Answer:
The velocity is 18.68m/s
Explanation:
Bernoulli's equation is applicable for stream line flow of a fluid. The flow must be steady and uniform flow. The Bernoulli's equation between inlet and outlet is written as:
P1/pg + V1/2g + Z1 = P2/pg + V2^2 + Z2
Where V1 and V2 are velocity of fluid at point 1 and 2b. The diameter of the tank too will be larger than that of the nozzle. Hence the velocity at point 1 will be 0.V1= 0
Substituting the values in to the equation
250 ×10^3/1000g + 0/g + 2.5 = 100×10^3/1000g + V2^2/2g + 0
250 + 2.5g = 100 + V2^2/2
250 + (2.5 × 9.8) = 100 V2^2/2
250 + 23.525- 100 = V2^2/2
174.525 = V2^2/2
Cross multiply
174.525 × 2 = V2^2
V2 = 349.05
V2 = Sqrt(349.05)
V2 = 18.68m/s
The initial discharge rate of water from the tank - 0.146 m/s.
According to Bernoulli's Equation when fluids are in an ideal state both pressure and density are inversely related.
Given:
diameter orifice D = 10-cm
water level above the outlet Z = 2.5 m
P1 of tank air pressure = 250 kPa
the atmospheric pressure P2 = 100 kPa
Solution:
From the Bernoulli Equation:
[tex]\frac{P_1}{\rho g}+\frac{V^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V^2}{2g} +Z_2\\\frac{P_1}{\rho g}+Z_1=\frac{P_2}{\rho g}+\frac{V^2}{2g}\\\frac{250\times 10^3}{1000(9.81)}+2.5=\frac{100\times 10^3}{1000(9.81)}+\frac{V^2}{2(9.81)}\\27.98=10.19+\frac{V^2}{19.62}\\(27.98-10.19) = \frac{V^2}{19.62}\\17.79=\frac{V^2}{19.62}\\V^2=349.03\\V = 18.68 \ m/s\\[/tex]
The initial discharge rate of water from the tank that is :
[tex]We\ know\ that\\\ V=\frac{Q}{A}\\18.68=\frac{Q}{\frac{\pi D^2}{4}}\\18.68 = \frac{Q}{\frac{\pi (10\times 10^{-2})^2}{4}}\\18.68 = \frac{Q}{7.85\times 10^{-3}}\\Q = 0.146 \ m^3/s[/tex]
Thus, The initial discharge rate of water from the tank - 0.146 m/s.
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