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Naya [18.7K]
3 years ago
14

Determine the brake horsepower required by a pump for a discharge of 0.2 m3/s and a total dynamic head of 20m, with an efficienc

y of 80%.
Engineering
1 answer:
GarryVolchara [31]3 years ago
4 0

Answer:

The BHP would be "65.659 HP".

Explanation:

The given values are:

Discharge,

Q = 0.2 m³/s

Dynamic head,

H = 20 m

Efficiency,

% = 80%

Now,

The Power will be:

⇒ P=\delta QH

On substituting the values, we get

⇒     =1000\times 9.81\times 0.2\times 20

⇒     =39240 \ W

The brake horse power will be:

⇒ BHP=\frac{100 Q H}{3960n}

On putting values, we get

⇒           =\frac{100\times 0.2\times 20}{3960\times 0.80}

⇒           =65.659 \ HP

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Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo
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Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

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Therefore, the total valve perimeter is shown to have increased by 60.8%

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Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

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b) There is a larger valve throat areas for the flow of gas

           

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