1.A mass of 5kg of Saturated water vapor at 300 kPa is heated at constant pressure until the temperature reaches 200C. calculate the work done during the process.

 

 

 

2.Steam enters a nozzle at 350C and 800 kPa with a velocity of 15 m/s, and leaves at 250C and 200 kPa for an inlet area of 750 cm^2. Determine the velocity of the steam at the nozzle exit.


This is the type of problems that will go out my exam. please help me answer this problem so i can study and understand this problem.

So it's about exam.... Could you show us a little work you have done.

1.A mass of 5kg of Saturated water vapor at 300 kPa is heated at constant pressure until the temperature reaches 200C. calculate the work done during the process.

 

 

 

2.Steam enters a nozzle at 350C and 800 kPa with a velocity of 15 m/s, and leaves at 250C and 200 kPa for an inlet area of 750 cm^2. Determine the velocity of the steam at the nozzle exit.

 

 

This is the type of problems that will go out my exam. please help me answer this problem so i can study and understand this problem.

for number 1

 

P1=300 kPa

Sat.vapor ------> V1=Vg=0.60582 m^3/kg

 

P2=300 kPa

T2=200 C ------->V2=0.71643 m^3/kg

 

Where V2 come from?

 

Wb=MP(V2-V1)

Wb=(5kg)(300KN/m^2)(0.71643-0.60582)m^3/kg

Wb=165.78 KJ

 

For question number 2 i don't have any idea.. please help

Edited October 17, 201312 yr by chuck14

For 1) You have an equation, or formula, basic to introductory thermodynamics, that nails down the relationship between Pressure, Volume, and Temperature. Given any two, you can calculate the third directly - good so far?

 

You have at hand the definition of Work - in this case, only one motion or displacement is involved, you see that.

 

You do not ask about the initial volume - you are not baffled by that, you got that from the mass, the word "saturated", and the Pressure/Temperature relation. That was the hard part, I would have thought.

 

The key information then is that the pressure is held constant - - - so you have your formula - - - .

 

2) This is the same idea, or basic situation, as 1. The key insight is that all the water gas coming into the nozzle is leaving - water in = water out - at the same time. No water is being condensed, chemically combined, or anything of that nature.

 

So instead of calculating from mass to volume to begin with, as in the first part of 1, you are calculating from volume (per second) to mass (per second). Then you have the same basic formula, the two situations (beginning and ending, entering and leaving) have to balance, - - - -

Edited October 17, 201312 yr by overtone