4m^3 of water is to be pumped to a height of 20 m and forced into a reservoir at a pressure of 2*10^5 Nm^2. The waoek done by the motor is(external pressure =10^5 Nm^2)
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Answered by
17
Work done is =
Volume of water pumped at a constant pressure * constant excess pressure over atmospheric pressure
P = excess pressure = 2 * 10⁵ N/m² - 1 * 10⁵ N/m² = 1 * 10⁵ N/m²
Volume of water pumped = V2 - V1 = V = 4 m³
W = P * V = 1 * 10⁵ N/m² * 4 m³ = 400 kJoules
Volume of water pumped at a constant pressure * constant excess pressure over atmospheric pressure
P = excess pressure = 2 * 10⁵ N/m² - 1 * 10⁵ N/m² = 1 * 10⁵ N/m²
Volume of water pumped = V2 - V1 = V = 4 m³
W = P * V = 1 * 10⁵ N/m² * 4 m³ = 400 kJoules
Answered by
29
Work done by the motor = change in gravitational potential energy + work done against external pressure
= mgh +¤pv
=V×rho ×gh +¤pv
=4×10^3×10×20 +[2×10^5 -10^5]×4
=8×10^5 + 4×10^5
=12×10^5 ::
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