A hot water stream at 80 oC enters a mixing chamber with mass flow rate of 3.6 kg/s and mixed with cold water at 20 oC. If the mixture temperature is 42 oC, determine the mass flow rate of cold stream in kg/min. Assume all the streams being at 250 kPa

Explanation:

The mixing chamber will be well insulated when steady operating conditions exist such that there will be negligible heat loss to the surroundings. Therefore, changes in the kinetic and potential energies of the fluid streams will be negligible and there are constant fluid properties with no work interactions.

    at 250 kPa =

    approx equal to at

              = 335.02 kJ/kg

    ≈ at

                      = 83.915 kJ/kg

and,     ≈ at = 175.90 kJ/kg

Therefore, mass balance will be calculated as follows.

And, energy balance will be given as follows.

As we are stating steady conditions,

      and cancel out to zero.

So,    

On combining the relations, we solve for as follows.

              =  

        = 0.865 kg/s

                       = 51.9 kg/min      (as 1 min = 60 sec)

Thus, we can conclude that the mass flow rate of cold stream is 51.9 kg/min.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s