No Downloads

Total Views

363

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

0

Comments

0

Likes

1

No embeds

No notes for slide

- 1. LECTURE UNIT NO. 7Fluid in Motion: The Energy EquationDevices that transfer energy in the fluid 1. Pumps – mechanically adds energy to the fluid 2. Hydraulic motors or turbines – mechanically remove energy from the fluid 3. Pipes, fittings, valves, filters and strainers – used to control the distribution, flow rate, pressure, and cleanliness of the fluid. These components inadvertently offer frictional resistance to flow and thus transfer some of the fluid’s pressure and kinetic energy into thermal energy (heat).Comments on energy transfer devices and frictional losses 1. Pumps – add pressure energy or kinetic energy to the fluid to shaft work. 2. Hydraulic motors or turbines – remove pressure energy or kinetic energy from the fluid via shaft work. 3. Pipes, fittings, valves, filters and strainers – transform elevation energy, pressure energy, and kinetic energy from one form to another and transform some of these energies into heat.THE ENERGY EQUATIONFor the Pump t s = td Qs = QdAssumptions: - Perfectly insulated - Isentropic Process or Reversible Adiabatic Process - Constant Volume Pumping Process
- 2. Water or Hydraulic PowerMass Density of Water 1. If tW = tS = td is given, use steam tables 2. If tW = tS = td is not given Use: Standard mass density of water ρw = 62.4 lb/ft3 = 1,000 kg/m3 = 8.33 lb/galTotal Dynamic HeadSource: Suction Lift TDH = potential head + velocity head + pressure headEnergy Equation From Law of Conservation of EnergySource: Suction Lift [Ein = Eout] - PEs + KEs + Us + Wfs + WP = PEd + KEd + Ud + Wfd + (hfs + hp + hfd) Us, Ud = 0, since Internal energy is a function of temperature and knowing that t W = ts = td . . . But: Vs = Vd = VW
- 3. Mass of WaterMass density of any working substance other than waterPump Specific Speed RPM required by the impellerEng’g Units:SI Units:Pump Affinity Laws
- 4. 1. Rate of discharge or rate of flow or volumetric flow rate Q = N D3 ρ a 2. Total Dynamic head H = N2 D2 ρ b 3. Water or hydraulic power P = N3 D5 ρ c Where: N = rotative speed D = impeller diameter ρ = mass density a, b ,c = constants A. Same Pumps (D1 = D2) 1st Condition 2nd ConditionPump Efficiency - Pump Efficiency - Motor Efficiency a.) Solve for the new rate of discharge
- 5. b.) Solving for new total dynamic headc.) Solving for new power inputB. Similar Pumps (N1 = N2) - but with different impeller diameter 1st Condition 2nd Conditiona.) Solving for the rate of dischargeb.) Solving for the new total dynamic head
- 6. c.) Solving for the new power inputSingle Impeller – Single Suction Centrifugal PumpSingle Impeller – Double Suction Centrifugal Pump2 Pumps in Parallel
- 7. Multi-Stage Centrifugal PumpPumps in SeriesProblems:
- 8. 1. Water enters a pump through a 250 mm diameter pipe at 35 kPa. It leaves the pump at 140 kPa through a 150 mm diameter pipe. If the flow rate is 150 liters / sec. Compute: a. the velocity of discharge pipe b. the energy added by the pump c. the horsepower delivered to the water by the pump. Assume suction and discharge sides of pump are at the same elevation 2. Water discharged through a nozzle having a diameter of the jet of 100 mm at a velocity of 60 m/s at a point 240 m below the reservoir. Compute a. the total head loss b. the horsepower produced by the jet c. the power lost in friction 3. A d-c motor driven pump running at 100 rpm delivers 30 liters per sec. of water at 40 °C against a total pumping head of 27 meters with a pump efficiency of 60%. Barometric pressure is 758 mm Hg abs. a. What speed and capacity would result if the pump rpm were increased to produce a pumping head of 36 meters assuming no change in efficiency? b. Can 15 kW motor be used under conditions indicated by (a) Local gravitational acceleration is 9.72 m/s2. 4. Water from a reservoir is pumped over a hill through a pipe 450 mm in diameter and pressure of 1.0 kg/cm2 is maintained at the summit. Water discharge is 30 m above the reservoir. The quantity pumped is 0.5 m3/s. Frictional losses in the discharge and suction pipe and pump is equivalent to 1.5 meters head loss. The speed of the pump is 800 rpm. a. What amount of energy must be furnished in the pump in kW? b. If the speed of the pump was increased to 1000 rpm, what are the values of discharge, head and power? 5. Water from a reservoir is pumped over a hill through a pipe 900 mm in diameter and a pressure of 1 kg/cm2 is maintained at the pipe discharge where the pipe is 85 meters from the pump center line. The pump have a positive suction head of 5 meters. Pumping rate of the pump at 1000 rpm is 1.5 m3/s. Friction losses is equivalent to 3 meters of head loss. a. What amount of energy must be furnished by the pump in kW? b. If the speed of the pump is 1200 rpm, what are the new values of the discharge, head and power?TurbinesTurbine is a rotary engine that converts the energy of a moving stream of water, steam, or gas intomechanical energy.The basic element in a turbine is a wheel or rotor with paddles, propellers, blades, or bucketsarranged on its circumference in such a fashion that the moving fluid exerts a tangential force that turnsthe wheel and imparts energy to it.Types: 1. Impulse (Pelton) Turbine Application: High head Specific speed range: 2.5 to 7 rpm Suitable for: 92 m – 1525 m Efficiency: 82 – 92% 2. Reaction (Francis) Turbine Application: Medium head Specific speed range: 20 to 100 rpm Suitable for: 12 m – 305 m Efficiency: 90 – 94%
- 9. 3. Propeller (Kaplan) Turbine Application: Low head Specific speed range: 80 to 180 rpm Suitable for: 3 m – 46 m Efficiency: up to 93%Design of Turbine wheel or Runner D n v = Ø 2gHE dj = jet diameter 1. Diameter of Turbine wheel 2. Turbine selectionExample: It is desired to develop a 3000 kW impulse turbine under an effective head of 280 m. The turbine efficiency is87% velocity coefficient of the nozzle is 98%. What should be the diameter of the wheel and the probable speed ofrotation in rpm.

Be the first to comment