Tuesday, October 11, 2011

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


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!


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