What Is the Mass Flow Rate at Exit in a Steady Flow System?

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SUMMARY

The discussion focuses on calculating the mass flow rate at the exit of a steady flow system involving air flowing through two inlets into a piping junction. Given the volumetric flow rates of 10 m³/s and 30 m³/s at temperatures of 20°C and 30°C respectively, and an outlet temperature of 55°C, the mass flow rate can be determined using the equation m1 + m2 = m3. The heat rate of the electric heater and the exit velocity of air are also derived using the first law of thermodynamics and the ideal gas properties, specifically with Cp = 1.0 kJ/(kg·K).

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Homework Statement



Air flows steadily into a well isnulated piping junction through the two pipes and is heated by an electric resistor at uknown rate befoe exiting through the pipe. Pressure renaubs approcimetaly constant
p=0.1 MPa, first inlet: volumetric flow rate = 10m^3/s, t1= 20 C, 2nd inlet: volumetric flow rate = 30m^3/s, t2= 30C. At the outlet: t3= 55C, cross section area = 2m^2. Assume the effects of kinetic and potential energy are negligible and air behaves as a perfect gas iwth Cp= 1,0kJ/(kgK). Find the : mass flow rate at exit, heat rate of electric heater and exite velocity of air.

Homework Equations

The Attempt at a Solution

:
m1+m2=m3
Q-W= m1*ht1 + m2*ht2 - m3*ht3
W=0
Q=m1ht1+m2ht2-(m1+m2)ht3
m=A*ro*velocity
[/B]
 
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madzior771 said:

Homework Statement



Air flows steadily into a well isnulated piping junction through the two pipes and is heated by an electric resistor at uknown rate befoe exiting through the pipe. Pressure renaubs approcimetaly constant
p=0.1 MPa, first inlet: volumetric flow rate = 10m^3/s, t1= 20 C, 2nd inlet: volumetric flow rate = 30m^3/s, t2= 30C. At the outlet: t3= 55C, cross section area = 2m^2. Assume the effects of kinetic and potential energy are negligible and air behaves as a perfect gas iwth Cp= 1,0kJ/(kgK). Find the : mass flow rate at exit, heat rate of electric heater and exite velocity of air.

Homework Equations

The Attempt at a Solution

:
m1+m2=m3
Q-W= m1*ht1 + m2*ht2 - m3*ht3
W=0
Q=m1ht1+m2ht2-(m1+m2)ht3
m=A*ro*velocity[/B]
This looks OK so far, except for the signs of the right hand side...they are all opposite to what they should be. Please show some more detail.
 

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