1. A fire hose nozzle has an orifice diameter of 1inch. According to some fire c

Question

1. A fire hose nozzle has an orifice diameter of 1inch. According to some fire codes, the nozzle must be capable of delivering at least 300gal/min If the nozzle is attached to a 3-inch diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this flow rate?

2.  Kerosene having a specific weight of 50.0 lb/ft3 is flowing at 10 gallons per minute from a standard 1-inch Schedule 40 steel pipe to a standard 2-inch Schedule 40 steel pipe. Calculate the difference in pressure in the two pipes.

3. A 4-inch diameter pipe carries 300 gal/min of water at a pressure of 60 psi. Determine:

a) The pressure head in feet of water

b) The velocity head

c) The total head with reference to a datum plane 20 ft below the pipe.

4. 8. Oil with a specific gravity of 0.90 is flowing downward through the venturi meter. If the pressure difference on the gauges is 2 psi , the diameters or 2" and 1", and the ele-vation difference from station 1 - 2 is 1 foot, calculate the volume flow rate of oil.

Explanation / Answer

1.

300 gal / min = 40.1 ft^3 / min = 0.6684 ft^3 /s

V1 = Q / A1 = 0.6684 / (3.14/4*(3/12)^2) = 13.62 ft/s

V2 = Q / A2 = 0.6684 / (3.14/4*(1/12)^2) = 122.61 ft/s

P1 + 1/2*rho*V1^2 = P2 + 1/2*rho*V2^2...............ignoring change in height

P1 + 1/2*62.4*13.62^2 = 0 + 1/2*62.4*122.61^2

P1 = 463255.8 lb/(ft-s^2) = 463255.8 / 32.2 lb/ft^2 = 14386.8 lb/ft^2 = 14386.8/12^2 psi = 99.91 psi

2.

For 1" sch 40 pipe, we have d1 = 1.049 "

For 2" sch 40 pipe, we have d2 = 2.067"

10 gal/min = 1.337 ft^3 / min = 0.02228 ft3/s

V1 = Q / A1 = 0.02228 / (3.14/4*(1.049/12)^2) = 3.714 ft/s

V2 = Q / A2 = 0.02228 / (3.14/4*(2.067/12)^2) = 0.9566 ft/s

P1 + 1/2*rho*V1^2 = P2 + 1/2*rho*V2^2...............ignoring change in height

P2 - P1 = -1/2*(50/32.2)*(0.9566^2 - 3.714^2)

P2 - P1 = 10 lb/ft^2

3.

300 gal/min = 0.6684 ft3/s

V = Q / A = 0.6684 / (3.14/4*(4/12)^2) = 7.663 ft/s

a) Pressure head = P / rho = (60*12^2) / 62.4 = 138.46 ft

b) Velocity head = V^2 / (2g) = 7.663^2 / (2*32.2) = 0.9118 ft

c) Total head = 138.46 + 0.9118 + 20 = 159.37 ft

4.

P1 + 1/2*rho*V1^2 + rho*g*z1 = P2 + 1/2*rho*V2^2 + rho*g*z2

P1 - P2 = 2 psi = 2*12^2 lb/ft^2 = 288 lb / ft^2

z1 - z2 = 1 ft

288 + (0.9*62.4*1) = 1/2*(0.9*62.4/32.2)*(V2^2 - V1^2)

V2^2 - V1^2 = 394.66 ft^2 / s^2....................1

Also, A1*V1 = A2*V2

3.14/4*2^2 *V 1 = 3.14/4*1^2 *V2

V2 = 4*V1.....................2

Solving 1 and 2, we get V1 = 5.13 ft/s

Vol. flow rate = A1*V1 = 3.14/4*(2/12)^2 *5.13 = 0.111 ft^3/s = 50.19 gpm

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.