Question 2.11: An insulated,electrically heated tank for hot water contains...
An insulated,electrically heated tank for hot water contains 190 kg of liquid water at 60°C. Imagine you are taking a shower using water from this tank when a power outage occurs. If water is withdrawn from the tank at a steady rate of \dot{m} = 0.2 kg·s^{−1}, how long will it take for the temperature of the water in the tank to drop from 60 to 35°C?
Assume that cold water enters the tank at 10°C and that heat losses from the tank are negligible. Here, an excellent assumption for liquid water is that C_{V} = C_{P} = C, independent of T and P.
This is an example of the application of Eq. (2.28) to a transient process for which \dot{Q} = \dot{W} = 0 .
\frac{d(mU)_{CV} }{dt} + \Delta (H\dot{m})_{fs}= \dot{Q } + \dot{W} (2.28)
We assume perfect mixing of the contents of the tank; this implies that the properties of the water leaving the tank are those of the water in the tank. With the mass flow rate into the tank equal to the mass flow rate out, mcv is constant; moreover, the differences between inlet and outlet kinetic and potential energies can be neglected. Equation (2.28) is therefore written:
m\frac{dU}{dt} + \dot{m}(H – H_{1}) = 0
where unsubscripted quantities refer to the contents of the tank (and therefore the water leaving the tank) and H1 is the specific enthalpy of the water entering the tank. With C_{V} = C_{P} = C ,
\frac{dU}{dt} = C\frac{dT}{dt} and H – H_{1} = C( T – T_{1} )
The energy balance then becomes, on rearrangement,
dt = – \frac{m}{\dot{m} } .\frac{dT}{T – T_{1} }
Integration from t = 0 (where T = T_{0}) to arbitrary time t yields:
t = \frac{m}{\dot{m} }1n\left(\frac{T – T_{1}}{T_{0} – T_{1}}\right)
Substitution of numerical values into this equation gives, for the conditions of this problem,
t = \frac{190}{0.2} 1n \left(\frac{35 – 10}{60 – 10} \right) = 658.5 s
Thus, the water temperature in the tank will drop from 60 to 35°C after about 11 minutes.