Question

A system consists of a liquid, considered incompressible with constant specific heat \(c\), filling a rigid tank whose surface area is A. Energy transfer by work from a paddle wheel to the liquid occurs at a constant rate. Energy transfer by heat occurs at a rate given by \(\dot{Q}=-\operatorname{ha}\left(T-T_{0}\right)\), where \(T\) is the instantaneous temperature of the liquid, \(T_{0}\) is the temperature of the surroundings, and \(\mathrm{h}\) is an overall heattransfer coefficient. At the initial time, \(t=0\), the tank and its contents are at the temperature of the surroundings. Obtain a differential equation for temperature \(T\) in terms of time \(t\) and relevant parameters. Solve the differential equation to obtain \(T(t)\).

Short answer

The differential equation is \( mc \frac{dT}{dt} = W_0 - hA(T - T_0) \). The solution is \( T(t) = \left( \frac{W_0}{hA} \right) + \left( T_0 - \frac{W_0}{hA} \right) e^{-\frac{t}{\tau}} \).

Step-by-step solution

Write the First Law of Thermodynamics

The First Law for a closed system can be written as \[\frac{dU}{dt} = \dot{Q} + \dot{W}\] where \(\frac{dU}{dt}\) is the rate of change of internal energy, \(\dot{Q}\) is the rate of heat transfer, and \(\dot{W}\) is the rate of work done by the paddle wheel.

Express Internal Energy in Terms of Temperature

For an incompressible liquid with constant specific heat, the change in internal energy can be expressed as \[\Delta U = mc\Delta T\]where \(m\) is the mass, \(c\) is the specific heat, and \(T\) is the temperature. Then, the rate of change of internal energy can be expressed as \[\frac{dU}{dt} = mc\frac{dT}{dt}\]

Substitute the Relations for \(\dot{Q}\) and \(\dot{W}\) into the First Law

Given the heat transfer rate \[\dot{Q} = -hA(T - T_0)\] and assuming the paddle wheel adds energy at a constant rate \(\dot{W}=W_0\), substitute these into the First Law equation: \[mc\frac{dT}{dt} = W_0 - hA(T - T_0)\]

Rearrange the Equation to Form a Standard Linear Differential Equation

Reorganize the terms to form: \[mc\frac{dT}{dt} + hA(T - T_0) = W_0\] Define the parameters \(\tau = \frac{mc}{hA}\) and \(C = \frac{W_0}{hA}\): \[\frac{dT}{dt} + \frac{1}{\tau}(T - T_0) = \frac{C}{\tau}\]

Solve the Differential Equation

This is a first-order linear differential equation. Solve it using the integrating factor method. The integrating factor is \[\mu(t) = e^{\int \frac{1}{\tau} dt} = e^{\frac{t}{\tau}}\] Multiply the differential equation by \(\mu(t)\): \[e^{\frac{t}{\tau}}\frac{dT}{dt} + \frac{1}{\tau}e^{\frac{t}{\tau}}T = \frac{C}{\tau}e^{\frac{t}{\tau}}\] This simplifies to: \[\frac{d}{dt}(e^{\frac{t}{\tau}} T) = \frac{C}{\tau} e^{\frac{t}{\tau}}\] Integrate both sides with respect to \(t\): \[e^{\frac{t}{\tau}} T = C e^{\frac{t}{\tau}} + D\] where \(D\) is a constant of integration. Solve for \(T\): \[T = C + De^{-\frac{t}{\tau}}\] Given initial condition at \(t=0\), \(T(0) = T_0\): \[T_0 = C + D\] Thus, \[D = T_0 - C\]. Substitute back to find \(T(t)\): \[T(t) = C + (T_0 - C)e^{-\frac{t}{\tau}}\]

Final Expression for Temperature Over Time

Since \(C = \frac{W_0}{hA}\), the final form of the temperature equation is: \[T(t) = \left(\frac{W_0}{hA}\right) + \left(T_0 - \frac{W_0}{hA}\right)e^{-\frac{t}{\tau}}\]

Key concepts

Incompressible Liquid

An incompressible liquid is a fluid whose density remains constant regardless of changes in pressure. This simplification makes calculations easier because we do not need to account for variations in volume or density when analyzing thermodynamic processes.
For the given exercise, we assume the liquid in the rigid tank is incompressible and has a constant specific heat capacity, which simplifies how we calculate changes in internal energy.

Specific Heat

Specific heat is a property that indicates the amount of energy required to raise the temperature of a unit mass of a substance by one degree Celsius. It is denoted by the symbol \(c\).
In mathematical terms, the change in internal energy \( \Delta U \) for an incompressible liquid with mass \( \ m \ \) and specific heat \( \ c \) can be expressed as:
\( \ \Delta U = mc\Delta T \)
This shows that the internal energy is directly proportional to the mass, specific heat, and change in temperature of the liquid.

Rate of Heat Transfer

The rate of heat transfer \( \ \dot{Q} \ \) represents the energy transferred per unit time. In engineering problems, heat transfer is often a function of the temperature difference between the system and its surroundings, and properties of the medium through which heat is transferred.
For the given exercise, heat transfer occurs between the liquid and its surroundings at a rate given by:
\( \ \dot{Q} = -hA(T - T_0) \)
where

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