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!
