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Question

(C) According to Bernoulli's equation for a horizontal pipe,the sum of pressure energy and kinetic energy per unit volume remains constant:$P_1 + \fra

Vedclass · Accepted Answer

$48.5$

Vedclass · Answer

(C) According to Bernoulli's equation for a horizontal pipe,the sum of pressure energy and kinetic energy per unit volume remains constant:$P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$Given:$P_1 = 50 \ kPa = 50 	imes 10^3 \ Pa$$v_1 = 1 \ m/s$$v_2 = 2 \ m/s$Density of water $ho = 1000 \ kg/m^3 = 10^3 \ kg/m^3$Rearranging the equation to solve for $P_2$:$P_2 = P_1 + \frac{1}{2} ho (v_1^2 - v_2^2)$$P_2 = 50 	imes 10^3 + \frac{1}{2} 	imes 10^3 	imes (1^2 - 2^2)$$P_2 = 50 	imes 10^3 + 0.5 	imes 10^3 	imes (1 - 4)$$P_2 = 50 	imes 10^3 - 1.5 	imes 10^3$$P_2 = 48.5 	imes 10^3 \ Pa = 48.5 \ kPa$

$A$ horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity $1 \ m/s$ when pressure is $50 \ kPa$ at a point. If the velocity of flow has to be $2 \ m/s$ at some other point,the pressure at that point should be (in $kPa$)

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