Name A portion of a fire protection system in which a pump draws water at 60°F f

Question

Name A portion of a fire protection system in which a pump draws water at 60°F from a reservoir and delivers it to a point B at the flow rate of 1500 gaVmin. 1. Neglecting any losses, if the design height h is 20 ft, what would the pump power be needed to supply the flow? 2. Considering friction losses with the design height h again being 20 ft, what would the pump power be needed to supply the flow 3. Using the results from part 2, what is the outlet pressure at point B? 4. Using the results from part 2, what is the inlet pressure at point A? 5. If the height in the tank h, was allowed to drop to 10 ft, what would the outlet pressure be at point B? By what percentage would doubling the diameter of the 45 ft long section of pipe change the pump power necessary? 6. 7. How long could the outlet pipe be with the same type pipe and have a positive pressure at B? Flow 2600-ft-long 8-in Schedule 40 steel pipe 25 ft Flow Pump 45-ft-long 10-in Schedule 40 steel pipe

Explanation / Answer

(1) Power, P (in W) = Q (in m3/s) x H (in m) x 9810

Q = 1500 gal/min = 0.0946 m3/s

H = 25-20 = 5 ft = 1.524 m

So, P = 0.0946 x 1.524 x 9810 = 1414.312 W

(2) Power, P (in W) = Q (in m3/s) x H (in m) x 9810

Q = 1500 gal/min = 0.0946 m3/s

Diameter of 10 inch sch 40 pipe = 273 mm

Diameter of 8 inch sch 40 pipe = 219.1 mm

Velocity of fluid in 10 inch sch 40 pipe = Q/A = 0.0946/(pi x 0.2732/4) = 1.616 m/s

Velocity of fluid in 8 inch sch 40 pipe = 0.0946/(pi x 0.21912/4) = 2.51 m/s

Dynamic viscocity of water = 0.00089 Pa

Roughness (e) for steel pipes = 0.045 mm

Friction factor (f) can be obtained from Moody's chart.

For 45 ft long 10 inch schedule 40 steel pipe,

Re = rho x v x D / mu = 1000 x 1.616 x 0.273 / 0.00089 = 495694

e/D = 0.045/273 = 0.000165

f = 0.015

For 2600 ft long 8 inch schedule 40 steel pipe,

Re = rho x v x D / mu = 1000 x 2.51 x 0.2191 / 0.00089 = 617911

e/D = 0.045/219.1 = 0.000205

f = 0.0155

Head loss due to friction = 0.015 x 13.716 x 1.6162/(2 x 9.81 x 0.273) + 0.0155 x 792.48 x 2.512/(2 x 9.81 x 0.2191)

                                        = 0.1 + 17.42 = 17.52 m

Total head loss = 1.524+17.52 = 19.044 m

So, P = 0.0946 x 19.044 x 9810 = 17673.33 W

(3) Applying Bernoulli's principle in reservoir inlet and outlet B,

1.6162/(2g) + 6.096 + 0 = 2.512/(2g) + (PB/(rho x g)) + 7.62 - 17.52

PB = 122.215 + 187.938 = 310.153 Pa

(4) Applying Bernoulli's principle in reservoir inlet and inlet A,

1.6162/(2g) + 6.096 + 0 = 1.6162/(2g) + (PA/(rho x g)) + 0 - 0.1

PA = 121.566 Pa

(5) Applying Bernoulli's principle in reservoir inlet and outlet B,

1.6162/(2g) + 3.048 + 0 = 2.512/(2g) + (PB/(rho x g)) + 7.62 - 17.52

PB = 62.413 + 187.938 = 250.351 Pa

(6) New diameter of 45 ft long pipe = 2 x 273 mm = 546 mm

New velocity of fluid in 45 ft long pipe = 0.0946/(pi x 0.5462/4) = 0.404 m/s

Re = rho x v x D / mu = 1000 x 0.404 x 0.546 / 0.00089 = 247847

e/D = 0.045/546 = 0.0000824

From Moody's chart, f = 0.016

Head loss due to friction in 10 inch sch 40 pipe = 0.016 x 13.716 x 0.4042/(2 x 9.81 x 0.546) = 0.0033 m

Total head loss due to friction = 0.0033 + 17.42 = 17.4233 m

Total head loss = 1.524+17.4233 = 18.947 m

Power = 0.0946 x 18.947 x 9810 = 17583.31 W

% decrease in power = (17673.33 - 17583.31) x 100 /17673.33 = 0.51 %

(7) PB = 0

Let length of outlet pipe be L.

Applying Bernoulli's principle in reservoir inlet and outlet B,

1.6162/(2g) + 6.096 + 0 = 2.512/(2g) + 0 + 7.62 - 0.1 - 0.0155 x L x 2.512/(2 x 9.81 x 0.2191)

L = 70.96 m = 232.82 ft

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

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