A well-insulated, full, water tank containing 167 kg of water initially at 46.1

Question

A well-insulated, full, water tank containing 167 kg of water initially at 46.1oC is being supplied with 18.2oC water at a rate of 4 kg/min. Water is exiting the tank at the same flow rate so that the mass of water in the tank remains constant. Circulation in the tank keeps the water at a uniform temperature. The specific heat of the water is given as C = 4.2 kJ/kg-K. Determine the time (minutes) at which the water temperature drops to 33.8oC.

Note: The energy of water in the tank is given as: E = mcv*C*T

The solution for an equation of the form: a*dT/dt + T +b = 0     is

T(t) = k1*exp(-t/a) + k2

For this problem, the given constraints are:

T(t=0) = Tinitial

T(t=infinity) = Tinlet

Hint: Use the given constraints to determine the constants k1 and k2

Explanation / Answer

Rate of temp rise dT/dt = Rate of energy in - energy out

dT/dt = mC*18.2 - mC*T

dT/dt = mC*(18.2 - T)

dT/dt + mC*T- mC*18.2 = 0

(1/mC)*dT/dt + T - 18.2 = 0

Comparing it with a dT/dt + T + b = 0 we get

a = 1/(mC) = 1/(4*4.2) = 0.0595

b = -18.2

Solution of DE: T = k1*exp(-t/0.0595) + k2

At t = 0, T = 46.1

and T = infinity, T = 18.2

Using these conditions, we get k1 + k2 = 46.1

and k2 = 18.2

Thus, k1 = 27.9

T = 27.9*exp(-t/0.0595) + 18.2

33.8 = 27.9*exp(-t/0.0595) + 18.2

Solving this, t = 0.0346 min

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