The Energy Equation for Control Volumes

Simplifications of the Energy Equation

Example Problem:

Given: An air compressor

Find: , how much heat is transferred into the compressor (in BTU/hr)? Note: We expect to be negative, since heat is given off by the compressor)]

Solution: Draw a C.V. around the entire thing (not shown on the sketch), with the shaft protruding out from the C.S.

Kinetic Energy Flux Correction Factor

As was done with the momentum equation, a correction factor for the velocity term (kinetic energy term) in the energy equation must be introduced to account for non-uniform inlets and exits. Consider again an actual versus equivalent one-dimensional outlet, as sketched:

"Head" Form of the Energy Equation

Efficiencies of Pumps and Turbines

Example Problem


A pump is used to draw water from a large reservoir as sketched. The water shooting out the nozzle at the pump outlet has uniform velocity V2, volume flow rate Q, and inner exit diameter D2.
Insert Image pumpabove.gif
Also known in the problem are pipe diameter D, pump efficiency , hlosses, and the elevations z0, z1, and z2. Note that hlosses represents irreversible head losses due to friction in the pipe and due to turbulent mixing and other irreversible losses in the pipe elbow and the inlet region of the pipe. At this point in our study, we do not know how to calculate these, but later on we will learn how to estimate these head losses.


(a) WHP (water horsepower) actually delivered to the water by the pump.

(b) BHP (brake horsepower) required to run the pump.


Example Problem

Given: A small water pipe flow rig:

Suppose we know that hlosses = 0.4 m = total head loss due to valve losses, pipe friction, elbow losses, etc. (Later on we will learn how to estimate these.)

Find: V2av (average velocity at pipe exit)