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n200080 [17]
3 years ago
12

A pump transfers water from a lake to a reservoir, which is located 29.2 m above the lake, at a rate of 11.5 L/s. Determine the

amount of power (in kW) required to pump water from the lake to the reservoir.
Engineering
1 answer:
Greeley [361]3 years ago
8 0

Answer:

P=3290.84KW

Explanation:

From the question we are told that:

Heighth=29.2m

Rate Q=11.5L/s=>11.5*10^{-3}

Generally the equation for Power is mathematically given by

P=\rho*g*Q*h

P=1000*9.81*11.5*10^{-3}*29.2

P=3290.84KW

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lbvjy [14]

Answer:

Technician B

Explanation:

Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.

5 0
3 years ago
Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?
zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

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4 0
1 year ago
The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint
snow_lady [41]

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = \frac{100}{100 - P} M kg  per day

put her value

mass of sludge produced = \frac{100}{100 - 78} 7240 kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge =  \frac{\rho sludge}{\rho water }

and as we know that

\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}

\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}

S sludge = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = \frac{mass sludge produce}{\rho sludge}

volume of biological sludge = \frac{32909.09}{1152}

volume of biological sludge = 28.566 m³ per day

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