Problem 1 Water enters the rotating sprinkler along the axis of rotation and leaves through three nozzles. How large is the resisting torque required to hold the rotor stationary for the angle that produces the largest torque? How fast will the rotor spin steadily if the resisting torque is reduced to zero and a) θ = b) θ = 3 c) θ = 6 d) θ = 9 Homework 6 n := 3 d jet := 3cm Q:= 5 L s θ r:= 5cm Solution 1 Q Q jet := ρ := 1 kg n m 3 πd jet A jet := 4 A jet = 7.69 cm V jet := Q jet A jet V jet = 3.579 m s ( ) := V jet V t θ, ( ) := ρ Q T θ, ( ) := ρ Q T θ, ( ) + r sin θ ( ) r V t θ, T( 9deg, ) 589 Nm r ( V jet sin( + r) = for the stationary case Now calculate the speed of rotation for the zero torque cases Solve for the speed of rotation ( T, (, 3.6 4.8 47 := = 1 s T + ( ρ Q r) Vjet ρ Q r r sin( (, 5. 39 = rev min θ := deg, 3deg.. 9deg rev := πrad
47. 45.3 θ := deg, 1deg.. 9deg 6 4 (, 1 3 4 5 6 7 8 9 θ deg T( 9deg, ) 589 Nm = The torque on the water is in the clockwise direction Problem Create a graph of torque (N m) and of power (Watts) that could be produced by the rotating sprinkler as a func of rpm assuming that θ = 9º and neglecting frictional effects. Use a range of rpm such that the entire range of is shown. Solution θ := 9deg The maximum rotational speed occurs when the torque is zero T:= ( T, T θ, ( ) Vjet ( ) T + ρ Q r sin θ := max := ( T, max = 47.157 1 ρ Q r r s ( V jet ) ( ) := ρ Q r sin( + r Change signs to get positive value since all we care about is the torque that will be used to turn a generator. We don't care about the direction of the torque. ( ) := ρ P θ, := 1 s Q r ( V jet sin( + r) max,.. 1 max
6 8 Nm 4 6 watts T( θ, ) 4 P( θ, ) 1 3 4 5 rev min Problem 3 Derive an equation for the angular velocity (r) at the maximum power production. (set dp/d=) What is the relationship between the jet velocity (relative to the nozzle) and the nozzle tip velocity (relative to stationary observer) at the maximum power output? Solution 3 P = ρ Q r ( V jet sin( + r) d d P = ρ Q r V jet sin( ρ Q r = ρ Q r V jet sin( ρ Q r maxpower := V jet sin( r max = 47.157 1 s maxpower = 3.579 1 s Jet velocity is V jet = 3.579 m s Nozzle tip velocity V tip := maxpower r V tip = 11.789 m s V tip V jet = sin( The nozzle tip velocity is 1/ of the jet velocity. Problem 4 Calculate the elevation head required to obtain the jet velocity (i.e. what elevation would a reservoir connected
jet have to have?). You may ignore frictional losses. Solution 4 Since we are ignoring frictional losses we can use the Bernoulli equation between the reservoir surface and th in the jet. In order to use the Bernoulli equation we can't have shaft work and thus the nozzle must be stationa the analysis. h V jet := where h is the difference in elevation between the reservoir and the jet g h = 8.345 m Reservoir elevation to get this water velocity Problem 5 Calculate the power required to lift water from the sprinkler up to the elevation calculated above and compare w maximum power generated by the rotating sprinkler. We assumed no frictional losses. So why does the sprinkle generate half of the power needed to return the water back to the reservoir feeding the sprinkler? Solution 5 P := ρ g Q h Power required to pump water up to this reservoir P = 13.9 kw Now let's calculate the maximum power produced by this simple turbine θ := 9deg maxpower = 3.579 1 s T := ρ Q r ( V jet sin( + maxpower r) T = 94.731 J P max := maxpower ( T) P max = 6.9 kw The maximum power is exactly half of the power required to pump the water back up to the reservoir. So this turbine is inefficient. The reason it is inefficient is because the water leaving the nozzle still has lots of kinetic energy. P = 13.9 kw required to return water to reservoir
Problem 6 A siphon is used to empty a tank of water. The mechanical energy losses from entering the siphon (minor losses) and from flowing through the siphon tube are () where V is the velocity of the water in the siphon. The siphon diameter, d is 1 cm. A) Calculate the siphon velocity (V) B) If the mechanical energy loss to the top of the siphon is 5% of the total head loss, what is the maximum value that h 1 can have such that the water doesn't boil in the siphon tube? Solution 6 p vapor := 34Pa h L h := 1cm h 1 = 1. V h g 3 := 5m d := 5cm cs 1 is on the water surface of the tank. cs is at the end of the pipe kpa := 1Pa α 1 V 1 g p 1 V + + z 1 + H p = α g p + + z + H t + h L V 1 := m s H t := m α 1 := 1 p 1 := kpa z 1 := H p := m p := kpa α := 1 z = h + h 3 ( ) V = 1 g V + ( h + h 3) + 1. g g h + h 3 V :=. ( ) A V = 6.7 m s B cs 1 is on the water surface of the tank. cs is at the high point of the pipe p peak of pipe r reservoir surface V r := m α s p := 1 z r := m z p = h 1 V r p r V p p p α r + + z g r + H p = α p + + z g p + H t + h L := 986 N m 3 p r V p = α p g p p V p + + h 1 +.6 g
Cavitation will occur when the absolute pressure in the pipe reaches the vapor pressure. We can either solve this problem by using absolute pressure as our pressure datum or by calculating the vapor pressure using atmospheric pressure as the datum. I will solve the problem using absolute pressure as the pressur datum. I will assume that the local atmospheric pressure is standard atmospheric pressure. p r := 11.3kPa p p := p vapor V p := V p r p p h 1 := V p 1.6 g B h 1 = 6.4 m Problem 7 How much water would be required to generate enough power for an average American? The efficiencies for the various processes are given below. The hydraulic efficiency is the ratio of userful hydraulic power delivered to the turbine to the total hydraulic power available. Note that this energy demand includes transportation. Reservoir Penstock Powerhouse G := 1 k := 1 River z := 5m e hydraulic :=.95 e generator :=.95 e turbine :=.85 The total energy consumption for an American is Solution 7 P electric := 35 GJ or P yr electric = 11 k W Draw control surfaces on the surface of the reservoir and the surface of the river. p 1 ρ g V 1 + + z g 1 = p ρ g V + + z g + h L + H t z 1 = z + h L + H t e hydraulic = H t z H t = e hydraulic z P electric = P water e turbine e generator
P water = ρ g H t Q ( ) P electric = ρ g e hydraulic z Q ( ) e turbine e generator Q := ρ g e hydraulic z P electric ( ) ( e turbine e generator) Q = 9.5 L s Problem 8 A pump moves oil at a refinery. Upstream of the pump where the pipe diameter is 9 mm, the pressure is 1 kpa. Downstream of the pump where the pipe diameter is 3 mm, the pressure is 4 kpa. Neglecting head loss, how much head does the pump add? ρ oil := 8 kg m 3 kpa := 1Pa p := 4kPa d := 3mm d 1 := 9mm p 1 := 1kPa z := 5m Solution 8 Q:= L s π d 1 Q A 1 := V 4 1 := A 1 V 1 =.314 m s π d Q A := V 4 := A V =.89 m s p 1 ρ oil g V 1 + + z g 1 + H p p ρ oil g V + + z g + h L = control surfaces at the pressure sensors. p p 1 H p := ρ oil g V V 1 + + z g H p = 41.1 m
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