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Goryan [66]
3 years ago
13

Savabuck University has installed standard pressure-operated flush valves on their water closets. When flushing, these valves de

liver 130.0 L min1 . If the delivered water costs $0.45 per cubic meter, what is the monthly cost of not repairing a broken valve that flushes continuously?
Engineering
1 answer:
Dvinal [7]3 years ago
7 0

Answer:

Cost = $2527.2 per month.

Explanation:

Given that

Discharge ,Q = 130 L/min

 So

Q=0.13\ m^3/min

Cost =  $0.45 per cubic meter

1 month = 30 days

1 days = 24 hr = 24 x 60 min

1 month = 30 x 24 x 60 min

1 month = 43,200 min

Lets xm^3\ water\ waste\ in\ a\ month

x = 0.13 x 43,200

x=5616\ m^3

So the total cost = 5616 x 045 $

Cost = $2527.2 per month.

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Salvage ethnography is the effort to ensure that ethnography remains an important part of anthropology. recording of linguistic
fenix001 [56]

Answer:

                D

Explanation:

                            D

6 0
3 years ago
What flow type occurs when the heat exchanger has multiple passes and baffles?
Vanyuwa [196]
I think that the answer would be B or C
4 0
3 years ago
The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i
koban [17]

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            b=25m/sin(60)\\=28.867

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N

Noting from the Bernoulli  equation that

Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0
3 years ago
Ok there........................................................................
Juliette [100K]

Answer:

ok THERE

Explanation:

4 0
3 years ago
A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4
Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0
3 years ago
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