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2 years ago

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A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets lo

cated on the perimeter of a wheel. Determine the power generation potential of this water jet.

1 answer:

snow_tiger [21]2 years ago

5 0

Answer:

216000 W or 216 kW

Explanation:

Power: This can be defined as the rate at which energy is consumed or used, The S.I unit of power is Watt (W)

Generally,

Power = Energy/time

P = E/t........................ Equation 1.

But

E = 1/2mv²..................... Equation 2

Where m = mass, v = velocity.

Substitute equation 2 into equation 1

P = 1/2mv²/t...................... Equation 3

Let flow rate (Q) = m/t

Q = m/t................ Equation 4

Substitute equation 4 into equation 3

P = Qv²/2........................ Equation 5

Where Q = flow rate, v = velocity, P = power.

Given: Q = 120 kg/s, v = 60 m/s

Substitute into equation 5

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus the power generation potential of the water jet = 216000 W or 216 kW

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Answer:

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Explanation:

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$Z_w = 18.015 g/mol$

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=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

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=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

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