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13 days ago

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A tank contains initially 2500 liters of 50% solution. Water enters the tank at the rate of 25 iters per minute and the solution

flows out at the rate of 50 liters per minute. Find the percentage of salt after 20 minutes

1 answer:

Kisachek [217]13 days ago

4 0

The solution's percentage is calculated to be 32.96%. Given there is initially a tank with a 50% solution, the total amount of solution is computed as 0.5 x 2500 = 1250 liters. If we denote the amount of solution at time t as A, we can proceed with the calculation. By integration, where C represents the constant, we observe that at t=0, A=1250. Therefore, at t=20 minutes, A=837.90. Thus, the percentage of solution after 20 minutes remains at 32.96%.

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1 day ago

The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re

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Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

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efficiency of the pump, $\eta = 0.74$

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cross-sectional area of output pipe, $A_o = 0.069 m^2$

cross-sectional area of input pipe, $A_i = 0.093 m^2$

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input height, $z_i = 0 m$

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$Q = \frac{60.6 kg/s}{1000 kg/m^3}$

$Q = 0.0606 m^3/s$

Next, compute the velocity (v) for both input and output

$v_o = \frac{Q}{A_o}$

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$v_o = 0.88 m/s$

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$v_i = \frac{0.0606 m^3/s}{0.093 m^2}$

$v_i = 0.65 m/s$

Subsequently, the total head (H) can be calculated

$H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}$

$H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}$

$H = 28.74m$

Finally, the computation of pump power is done as follows

$P = \frac{Q \, \rho \, g \, H}{\eta}$

$P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}$

$P = 23.09 kW$

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25 days ago

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t = KW/(ρA(CPR))

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Answer:

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21 hour ago