Question

A hot-water stream at 80°C enters a mixing chamber with a mass flow rate of 0.56 kg/s where it is mixed with a stream of cold water at 20°C. If it is desired that the mixture leave the chamber at 42°C, determine the mass flow rate of the cold-water stream. Assume all the streams are at a pressure of 250 kPa. The enthalpies are 335.02 kJ/kg, 83.915 kJ/kg, and 175.90 kJ/kg. The saturation temperature at a pressure of 250 kPa is 127.41°C.

Answer 1

Answer:

0.96kg/s

Explanation:

Hello! To solve this exercise we must use the first law of thermodynamics, which states that the sum of the energies that enter a system is the same amount that must go out. We must consider the following!

state 1 : is the first   flow in the input of the chamber

h1=entalpy=335.02KJ/kg

m1=mass flow=0.56kg/s

state 2 : is the second  flow in the input of the chamber

h2=entalpy=83.915KJ/kg

state 3:is the flow that comes out

h3=entalpy=175.90 kJ/kg

now use the continuity equation that states that the mass flow that enters is the same as the one that comes out

m1+m2=m3

now we use the first law of thermodynamics

m1h1+m2h2=m3h3

335.02m1+83.915m2=175.9m3

as the objective is to find the cold water mass flow(m2) we divide this equation by 175.9

1.9m1+0.477m2=m3

now we subtract the equations found in the equation of continuity and first law of thermodynamics

m1     +       m2    =  m3

-

1.9m1 +   0.477m2=m3

----------------------------------

-0.9m1+0.523m2=0

solving for m2

$m2=m1\frac{0.9}{0.523} .\\m2=(0.56)\frac{0.9}{0.523} =0.96kg/s$

the mass flow rate of the cold-water is 0.96kg/s