CHAPTER 7 ENERGY PRINCIPLE Dr . Ercan Kahya Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

General Energy Consideration Z: the position P/g: the pressure head V2/2g: the velocity head hp: head supplied by pump ht :head given up to a turbine hf : head loss • local losses (bends, expansions, valves) • frictional losses (function of pipe type, length) a : Velocity coefficient and can be set to unity for regular & symmetrical cross-section like pipe

Bernoulli vs. Energy Z is the position P/g is the pressure head V2/2g is the velocity head Relates velocity and piezometric pressure along a streamline, steady, incompressible, inviscid flow. Relates energy at two points for viscous, incompressible flow in a pipe, with accounting for additional energy addition / extraction

Energy Principle • So far, mechanical forces on a fluid – Pressure – Gravity – Shear Stress • Considering Energy, we can solve: – Power required to move fluids – Effects of pipe friction – Flow rates of fluids moving through pipes & orifices – Effects of obstacles, bends, and valves on flow

First Law of Thermodynamics • E = energy of a system • Q = heat transferred to a system in a given time t • W = work done by the system on its surroundings during the same time • Energy forms: Kinetic and Potential energy of a system as a whole and energy associated with motion of the molecules (atomic structure, chemical energy, electrical energy) E = Ek + Ep + Eu + heat transferred to the system + work done by the system - heat transferred from the system ‐ work done on the system Involves sign convention:

First Law of Thermodynamics

Derivation of Energy Equation Reynolds Transport Theorem applied to First Law of Thermodynamics E: extensive property of the system e: intensive (energy per unit mass)

Flow Work Work is classified as: (work) = (flow work) +(shaft work) Flow Work: Work done by pressure forces as the system moves through space Force (F) = p A Work = F Dl = (pA) (V Dt) At section 2, work rate done on surrounding fluid is → V2 p2 A2 At section 1, work rate done by surrounding fluid is → - V1 p1 A1

Shaft Work (any work not associated with a pressure force!) • Work done on flow by a pump – increases the energy of the system, thus the work is negative • Work done by flow on a turbine – decreases the energy of the system, thus the work is positive

Reynolds Transport Theorem: Simplified form If the flow crossing the control surface occurs through a number of inlet and outlet ports, and the velocity v is uniformly distributed (constant) across each port; then Steady-Flow Energy Equation . Q = rate of heat transfer TO the system (input) W = rate of work transfer FROM the system (output) m = rate of mass flow h = specific enthalpy (h = u + p/ρ)

Example 7.2: If the pipe is 20cm and the rate of flow 0.06m3/s, what is the pressure in the pipe at L=2000m? Assume hl=0.02(L/D)V2/2g This energy equation assumes steady flow & constant density

Let’s relate “head” to “power & efficiency” Power Equation Let’s relate “head” to “power & efficiency” Pump power: Power delivered to turbine: Both pump & turbine lose energy due to friction which is accounted for by the “efficiency” defined as the ratio of power output to power input. If mechanical efficiency of the turbine is ηt , the output power supplied by the turbine:

Example 7.4: Power produced by a turbine Discharge Q = 14.1 m3/s ; Elevation drop = 61 m Total head loss = 1.5 m ; Efficiency = 87% Power = ? Evaluations: V1 = V2 = 0 p1 = p2 = 0 z1-z2 = 61m ht = (z1-z2) - hL = 61 – 1.5 = 59.5 m Power equation: = (9810 N/m3) (14.1 m3/s) (59.5m) = 8.23 MW Efficiency equation: = 0.87(8.23 MW) = 7.16 MW

Application of the Energy, Momentum and Continuity Principles in Combination Neglecting the force due to shear stress Sudden expansion head loss

Example 7.5: Force on a contraction in a pipe Find horizontal force which is required to hold the contraction in place if P1=250kPa ; Q=0.707m3/s & head loss through the contraction Assume α1= α2 = 0 (kinetic energy correction factor)

SOLUTIONS: Fx and p2 : unknown To obtain unknown p2: Q , p1, V1 and V2 : known Fx and p2 : unknown To obtain unknown p2:

HYDRAULIC & ENERGY GRADE LINES

GRADE LINE INTERPRETATION - PUMP

GRADE LINE INTERPRETATION TURBINE

GRADE LINES - NOZZLE

GRADE LINES - PIPE DIAMETER CHANGE

GRADE LINES - SUB-ATMOSPHERIC PRESSURE

CLASS EXERCISE: Q7.32 Find the head loss btw the reservoir surface and point C. Assume that the head loss btw the reservoir surface and point B is three quater of the total head loss.

CLASS EXERCISE: Q7.36

CLASS EXERCISE: Q7.60

CLASS EXERCISE: Q7.71