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Stella [2.4K]
3 years ago
9

From a lake, water is pumped at a rate of 67 L/s to a storage tank positioned 14 m above while consuming 16.4 kW of electrical p

ower. (Assuming no frictional losses or kinetic energy change taking place), determine the overall efficiency of the pump-motor unit. (Take gravitational acceleration as 10 m/s2).
Physics
1 answer:
LenaWriter [7]3 years ago
3 0

Answer:

57 %

Explanation:

input power = 16.4 kW = 16.4 x 10^3 W = 16400 W

Water pumped per second = 67 L/s

Mass of water pumped per second, m = Volume of water pumped epr second x density of water

m = 67 x 10^-3 x 1000 = 67 kg/s

height raised, h = 14 m

Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W

efficiency = output power / input power = 9380 / 16400 = 0.57

% efficiency = 57 %

thus, the efficiency of the pump is 57 %.

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a = 2 m/s2

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we know from newtons 2nd law

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kx = ma,

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

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In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

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applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

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T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
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