Flow Process is meant for open system. It is one in which mass is entering the system. If the mass entering is equal to the mass leaving then the process is said to be the steady flow Process. If the mass entering is not equal to the mass leaving then it is said to be a non- steady flow process.
(i). Steady Flow Energy Equation(SFEE) :
Consider a Control Volume as shown in the Figure. Assume that a steady flow occurs across the control volume. Let “m” be the mass of the fliud entering and leaving the system. “Q” is the heat supplied and “W” be the work delivered 1-1 is the inlet section. 2-2 is the outlet section
Section 11
u1 is the internal energy/ unit mass
P1v1 is the flow energy /unit mass
gz1 is the potential energy /unit mass
V12 /2 is the kinetic energy/unit mass
Section 22
u2 is the internal energy/ unit mass
P2v2 Flow energy/unit mass
gz2- Potential energy/unit mass
V22 /2 is the kinetic energy/unit mass
Energy entering =Energy leaving
The equation is
m[u1 + P1v1/J + V12/2J + gZ1/J] + Q = m[u2 + P2v2/J + V22/2J + gZ2/J] + W/J
m[h1 + (V12/2J) + gZ1/J] + Q = m[h2 + (V22/2cJ) + gZ2/J] + W/J
m[(h2-h1) + ( V22-V12)/2J + g(Z2-Z1)/J] = Q - W/J --------(a)
(ii). Application of Steady Flow Energy Equation
Case 1: Let ∆KE = 0 ; ∆PE = 0; W = 0
From (a) we get
m(h2-h1) = Q
eg.: Boiler
Case 2: Let ∆KE = 0 ; ∆PE = 0; Q = 0
From (a) we get
m(h1-h2) = W/J e.g.: Turbine
Case 3: Let ∆PE = 0; Q = 0; W=0
From (a) we get
(h1-h2) = ( V22-V12)/2J
e.g.: Nozzle
Case 4: Let ∆KE = 0; ∆PE = 0; Q = 0; W=0
From (a) we get
(h1-h2) = 0
h1 = h2
e.g.: Throttle Valve
Throttling process: It is process in which enthalpy remains constant
consider a flow of fluid with steady mass of m through a duct as shown in fig.
net work is done as the fluid passes through the duct .Duct is well insulated so, that
there is no heat transfer. The high pressure fluid at pressure p1 pass through porous plug
and comes out at reduced pressure p2 V1= V2 as there is no heat transfer.
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