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r-ruslan [8.4K]
3 years ago
12

Description: An apartment building has a pumped ring main that circulates hot water at 75 C through insulated 40 mm pipe with he

at loss of 17 Watts/metre. At each apartment, this hot water is mixed with cold water to a delivery temperature of 50C? One of the sub-questions provides extra info: It takes 0.00119 kWh/degree/litre to heat water from 15C to 50C.
My problem:

With typical water-efficient shower and taps, typical Melbourne usage of hot water is 44 litres/day/person. For typical occupancy of 1.8 people, what is the effective efficiency of this hot water distribution system (for simplicity assume the existing heating system feeding the distribution system is 100% efficient).? Note that this efficiency calculation ignores the efficiency of production of hot water.

If anyone can help me or if anyone has any idea that would be great!

Thanks :)
Engineering
1 answer:
Ratling [72]3 years ago
7 0

Answer:

hfhhfhff

Explanation:

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Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Di
Tresset [83]

Answer:

0.556 Watts

Explanation:

w = Weight of object = 762 N

s = Distance = 5 m

t = Time taken = 29 seconds

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=\frac{2\times (s-ut)}{t^2}\\\Rightarrow a=\frac{2\times (5-0)}{29^2}=\frac{10}{481}

Mass of the body

m=\frac{w}{g}=\frac{762}{9.81}

Force required to move the body

F=ma\\\Righarrow F=\frac{762}{9.81}\times \frac{10}{481}

Velocity of object

v=u+at\\\Rightarrow v=0+\frac{10}{481}\times 29\\\Rightarrow v=\frac{10}{29}

Power

P=Fv\\\Rightarrow P=\frac{762}{9.81}\times \frac{10}{481}\times \frac{10}{29}=0.556\ W

∴ Amount of power required to move the object is 0.556 Watts

7 0
4 years ago
Study the graphs below what type of trend and pattern in observed .
hichkok12 [17]

Answer:

negative and linear

Explanation:

its a straight line going down

8 0
3 years ago
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What is the the tv series friends about
Fed [463]

Answer:

its about a lot of stuff, just a group of friends who have funny moments and problems, each episode can be different

Explanation:

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3 years ago
Please helpppp TY SO MUCH have a beautiful day :)
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3 years ago
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A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a)
Bess [88]

Answer:

The correct answer is option 'a': 0.046 meters.

Explanation:

We know that the exit velocity of a jet of water is given by Torricelli's law as

v=\sqrt{2gh}

where

'v' is velocity of head

'g' is acceleration due to gravity

'h' is the head under which the jet falls

Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.

Applying the values in above equation we get the exit velocity as

v=\sqrt{2\times 9.81\times 90}=42.02m/s

now we know the relation between discharge and velocity as dictated by contuinity equation is

Q=V\times Area

Applying values in the above equation and solving for area we get

Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}

The circular area is related to diameter as

Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m

Thus the diameter of the nozzle is 0.246 meters

7 0
3 years ago
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