added to the room through condensation (with \(\dot{m}_{\text{s}}\) being the mass flow rate of the condensing steam and \(h_{\text{fg}}\) its heat of vaporization), and \(\dot{Q}\) accounts for the heat loss towards outside. Equating the given value of \(\dot{Q}\), you get:\[ \frac{d\left(nCvT\right)}{dt} = \dot{m}_{\text{s}}h_{\text{fg}} - 30.0\left(T-T_{0}\right) \]Here, \(t\) is the time, \(n\) is the quantity of air in the room, \(Cv\) is the heat capacity of air, \(T\) is the room temperature, and \(T_{0}\) is the outside temperature.

02

Write the steady-state energy balance and calculate steam condensation rate

The steady state is reached when the room temperature doesn't change, indicating that the rates of heat entering and displacement are equal. Therefore, \( \frac{d\left(nCvT\right)}{dt} = 0 \), so the energy balance equation becomes:\[ \dot{m}_{\text{s}}h_{\text{fg}} = 30.0\left(T_{\text{ss}}-T_{0}\right) \]This equation lets you find the steam condensation rate, \( \dot{m}_{\text{s}} \), required to maintain a constant room temperature (\( T_{\text{ss}} = 24°C \)). You would divide the right side of the equation by \( h_{\text{fg}} \).

03

Calculate the time for room temperature to rise by 99%

To find the time required (\( t_{\text{99}} \)) for the room temperature to rise by 99% of the interval from its initial value to its steady-state value, integrate the transient energy balance equation over time and temperature. Rearrange the equation by isolating \( dt \) and integrate it from \( t = 0 \) to \( t = t_{\text{99}} \) and \( T = T_{\text{i}} \) to \( T = 0.99 * (T_{\text{ss}} - T_{\text{i}}) + T_{\text{i}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Steady-State Energy Balance

In the context of chemical processes, a steady-state energy balance is fundamental for understanding how systems reach thermal equilibrium. When a system arrives at a steady-state condition, it means that all the temperatures within the system are constant over time—even though energy may still flow in and out. This is a common scenario in industrial processes, where continuous operations aim for a consistent output.

Applying the steady-state assumption to the steam radiator scenario simplifies the problem significantly. The energy balance equation \( \dot{m}_{\text{s}}h_{\text{fg}} = 30.0\left(T_{\text{ss}}-T_{0}\right) \) reveals exactly how much steam must condense to maintain the desired room temperature. The term \( \dot{m}_{\text{s}} \) represents the condensation rate of steam within the radiator, \( h_{\text{fg}} \) is the latent heat released during condensation, and \( 30.0\left(T_{\text{ss}}-T_{0}\right) \) is the rate of heat loss from the room to the environment. Hence, the condensation rate can be determined to ensure that heat inflow matches the outflow, keeping the room temperature constant at \( 24^\circ C \).

Transient Energy Balance

Contrary to the steady-state, a transient energy balance considers the time-dependent aspects of a thermal system. This balance is crucial when analyzing how a system's temperature changes over time before reaching the steady state. Transience can complicate calculations because it requires integrating variables over a period until equilibrium is achieved.

In our radiator example, the transient energy balance comes into play right when the radiator begins to warm the room. Mathematically, this involves integrating the differential equation \( \frac{d(nC_vT)}{dt} = \dot{m}_{\text{s}}h_{\text{fg}} - 30.0(T-T_{0}) \). The left side of this equation expresses the rate at which room air energy changes due to thermal interactions. By integrating, we can predict how long it will take for the room to reach a certain percentage of its final steady-state temperature, considering the rate of steam condensation determined in the steady-state energy balance. For instance, to rise by 99% toward steady-state affairs, one would solve for \( t_{\text{99}} \) in the context of this equation.

Heat Capacity

Heat capacity is an intrinsic property that measures how much thermal energy a substance can store. It plays a significant role in calculating energy balances as it helps us understand how much energy is required to raise the temperature of a specific amount of matter by one-degree Celsius (or one Kelvin). Specifically, a substance with high heat capacity can absorb a lot of heat without experiencing a significant rise in temperature.

In our exercise, the heat capacity of the room air \( (C_v) \) is a crucial component of the differential energy balance. It influences how quickly the room heats up, given a certain rate of steam condensation. For example, because air has a relatively low heat capacity, a given amount of energy can raise the room temperature fairly quickly. This is represented by the constant \(20.8 \mathrm{J} / (\mathrm{mol} \cdot^\circ \mathrm{C})\) provided in the problem, allowing us to calculate the rate of temperature change over time given varying conditions and energy input rates.