# Fluid Flow/Heat Transfer question struggle

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1. Jul 31, 2016

### williamcarter

1. The problem statement, all variables and given/known data
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 !

2. Relevant equations

http://i.stack.imgur.com/xSsYu.jpg
http://i.stack.imgur.com/aiO1J.jpg

3. The attempt at a solution
Struggling with it.

2. Jul 31, 2016

### 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?

3. Jul 31, 2016

### williamcarter

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.

4. Jul 31, 2016

### 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?

5. Jul 31, 2016

### williamcarter

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

6. Jul 31, 2016

### billy_joule

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

7. Jul 31, 2016

### williamcarter

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)

8. Jul 31, 2016

### 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?

9. Jul 31, 2016

### 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

10. Jul 31, 2016

### Staff: Mentor

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

11. Jul 31, 2016

### williamcarter

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

12. Jul 31, 2016

### Staff: Mentor

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

13. Jul 31, 2016