PHYSICS! Estimate the air pressure at the center of a category 5 hurricane with

Question

PHYSICS!

Estimate the air pressure at the center of a category 5 hurricane with a wind speed of 300 km/hr. Water at a pressure of 3.80 atmospheres (overpressure or gauge pressure) at street level flows into an office building at a speed of 0.600 m/s through a 5.0 cm diameter pipe. The pipes taper down to 2.60 cm in diameter by the top floor, 20.0 m above. Calculate the flow velocity and the pressure in such a pipe on the top floor (in atmospheres of overpressure). The overpressue in a fire hose of diameter 6.40 cm is 3.50 times 105 Pa and the speed of flow is 4.00 m/s. The fire hose ends in metal tip of diameter 2.50 cm. What are the pressure and speed of the water in the tip. What is the speed and pressure of the water just OUTSIDE the tip of the fire hose ? (The pressure outside the hose is zero overpressure, i.e. 1.00 atmosphere absolute). To fight a fire on the fourth floor of a building, firemen want to use a hose of diameter 6.35 cm to shoot 950 liters/minute to a height of 12.0 m. With what minimum speed must the water leave the nozzle of the fire hose if it is to ascend 12.0 m? What pressure must the water have inside the fire hose? Poiseuille's Equation section: Q = piR4(DeltaP)/8etaL, where R is the radius of a cylindrical tube in mcters, DeltaP is the difference in pressure between the two ends in Pascals, eta is the viscosity in Pascal-seconds, L is the length of the tube in meters, and Q is the flow rate in m3/s. Engine oil (viscosity 200 times 10-3 Pa-s) passes through a 1,80-mm-inside diameter tube in a prototype engine. The tube is 5.5 cm long. What pressure difference is needed to maintain a flow rate of 5.60 milliliters/second? Blood from an animal (density = 1050 kg/m3) is placed in a bottle 1.70 m above a 3.80 cm-long needle of inside diameter 0.400 mm from which it flows at a rate of 4.10 cm3/minute. What is the viscosity of the blood?

Explanation / Answer

1.

P1 + 1/2* rho*V1 ^2 = P2 + 1/2 *rho*v2 ^2

We have

P1 = P_atm = 101325 Pa

v1 = 0

v2 = 300 km / hr = 300 *1000 / (60*60) m / s = 83.33 m/s

rho = 1.2 kg / m^3 (air density)

Putting values, we get

P2 = 97158 Pa = 97.158 kPa

2.

A1*v1 = A2*v2

pi/4 *d1 ^2 *v1 = pi/4 *d2 ^2 *v2

d1 ^2 *v1 = d2 ^2 *v2

v2 = v1 *(d1 / d2)^2

v2 = 0.6*(5 / 2.6)^2

v2 = 2.219 m/s

P1 = 3.8 atm gage = 3.8 + 1 atm abs = 4.8 atm abs = 4.8*101325 Pa abs = 486360 Pa

P1 + 1/2*rho*v1 ^2 + rho*g*h1 = P2 + 1/2*rho*v2 ^2 + rho*g*h2

486360 + 1/2 *1000*0.6^2 + 0 = P2 + 1/2*1000*2.219^2 + 1000*9.81*20

P2 = 287878 Pa = 287878 / 101325 atm = 2.841 atm abs

P2 = 2.841 - 1 atm gage

P2 = 1.841 atm gage

3.

a)

A1*v1 = A2*v2

pi/4 *d1 ^2 *v1 = pi/4 *d2 ^2 *v2

d1 ^2 *v1 = d2 ^2 *v2

v2 = v1 *(d1 / d2)^2

v2 = 4*(6.4 / 2.5)^2

v2 = 26.2144 m/s

P1 = 3.5*10^5 Pa gage = (3.5*10^5) + 1.01325*10^5 abs = 4.51325*10^5 Pa abs

P1 + 1/2*rho*v1 ^2 + rho*g*h1 = P2 + 1/2*rho*v2 ^2 + rho*g*h2

4.51325*10^5 + 1/2*1000*4^2 + 0 = P2 + 1/2*1000*26.2144^2 + 0

P2 = 115727.6 Pa abs

P2 = 115727.6 - 101325 Pa gage

P2 = 14402 Pa gage

b)

Since water comes out as a jet, P will be 14402 Pa gage and v will be 26.2144 m/s.

4.

a)

Flowrate Q = 950 lt / min = 950*10^-3 m^3 / 60 sec = 0.015833 m^3 /s

Q = A1*v1 = A2*v2

v1 = Q / A1 = 0.015833 / (3.14 / 4 *(6.35*10^-2)^2) = 1.25 m/s

b)

P1 + 1/2*rho*v1 ^2 + rho*g*h1 = P2 + 1/2*rho*v2 ^2 + rho*g*h2

Since hose area is constant, v1 = v2. Also, P2 = 0 gage. Hence,

P1 + rho*g*h1 = rho*g*h2

P1 = rho*g*(h2 - h1)

P1 = 1000*9.81*(12 - 0)

P1 = 117720 Pa gage

5.

Q = 5.6 mL/s = 5.6*10^-3 L/s = 5.6*10^-6 m^3 /s

R = d/2 = 1.8 / 2 = 0.9 mm = 0.0009 m

L = 5.5 cm = 0.055 m

Putting in formula,

5.6*10^-6 = 3.14*0.0009^4 *(dP) / (8*200*10^-3 *0.055)

dP = 239205.4 Pa

6.

Flowrate Q = 4.1 cm^3 /min = 4.1*10^-6 / 60 m^3/s = 6.833*10^-8 m^3/s

dP = rho*g*h = 1050*9.81*1.7 = 17510.9 Pa

R = d/2 = 0.4 / = 0.2 mm = 0.0002 m

L = 3.8 cm = 0.038 m

Putting in formula

6.833*10^-8 = 3.14*0.0002^4 * 17510.9 / (8*eta*0.038)

eta = 0.00423 Pa-s

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