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Nesterboy [21]
3 years ago
8

water at 20c discharges from a stroke tank through a 150m length of horizontal pipe. The pipe is smooth and has a diameter of 75

mm. The entrance of the pipe is square edged. What is depth of water needed to produce a volumetric flow rate of .1m3/s
Engineering
1 answer:
inessss [21]3 years ago
7 0

Answer:

The dept of the water needed is 26.11m

Explanation:

Given the following parameters:

Length of the pipe = 150m

Diameter of the pipe = 75mm = 0.75m

Volumetric flow rate 1m^{3}/s

Knowing that:

Volumetric flowrate = area (A) * Velocity(V)

==> 0.1 = A*V                              *Knowing that V = \sqrt{2gh}

==> 0.1  = \frac{\pi }{4}* (0.075)^{2} * \sqrt{2gh}

==> h = 26.11m

Hence, the dept of water needed to produce a volumetric flow rate of 1m^{3}/s is 26.11m.

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professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ \frac{1}{\mu(1-\frac{\lambda}{\mu})}

thus,

on substituting the respective values, we get

Average time spent by the vehicle = \frac{1}{60(1-\frac{30}{60})}

or

Average time spent by the vehicle = \frac{1}{60(1-0.5)}

or

Average time spent by the vehicle = \frac{1}{60(0.5)}

or

Average time spent by the vehicle = \frac{1}{30} hr/veh

or

Average time spent by the vehicle = \frac{1}{30}\times60 min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

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What are the philological elements of interior design most like?
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Ea public address glven via the intercom system of a large buildingxplanation:

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What type of building project is most like an enhanced skills occupation?
enot [183]

Answer:

4.modifying a building to achieve an entirely new purpose

8 0
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Explain 3 ways that people in sports use engineering to increase their performance?
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7 0
3 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
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