Known Liquid water flows into and out of a well-stirred tank with equal mass flow rates as the water in the tank is cooled by a cooling coil.
Find Plot the variation of water temperature with time.
Schematic and Given Data:
Engineering Model
1. The control volume is defined by the dashed line on the accompanying diagram.
2. For the control volume, the only significant heat transfer is with the cooling coil. Kinetic and potential energy effects can be neglected.
1 3. The water temperature is uniform with position throughout and varies only with time: T = T(t).
4. The water in the tank is incompressible, and there is no change in pressure between inlet and exit.
Analysis The energy rate balance, Eq. 4.15, reduces with assumption 2 to
\frac{d E_{ cv }}{d t}=\dot{Q}_{ cv }-\dot{W}_{ cv }+\sum_{i} \dot{m}_{i}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g z_{i}\right)-\sum_{e} \dot{m}_{e}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g z_{e}\right) (4.15)
\frac{d U_{ cv }}{d t}=\dot{Q}_{ cv }-\dot{W}_{ cv }+\dot{m}\left(h_{1}-h_{2}\right)
where \dot{m} denotes the mass flow rate.
The mass contained within the control volume remains constant with time, so the term on the left side of the energy rate balance can be expressed as
\frac{d U_{ cv }}{d t}=\frac{d\left(m_{ cv } u\right)}{d t}=m_{ cv } \frac{d u}{d t}
Since the water is assumed incompressible, the specific internal energy depends on temperature only. Hence, the chain rule can be used to write
\frac{d u}{d t}=\frac{d u}{d T} \frac{d T}{d t}=c \frac{d T}{d t}
where c is the specific heat. Collecting results
\frac{d U_{ cv }}{d t}=m_{ cv } c \frac{d T}{d t}
With Eq. 3.20b the enthalpy term of the energy rate balance can be expressed as
h_{2}-h_{1}=c\left(T_{2}-T_{1}\right)+\underline{v\left(p_{2}-p_{1}\right)} (3.20b)
h_{1}-h_{2}=c\left(T_{1}-T_{2}\right)+v\left(p_{1}-p_{2}\right)^{\nearrow 0}
where the pressure term is dropped by assumption 4. Since the water is well mixed, the temperature at the exit equals the temperature of the overall quantity of liquid in the tank, so
h_{1}-h_{2}=c\left(T_{1}-T\right)
where T represents the uniform water temperature at time t.
With the foregoing considerations the energy rate balance becomes
m_{ cv } c \frac{d T}{d t}=\dot{Q}_{ cv }-\dot{W}_{ cv }+\dot{m} c\left(T_{1}-T\right)
As can be verified by direct substitution, the solution of this first-order, ordinary differential equation is
T=C_{1} \exp \left(-\frac{\dot{m}}{m_{ cv }} t\right)+\left(\frac{\dot{Q}_{ cv }-\dot{W}_{ cv }}{\dot{m} c}\right)+T_{1}
The constant C_{1} is evaluated using the initial condition: at t = 0, T=T_{1}. Finally,
T=T_{1}+\left(\frac{\dot{Q}_{ cv }-\dot{W}_{ cv }}{\dot{m} c}\right)\left[1-\exp \left(-\frac{\dot{m}}{m_{ cv }} t\right)\right]
Substituting given numerical values together with the specific heat c for liquid water from Table A-19
T=318 K +\left[\frac{[-7.6-(-0.6)] kJ / s }{\left(\frac{270}{3600} \frac{ kg }{ s }\right)\left(4.2 \frac{ kJ }{ kg \cdot K }\right)}\right]\left[1-\exp \left(-\frac{270 kg / h }{45 kg } t\right)\right]
=318-22[1-\exp (-6 t)]
where t is in hours. Using this expression, we construct the accompanying plot showing the variation of temperature with time.
1 In this case idealizations are made about the state of the mass contained within the system and the states of the liquid entering and exiting. These idealizations make the transient analysis manageable.
Skills Developed
Ability to…
• apply the time-dependent mass and energy rate balances to a control volume.
• develop an engineering model.
• apply the incompressible substance model for water.
• solve an ordinary differential equation and plot the solution.
Quick Quiz
What is the water temperature, in °C, when steady state is achieved? Ans. 23°C.
