lesson 14 problem set

What is the laminar flow rate of water in a pipe with an internal radius of r and a length of L m if the pressure difference across the pipe is 2.0 kPa? The water in the pipe has a viscosity of approximately 0.6 mPa-s. (4 pts) Radius r = 5.0 cm Length L = 5.0 m c. Describe the effect on the flow rate given each of the following changes. Each change is to the original system and all other variables are held constant. (1 pt each) i. The radius decreases ii. The pipe length decreases iii. The fluid viscosity decreases iv. The pressure difference decreases

Q = ΔPπr4 / (8ηL) Q = (2.0E3 Pa)π(0.05 m)4 / (8(0.6E-3 Pa-s)(5.0 m)) Q = 0.039 / 0.024 Q = 1.64 m3/s c. i. Flow rate decreases. ii. Flow rate increases. iii. Flow rate increases. iv. Flow rate increases.

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from the bottom of the tank. The water flows out of the small hole into an empty, 0.2 m-diameter container placed below the tank. The student uses a garden hose to keep the water level in the tank at the original height. a. What is the velocity of the water as it leaves the hole near the bottom of the tank? (4 pts) Height h1 = 0.75 m Height h2 = .10 m b. What is the volume flow rate of water from the garden hose into the tank? (Hint: What is the flow rate out of the tank?) (4 pts) c) After 60 seconds of operation, what is the water pressure at the bottom of the container that is being used to catch the water leaving the tank? (6 pts)

a) Gh1 = 1/2v2^2 + gh2 (9.8N/kg)(0.75m) = ½(v^2) + (9.8N/kg)(0.10m) V = 3.6m/s b) Q = πr^2v Q = π(0.01m)^2(3.6m/s) Q = 0.001m3/s c)Q = V/t = Ah/t (1.13E-3 m3/s) = π(0.2 m)2h/(60 s) h = 0.54 m P = ρgh P = (1000 kg/m3)(9.8 N/kg)(0.54 m) P = 5292 Pa

A student turns on a 1.27 cm diameter faucet in a lab class and measures the flow rate of water at a specific setting to be Q. a. What is the velocity of the water as it exits the faucet? (3 pts) b. A student attaches a nozzle with an exit radius that is n times smaller than the faucet radius. i. How will this affect the flow rate? Be specific in your explanation. (3 pts) ii. How will this affect the velocity of the water? Be specific in your explanation. (3 pts)

a) Let Q = 7.0E-4 m3/s radius= 1.27 cm/2 = 0.635 cm= 0.00635 m Q = Av (7.0E-4 m3/s) = π(6.35E-3 m)2v v = 5.5 m/s b) i. The flow rate is constant. The fluid is firm, so the volume and mass entering the pipe length every second must equal the volume and mass leaving the pipe length. This process is known as conservation of mass. ii. The velocity is not constant. Instead, the velocity increases to adjust to the changing diameter to keep the flow rate constant.

A pipe has an initial radius of r1, which then changes to a radius of r2. The velocity of the water in the initial part of the pipe is v1. The segment of the pipe with the smaller radius is also 0.5 m higher than the initial pipe segment. What is the pressure difference between the two parts of the pipe? (6 pts) Radius r1 = 6.0 cm Radius r 2 = 3.0 cm Velocity v1 = 32 cm/s b) What is the volume flow rate in the pipe? (3 pts)

πr2v = πr2v π (0.06 m)2 (0.32 m/s)= π (0.03 m/s)2 v v= 1.28 m/s 1/2ρv12+ρgh1+P1=1/2ρv22+ρgh2+P2 (0.5) (1000 kg/m3) (0.32 m/s)2 + (1000 kg/m3) (9.8 N/kg) (0m)+P1 = (0.5) (1000 kg/m3) (1.28 m/s)2 + (1000 kg/m3) (9.8 N/kg) (0.5 m) + P2 51.2 kg/ms2+0 + P1= 819.2 kg/ms2 +4900 N/m2 + P2 P1-P2= 5668 Pa b) Q = Av Q = (pie)(0.060m)^2(.032m/s) Q = 3.61*10^-4m^3/s