Fluid Flow/Heat Transfer question struggle

[williamcarter]

Homework Statement

I would really appreciate if someone could help me out with this exercise in fluid flow/heat transfer ,because I really struggling with it.Below I am attaching photo of problem+table with values. Thank you in advance !

Homework Equations

Question link:[/B]
http://i.stack.imgur.com/xSsYu.jpg
Table link:
http://i.stack.imgur.com/aiO1J.jpg

The Attempt at a Solution

Struggling with it.[/B]

 

[billy_joule]

Do you know the density of water? Are you familiar with the continuity equation?
Q =vA
What do you know about Reynolds number?

 

[williamcarter]

Hello, thank you for your quick reply.
I know Reynolds number has 2 formulas
#1 Re=Diameter*velocity*density/viscosity=D*u*ro/Mew
#2Re=4*mass flowrate/Pi*Diameter*Viscosity=4*m/pi*D*Mew

Reynolds number is dimensionless and can show us 3 types of flow:Laminar(Re<2100),Turbulent(2100<Re<10000) and Ideal(frictionless)

Continuity eq:mass flowrate m=Density1*Velocity1*Area1=Density2*Velocity2*Area2
=>m=ro1*u1*A1=ro2*u2*A2
For incompressible flow such as liquids and gasses at low Press, the density is the same, hence q which is vol flowrate has the formula
q=u1A1=u2A2

Density is 1/specific volume, we have all the values in the above steam table , but specific volume is in mg/m^3 hence density will be 10^3/specific volume.
First column in each field of the table is for liquid and 2nd for vapour.

 

[billy_joule]

Sounds like you have some understanding. Where exactly are you stuck? Can you find the density of water at point 1 and 2?
Can you apply the continuity equation to find the velocities?

 

[williamcarter]

Thank you again for your reply , really appreciated.
we know mass flowrate m=720kg/h.However we need to convert it to kg/s
The mass flowrate is constant m in =m out , no accumulation term,is at steady flow.
Thus 720kg/h=720kg/3600s=0.2kg/s.
A1=100cm^2=0.01m^2
A2=25cm^2=0.0025m^2

At point 1 at T=10 deg Celsius we have :
Density1=10^3/1.000 =1000kg/m^3
A1=0.01m^2
Viscosity Mew1=1304*10^(-6)
From continuity u1=mass flowrate/density1*Area1=0.2/10^3*0.01
=>u1=0.02m/s
Re=D*u*ro/mew
We have A1 ,we need D1;
Diameter1=sqrt(4*0.01/pi)
Hence D1=0.1128 m
so Re=D*u*ro/mew
Re1=0.1128*0.02*10^3/1304*10^(-6)=1730.06

So at point 1 we calculated:
u1=0.02m/s and Re1=1730.06

Same for point 2 at outlet where T2=80 degrees Celsius

 

[billy_joule]

That looks correct, good work.
And what about the duty? Are you familiar with this equation?
Q =m_dot * Cp * deltaT

 

[williamcarter]

Thank you for your reply, I am glad I answered correctly.

As for c)with Heat Duty required we are supposed to use the equation Q=mass flowrate*specific heat*delta T
in our case heat duty transfer rate Q=m*c*(Tout-Tin)

We will take the cp as the cp of (10+80) degrees /2 so we will look in table for cp of 45 degrees celsius
As the value for cp is not there , and it is for values such as 44 and 46 degrees celsius we will have to linear interpolate.

Hence cp at 45 deg Celsius=4179.5(I could have estimated it as 4180, but I wanted to be more precise)
Furthermore it is not required to transform Celsius to Kelvin because the difference(delta T) would be the same , so I would leave it in Celsius.

so we have everything as it follows:
m=0.2 kg/s
cp=4179.5 J/Kg*K
Tout=T2=80 degrees Celsius
Tin=T1=10 degrees Celsius

We know Q=m*c*(Tout-Tin)
hence Q=0.2*4179.5*(80-10)
Q=58513 J (Which is the heat duty required to raise water from 10 degrees Celsius to 80 degrees Celsius)

 

[billy_joule]

That's right, except the 'per second' from the mass flow rate went missing. the answer should be joules per second (aka Watts).
Any questions about the next questions?

 

[williamcarter]

Thank you for your answer, that is correct it is in Watts(J/s), I do not know why I forgot about seconds,I did this probably because I was thinking of Heat which is in Joule and not Power which is Watt, and I forgot about seconds from the mass flowrate.
As for d) I cannot completely understand what I am supposed to do
Thank you in advance.

 

[Chestermiller]

Thank you for your reply, I am glad I answered correctly.

As for c)with Heat Duty required we are supposed to use the equation Q=mass flowrate*specific heat*delta T
in our case heat duty transfer rate Q=m*c*(Tout-Tin)

We will take the cp as the cp of (10+80) degrees /2 so we will look in table for cp of 45 degrees celsius
As the value for cp is not there , and it is for values such as 44 and 46 degrees celsius we will have to linear interpolate.

Hence cp at 45 deg Celsius=4179.5(I could have estimated it as 4180, but I wanted to be more precise)
Furthermore it is not required to transform Celsius to Kelvin because the difference(delta T) would be the same , so I would leave it in Celsius.

so we have everything as it follows:
m=0.2 kg/s
cp=4179.5 J/Kg*K
Tout=T2=80 degrees Celsius
Tin=T1=10 degrees Celsius

We know Q=m*c*(Tout-Tin)
hence Q=0.2*4179.5*(80-10)
Q=58513 J (Which is the heat duty required to raise water from 10 degrees Celsius to 80 degrees Celsius)

Your steam tables give values for the enthalpy of liquid water at 80 C and 10 C. Why didn't you just use this?

 

[williamcarter]

Hello, and thank you for your answer.
I did the arithmetic mean between the 2 temperatures in order to get the mean cp (in my case at 45 degrees celsius) and use it in Q=m*c*delta T to get the duty as a power in Watts.The reason I did that was because I wanted to be consistent with the formula I was given in the course, however your path seems easier, cheers for that.

I could also ,as you said use the enthalpies and calculate the Q as following:
Q=m_dot*[(enthalpy h at 80 degrees Celsius of water)-(enthalpy h at 10 degrees Celsius of water)]

This will be:Q=0.2kg/s*(335.0*10^3 J/Kg -42.03*10^3 J/Kg)
=>Q=58594 Watt

 

[Chestermiller]

For part (d), what is the change in enthalpy per kg in going from saturated steam at 130C to saturated liquid water at 130C?

 

[williamcarter]

Thank you for your reply!
For part d)
I would say Qdot on water side= Qdot on steam side.
Q on water side was calculated at c) and it is 58594 Watt
Q on steam side is : Q=m*lat heat ;but Qwater=Qsteam=58594(J/s)=58594 Watt
and mdot=Qdot/latent heat
From tables latent heat of steam at 130 degrees is 2174 kJ / Kg
=>mdot=58594 Watt /2174*10^3 J/Kg
mdot=0.0269 kg/s