Thermodynamics problem 1st law

In summary, the problem involves steady air flow into a well-insulated piping junction, with two inlets and one outlet. The air is heated by an electric resistor at an unknown rate, and the pressure remains constant at 0.1 MPa. The first inlet has a volumetric flow rate of 10m^3/s and a temperature of 20 C, while the second inlet has a volumetric flow rate of 30m^3/s and a temperature of 30C. At the outlet, the temperature is 55C and the cross section area is 2m^2. It is assumed that kinetic and potential energy effects are negligible, and that the air behaves as a perfect gas with Cp= 1,
  • #1
madzior771
1
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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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  • #2
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.
 

1. What is the 1st law of thermodynamics?

The 1st law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, but it can be transferred from one form to another.

2. How is the 1st law of thermodynamics applied to thermodynamic problems?

In thermodynamic problems, the 1st law is used to analyze the energy flow in a system. It helps determine the changes in internal energy, heat, and work of a system.

3. What is the difference between internal energy and heat?

Internal energy is the total energy of a system, including its kinetic and potential energy. Heat, on the other hand, is the transfer of energy from a hotter object to a cooler object.

4. How is the 1st law of thermodynamics related to the concept of entropy?

The 1st law of thermodynamics states that energy cannot be created or destroyed, but it can be transferred. Entropy, on the other hand, is a measure of the disorder or randomness in a system. The 1st law of thermodynamics helps to explain the relationship between energy and entropy in a system.

5. How can the 1st law of thermodynamics be used to improve energy efficiency?

By understanding the 1st law of thermodynamics, we can identify ways to improve energy efficiency in various systems. This can include reducing heat loss, optimizing energy transfer, and using more efficient processes to convert energy from one form to another.

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