What is the exit steam velocity?

AI Thread Summary
The discussion revolves around calculating the exit steam velocity from a nozzle given specific conditions, including steam enthalpy and heat loss. The initial enthalpy of the steam is identified as 2789.88 kJ/kg, and the specific volume is noted as 0.132. Participants clarify that the problem can be solved per unit mass, eliminating the need to determine mass or area. The heat loss of 25 kJ/kg is emphasized as already being per unit mass, simplifying the calculations. Ultimately, the original poster successfully solves the problem after receiving guidance.
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Homework Statement


Steam, at 15 bar and 280oC, enters a nozzle with an initial velocity of 125 m/s. The steam enthalpy at the exit section is 2800 kJ/kg and the heat loss is 25 kJ/kg. What is the exit steam velocity?.

Homework Equations


1)Qin - Qout = m [h2-h1+((c22-c12))/2]
2)h = Pv + u
3)A1c1/ρ1 = A2c2/ρ2

The Attempt at a Solution


0-25k = m((2800-h1)+-125^2/2)

Steam is not an ideal gas so I can't use Pv=mRT. I have two unknowns and I don't know where to start. I hope you can help.
 
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For steam at 15 bar and 280 C, from your steam tables, what is the specific volume and the specific enthalpy?
 
Nemo's said:

Homework Statement


Steam, at 15 bar and 280oC, enters a nozzle with an initial velocity of 125 m/s. The steam enthalpy at the exit section is 2800 kJ/kg and the heat loss is 25 kJ/kg. What is the exit steam velocity?.

Homework Equations


1)Qin - Qout = m [h2-h1+((c22-c12))/2]
2)h = Pv + u
3)A1c1/ρ1 = A2c2/ρ2

The Attempt at a Solution


0-25k = m((2800-h1)+-125^2/2)

Steam is not an ideal gas so I can't use Pv=mRT. I have two unknowns and I don't know where to start. I hope you can help.
1. What's the enthalpy of the steam in the nozzle inlet?
2. What have you done with the heat loss of 25 kJ/kg which occurs while the steam passes thru the nozzle?
 
Chestermiller said:
For steam at 15 bar and 280 C, from your steam tables, what is the specific volume and the specific enthalpy?
Thanks for letting me pay attention to the tables.
h1=2789.88
v=0.132
Now if I can only get the area I can use m= A*c1/ρ1 to get m and then substitute in 1 to get c2.
 
Nemo's said:
Thanks for letting me pay attention to the tables.
h1=2789.88
v=0.132
Now if I can only get the area I can use m= A*c1/ρ1 to get m and then substitute in 1 to get c2.
You don't need to determine m or A, because you are going to solving the problem per unit mass. The 25 kJ/kg is already per unit mass. So the m should not be in your equation.
 
Chestermiller said:
You don't need to determine m or A, because you are going to solving the problem per unit mass. The 25 kJ/kg is already per unit mass. So the m should not be in your equation.
O.k thanks a lot
 
I solved it . Thank you anyway
 
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