Ask question

Ask question

All categories

2 years ago

7

A residential subdivision encompasses 1100 acres with a housing density of four houses per acre. Assume that a high-value reside

nce uses 800 g/day/house and has a fire flow requirement of 1000 gpm. If the water system has a daily peaking factor = 1.64 and an hourly peaking factor = 2.5, determine (a) the average daily demand of this subdivision and (b) the design-demand used to design the distribution system.

1 answer:

aleksandrvk [35]2 years ago

8 0

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

$\text{Number of house in 1100 acres} = 4\times1100$

$\text{Number of house in 1100 acres} = 4400$

Per house water demand = 800 g/day/house

(a). We need to calculate the average daily demand of this subdivision

Using formula for average daily demand

$\text{average daily demand}=house\times\text{Per house water demand}$

$\text{average daily demand}=4400\times800\ gallon/day$

$\text{average daily demand}=3520000\ gallon/day$

$\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min$

$\text{average daily demand}=2444.44\ gallon/min$

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

$\text{design demand}=(Q_{max})daily\times\text{fire flow}$

$\text{design demand}=1.64\times2444.4\times1000$

$\text{design demand}=4008816\ gallon/m$

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

You might be interested in

A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 n

RSB [31]

Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

Hope it helps

3 0

2 years ago

A,b, e are complete. Help on the others would be so appreciated!!

bixtya [17]

Answer:

serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5

1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

1 / Ceq = 1,147

Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

Dv = Q / Ceq

Q = DV Ceq

Q = 14 0.678 10⁻⁶

Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

ΔV = Q / C5

ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

ΔV = 1,725 V

d) The energy stores is

U = ½ C V²

U = ½ 0.678 10-6 14²

U = 66.4 10⁻⁶ J

e) Parallel system

Ceq = C1 + C2 + C3 + C4 + C5

Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

Q5 = C5 V

Q5 = 5.5 10⁻⁶ 14

Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

U = ½ C V²

U = ½ 22.7 10-6 14²

U = 2.2 10⁻³ J

8 0

3 years ago

A 2 L balloon filled with gas is warmed from 280 K to 700 K. What is the volume of the gas after it is heated?

irakobra [83]

Answer:

New volume, v2 = 0.8L

Explanation:

<u>Given the following data;</u>

Original Volume = 2L

Original Temperature = 280K

New Temperature = 700K

To find new volume V2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$VT = K$

$\frac{V1}{T1} = \frac{V2}{T2}$

$\frac{V1}{T1} = \frac{V2}{T2}$

Making V2 as the subject formula, we have;

$V_{2}= \frac{V1}{T1} * T_{2}$

$V_{2}= \frac{2}{700} * 280$

$V_{2}= 0.0029 * 280$

V2 = 0.8L

Therefore, the volume of the gas after it is heated is 0.8L.

7 0

3 years ago

Select the correct answer.

Alexeev081 [22]

For the scale od the paraboliczne curviczne Ur

3 0

2 years ago

Read 2 more answers

Which physical property is used to test the purity of metal ?

dimulka [17.4K]

The physical and chemical properties of metal can differ between the elements but if we try and generalize.

Physical properties are that they are hard, dense, shiny, malleable (can be bent), ductile (can be pulled into wires), a good conductor of heat and have high melting points meaning that they are solid at room temperature

Their chemical properties include being good electrical conductors, they make metallic bonds (ionic bonds with non-metals) and their atoms are arranged in a lattice. They release hydrogen when reacted with acids and are usually cations (positive ions).

4 0

2 years ago