MrsDrPoe: The First Law of Thermodynamics, Part 6

Hello and welcome to Thesis Tuesday on the blog!  As promised, today we will be finishing our look at the first law of thermodynamics by applying the differential form of the first law to an example problem.

Problem: A liquid is flowing downward along an inclined plane surface, as shown in the figure:

The free liquid surface (y = h) is maintained at temperature Th, and the solid surface (y = 0) is maintained at To.  Determine an expression for the temperature distribution in the film, recognizing that the viscous heating effects can be ignored.

Given: non-isothermal film flow, T(0) = T0 and T(h) = Th
Find: An expression for the temperature distribution in the fluid film
Assumptions: steady, laminar, incompressible, Newtonian fluid, ignore viscous heating effects, constant thermal conductivity, assume T = f(y), constant properties
Solution: 


Starting with the general form of the differential energy equation for an incompressible Newtonian fluid:

rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz)) + mu*Phiv

Steady flow eliminates the first term on the left side; the last term on the right side is eliminated because we are ignoring viscous heating effects.

rho*cp*(u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz))

Our assumption of laminar flow means that v = w = 0; the first term on the left side is also zero since T is in not a function of x:

0 = (d/dx(qx) + d/dy(qy) + d/dz(qz))

The flux terms can be related to the temperature gradient using Fourier's law:

qx = k*d/dx(T), qy = k*d/dy(T), and qz = k*d/dz(T)

Furthermore, since the thermal conductivity (k) is constant, our energy equation becomes:

0 = k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))

We can divide both sides by k; the first and third terms cancel since T is not a function of x or z:

0 = d/dy(d/dy(T))

If we separate and integrate this equation twice, we end up with the expression:

T(y) = c1*y + c2

We can solve for c1 and c2 by applying the given boundary conditions at 0 and h:

c2 = T0 and c1 = (Th - T0)/h

So we now know the expression that shows us the temperature distribution in the fluid film:

T(y) = (Th - T0)/h*y + T0

Not so bad, huh?  Next week we'll begin anew with a fresh subject, but until then, happy studying!

MrsDrPoe: The First Law of Thermodynamics, Part 5

Happy Thesis Tuesday to you all!  Today we'll be continuing our discussion of the First Law of Thermodynamics by examining its final form- the differential form.  For an incompressible, Newtonian fluid in Cartesian coordinates, this is:

rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz)) + mu*Phiv

where

qx = d/dx(k*T), etc.

Phiv = 2*((d/dx(u))*(d/dx(u)) + (d/dy(v))*(d/dy(v)) + (d/dz(w))*(d/dz(w))) + (d/dx(v) + d/dy(u))*(d/dx(v) + d/dy(u)) + (d/dy(w) + d/dz(v))*(d/dy(w) + d/dz(v)) + (d/dx(w) + d/dz(u))*(d/dx(w) + d/dz(u)) - (2/3)*(d/dx(u) + d/dy(v) + d/dz(w))*(d/dx(u) + d/dy(v) + d/dz(w))

Typically, Phiv is neglected.  It is only important for flows with large velocity gradients or very large viscosities.  Furthermore, it should be noted that if the fluid has a constant thermal conductivity, the first term on the right-hand side of the equation becomes:

(d/dx(qx) + d/dy(qy) + d/dz(qz)) = k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))

So the final equation that we will be dealing with is:

 

rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))

In these equations, T is the temperature of the fluid and q is the heat flux.  Next week we'll conclude our look at the first law with an example employing this differential equation. 

 

MrsDrPoe: The First Law of Thermodynamics, Part IV

Happy Thesis Tuesday!  Today we'll continue our discussion of the head form of the First Law of Thermodynamics with an example problem:

Water is moved from one large reservoir to another at a higher elevation.  The loss of available energy associated with 2.5 ft*ft*ft/s being pumped from sections (1) to (2) is 30.5 |V|*|V|/g, where |V| is the average velocity of water in the 8 in inside diameter piping involved.  Determine the amount of shaft power required to pump the water.

Given: Q = 2.5 ft*ft*ft/s, losses = 30.5 *|V|*|V|/g, zdelta = 50 ft, D = 8 in

Find: shaft power required

Assumptions: steady, laminar, 1D, incompressible, constant average velocity, and constant properties

Solution:

For this case, we do not need to consider the continuity equation if we set our problem up correctly.  So we will start with the head form of the energy equation:

(pout/gamma) + 0.5*|V|*|V|/g + zout = (pin/gamma) + 0.5*|V|*|V|/g + zin + hP - hL

We will chose "in" as a point at the surface of the lower tank and "out" as a point at the surface of the upper tank, which tells us:

pin = 0, pout = 0 since both surfaces are exposed to the atm

vin = 0, vout = 0 since we can assume that the water on the surfaces is still

zin = 0, zout = zdelta setting z = 0 at the surface of the bottom tank

In putting these values into the equation, we have:

zout = hP - hL where hL = losses

Since our equation for the losses is given as a function of the average velocity, we know that these losses are primarily due to the friction in the pipe.  We can determine this velocity since we know both the flow rate and the pipe diameter.  NOTE: Q = |V|*A is only valid for use with uniform velocity or average velocity.  We know that since there are friction losses, mu is not zero, our velocity is not uniform, so in order to use this equation, we must use average velocity, which we are given.

|V| = 4*Q/(pi*D*D) = 7.162 ft/s

hL = 30.5*|V|*|V|/g = 48.625 ft

Solving for hP:

hP = zout + hL = 98.625 ft 

(since this answer is positive, work is introduced INTO the flow)

Now we can determine the necessary power from the pump head:

Wreq = hP*gamma*Q = 20,850 W = 27.958 hp

And that's the head form of the equation.  Next week, we'll look at the differential form of the First Law!

MrsDrPoe: The First Law of Thermodynamics, Part 3

Once again Thesis Tuesday is upon us, and today we'll continue our discussion of the First Law of Thermodynamics.

If we examine the integral form of the first law that we derived at the beginning of our discussion and make the assumptions of steady, laminar, and incompressible flow with uniform properties, pressure and constant average velocity at our inlets and outlets, our equation transforms as follows:

rho*int((u + (p/rho) + 0.5*|V|*|V| + g*z)*|V|)dAout - rho*int((u + (p/rho) + 0.5*|V|*|V| + g*z)*|V|)dAin = Qdot,net_in + Wdot,shaft_in

rho*(u + (p/rho) + 0.5*|V|*|V| + g*z)*|V|int(1)dAout - rho*(u + (p/rho) + 0.5*|V|*|V| + g*z)*|V|int(1)dAin = Qdot,net_in + Wdot,shaft_in

rho*(u + (p/rho) + 0.5*|V|*|V| + g*z)*|V|*Aout - rho*(u + (p/rho) + 0.5*|V|*|V| + g*z)*|V|*Ain = Qdot,net_in + Wdot,shaft_in 

mdot*(u + (p/rho) + 0.5*|V|*|V| + g*z)out - mdot*(u + (p/rho) + 0.5*|V|*|V| + g*z)in = Qdot,net_in + Wdot,shaft_in  

Furthermore, if we consider this equation per unit mass:

((p/rho) + 0.5*|V|*|V| + g*z)out - ((p/rho) + 0.5*|V|*|V| + g*z)in = qdot,net_in + wdot,shaft_in

Since the flow is steady, we can combine uout - uin - qnet,in into a losses term so that:

((p/rho) + 0.5*|V|*|V| + g*z)out = ((p/rho) + 0.5*|V|*|V| + g*z)in + wdot,shaft_in - losses

This form of the equation is known as the mechanical energy equation or the extended Bernoulli equation.  Each of the terms in this equation are of the form energy per unit mass.


If we divide the mechanical energy equation by the gravitational constant, g, we obtain:

((p/gamma) + 0.5*|V|*|V|/g + z)out = ((p/gamma) + 0.5*|V|*|V|/g + z)in + wdot,shaft_in/g - losses/g

The term wdot,shaft_in/g has the dimensions of energy per unit weight, which simplifies to the unit of height or length and can thus be expressed as hs (shaft head).  This term becomes hT (turbine head) if a turbine is present in the system or hP (pump head) if a pump is present in the system.  The term losses/g has the same units and can be written as hL.  So our final equation is:

((p/gamma) + 0.5*|V|*|V|/g + z)out = ((p/gamma) + 0.5*|V|*|V|/g + z)in + hs - hL

which is the head form of the energy equation.  Each term has units of head (length).  This particular form is used extensively in pipe flow applications, which we'll look at briefly next week.  Until then, happy studying!

MrsDrPoe: The First Law of Thermodynamics, Part II

Happy Thesis Tuesday to you!  Last week, we started talking about the integral form of the first law of thermodynamics.  Today we'll look at a simple example of how this equation is applied.

Problem: Suppose water is flowing through a 0.4 m*m pipe (1) with a velocity of 25 m/s and a static pressure of 940 kPa.  The pipe splits into two branches, one that is 0.18 m*m (3) and one that is 0.28 m*m (2).  At section (3), the static pressure is measured to be 570 kPa, and the velocity is 30 m/s; at section (2), the static pressure is 1140 kPa, and the velocity is unknown.  Determine the amount of available power lost in this horizontal y-connection.

Given: we know the areas and pressures at each location, the velocity at two locations, and the fluid properties.
Find: amount of available power lost
Assumptions: steady, incompressible, uniform properties and pressure across the pipe cross-section, horizontal pipe, no pumps/turbines, constant average velocity

Solution: 

After making our assumptions and drawing our control volume, we must apply the continuity equation:

d/dt(int(rho)dV) + int(rho*(V.n))dA = 0

For steady, incompressible flow with three surfaces where flow is crossing the CS:

rho*int(V1.n1)dA1 + rho*int(V2.n2)dA2 + rho*int(V3.n3)dA3 = 0

Since (1) is an inlet, (2) and (3) are outlets, and the velocities are constant averages, and dividing out density:

-|V1|*A1 + |V2|*A2 + |V3|*A3 = 0 (I)

|V2| = (|V1|*A1 - |V3|*A3)/A2 = 16.429 m/s

Now that we have determined the velocity at (2), we can apply the first law:

d/dt(int(e*rho)dV) + int((u + (p/rho) + 0.5*|V|*|V| + g*z)*rho*(V.n))dA = Qdot,net_in + Wdot,shaft_in

For steady, incompressible flow with no pumps or turbines, and three surfaces where flow crosses the CS:

int((u1 + (p1/rho) + 0.5*|V1|*|V1| + g*z1)*rho*(V1.n1))dA1 + int((u2 + (p2/rho) + 0.5*|V2|*|V2| + g*z2)*rho*(V2.n2))dA2 + int((u3 + (p3/rho) + 0.5*|V3|*|V3| + g*z3)*rho*(V3.n3))dA3= Qdot,net_in

Since (1) is an inlet, (2) and (3) are exits, and we have uniform properties and constant average velocities:

-rho*(u1 + (p1/rho) + 0.5*|V1|*|V1| + g*z1)*|V1|*A1 + rho*(u2 + (p2/rho) + 0.5*|V2|*|V2| + g*z2)*|V2|*A2 + rho*(u3 + (p3/rho) + 0.5*|V3|*|V3| + g*z3)*|V3|*A3 = Qdot,net_in

Consider the terms:

-rho*g*z1*|V1|*A1 + rho*g*z2*|V2|*A2 + rho*g*z3*|V3|*A3

Since the y is horizontal, there is no difference in z1, z2 or z3:

rho*g*z*(|V2|*A2 + |V3|*A3 - |V1|*A1) (II)

The terms in the parentheses above is the same as those in expression (I), which we know is equal to zero; therefore, (II) is equal to zero.

Our losses for this problem are defined by:

-Qdot,net_in - rho*u1*|V1|*A1 + rho*u2*|V2|*A2 + rho*u3*|V3|*A3 

So we end up with:

losses = rho*((p1/rho) + 0.5*|V1|*|V1|)*|V1|*A1 -rho*((p2/rho) + 0.5*|V2|*|V2|)*|V2|*A2 -rho*((p3/rho) + 0.5*|V3|*|V3|)*|V3|*A3

losses = 1.152 x 10^6 W 

And there ya go.  Next week we'll look at the head form of the first law, but until then, happy studying!

MrsDrPoe: The First Law of Thermodynamics, Part I

It's another Thesis Tuesday here at the blog, and today we begin discussion of the third governing equation of fluid mechanics- the first law of thermodynamics.  (In case you've forgotten, the other governing equations are the conservation of mass equation and the conservation of momentum equation.)

The first law of thermodynamics (also called the first law and the conservation of energy equation) simply tracks the energy in a system- just like mass and momentum, energy is conserved and every bit of it in every system/process can be accounted for at any instant in time.  In mathematical speak, the first law states that, "the time rate of increase of the total stored energy of the system is equal to the sum of the net time rate of energy addition by heat transfer into the system, and the net time rate of energy addition by work transfer into the system" or:

D/Dt(int(E*rho)dVsys) = (sum(Qdot)in - sum(Qdot)out)sys + (sum(Wdot)in - sum(Wdot)out)sys

This form of the equation is useful for Lagrangian viewpoints; recall that in order to transform it into a form useful for Eulerian viewpoints, we must employ the Reynolds Transport Theorem to the term on the left side:

D/Dt(int(E*rho)dVsys) = d/dt(int(e*rho)dV) + int(e*rho*(V.n))dA

meaning, "the time rate of increase of the total stored energy of the system is equal to the sum of the time rate of increase of the total stored energy of the contents of the control volume and the net rate of flow of the total stored energy out of the control volume across the control surface.  Our first law equation then becomes:

d/dt(int(e*rho)dV) + int(e*rho*(V.n))dA = Qdot,net in + Wdot, net in

In general, we will denote both work and heat coming into a system as positive and going out of a system as negative.

In the equations above, e is the sum of the potential, kinetic, and internal energies in the fluid (e = u + (|V|*|V|)/2 + g*z).  The heat transfer rate, Qdot, denotes all the ways energy is exchanged between the control volume and its surroundings due to a temperature difference.  You may recall from other classes that making the adiabatic assumption means that there is no heat transfer to or from the system.  This net rate of heat transfer is also zero if the heat transfer into the control volume is equal to the heat transfer out of the control volume.  

The work rate (or power), Wdot, denotes the work done on the control volume by the surroundings (when positive).  This work can be done by a shaft at the control surface, such as in a piston/cylinder arrangement, which calculated by multiplying the torque of the shaft by the shaft's angular velocity.  The work also occurs due to the normal stresses present on the surfaces of the control volume due to pressure and tangential stresses due to shearing.  Typically, the control volume is chosen in such a way that the tangential stress work is zero; the normal stress work is calculated by:

Wdot, norm = int(-p*(V.n))dA

If we put our expanded work terms into the energy equation derived earlier, we are left with:

d/dt(int((u + (|V|*|V|)/2 + g*z)*rho)dV) + int((u + (|V|*|V|)/2 + g*z)*rho*(V.n))dA = Qdot,net in + Wdot, net in - int(p*(V.n))dA

or:

d/dt(int((u + (|V|*|V|)/2 + g*z)*rho)dV) + int((u + (|V|*|V|)/2 + g*z)*rho*(V.n))dA = Qdot,net in + Wdot, shaft in

And that's the integral form of the first law.  Next week we'll look at a brief example of its use, but until then, happy studying!

MrsDrPoe: The Second Law of Thermodynamics as It Applies to Christianity

This Theology Thursday, I invite you to open up your Bibles with me as we take a look at a portion of God's word.  Today we'll be taking a bit of a different approach to some passages, examining how the second law of thermodynamics applies to Christianity.

The second law essentially states that the total amount of entropy in the universe is always increasing or every existing process is irreversible.  This wording may not mean much to you if you're not in some type of science or engineering field, but it definitely effects you.  This law in layman's terms tells us that no process is 100% efficient i.e. you put in more energy than you get out in useful energy or work.  

For instance: A tank of gasoline has x amount of chemical energy.  Engines use the gas to power your vehicle, but they only 'see' y amount of the energy available- the rest (x-y = z) is 'lost' to friction, heat dissipation, etc.  The fact that there is a lot of chemical energy that does not go directly into running your car means that the process is much less than 100% efficient.  

While we know that energy can never be created or destroyed (the first law), we should now also know that not all of the energy we put into something produces useful results.  They key in much of engineering is to design processes which eliminate as much of this 'wasted energy' as we can, ensuring that these processes use energy as efficiently as possible.


So what in the WORLD does this have to do with Christianity?  Look at these verses from the new testament:

"And do not be drunk with wine, in which is dissipation; but be filled with the Spirit" (Eph 5:18)

"For we have spent enough of our past lifetime in doing the will of the Gentiles- when we walked in lewdness, lusts, drunkenness, revelries, drinking parties, and abominable idolatries.  In regard to these, they think it strange that you do not run with them in the same flood of dissipation, speaking evil of you." (1 Pet 4:3-4)

The word dissipation in these verses is actually the Greek word asotia, which means prodigality.  In Luke 15, we read of Jesus' parable of the prodigal son, so most of us are probably familiar with the word 'prodigal'...but do we truly know what it means?  Prodigal is defined most frequently as 'excessive wastefulness.'  Thus in the above passages, these sinful acts are called wasteful, which means that in committing them we are being tremendously inefficient with the energy given to us by God.

But if we say to ourselves, "well this doesn't apply to me, because I don't do any of these things," we need to realize that "...all have sinned and fall short of the glory of God..." (Rom 3:23).  Like the car engine, much of the energy and things we have are wasted in that they do not serve a useful purpose to the Lord by glorifying His name.  While we can never be perfect as Christ was, His model of life on this earth should be what we're striving to match every day- we should all be 'engineers' trying through the grace of God to make our lives as 'efficient' as possible by ridding them of sin.