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Pavel
16 days ago
10

For flow over a plate, the variation of velocity with vertical distance y from the plate is given as u(y) = ay − by2 where a and

b are constants. Choose the correct relation for the wall shear stress in terms of a, b, and μ.
Engineering
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You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
pantera1 [306]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

The half-life for the specified RC circuit can be expressed as

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.

Thus, the new half-life is

t'_{1\2} =R'Cln2

Now, divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

After substituting all values, we can calculate the revised half-life

t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
3 months ago
A distância entre duas retas reversas é a medida de um segmento orientado que *?
alex41 [359]
Fatec – SP) Let A be a point on line r, which is contained within plane α. It holds true that: a) there is exactly one line that is perpendicular to line r at point A. b) there exists one unique line, not lying in plane α, that is parallel to line r. c) there are infinitely many distinct planes that are parallel to plane α and contain line r. d) there are infinitely many distinct planes that are perpendicular to plane α and contain line r. e) there are countless distinct lines contained within plane α that are parallel to line r.
8 0
2 months 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
mote1985 [299]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, g = 9.81 m/s^2

mass flow rate, \dot{m} = 60.6 kg/s

flow density, \rho = 1000 kg/m^3

efficiency of the pump, \eta = 0.74

output gauge pressure, p_o = 344.75 kPa

input gauge pressure, p_i = 68.95 kPa

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

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

height of discharge, z_o = 1.22 m - 0.61 m = 0.61 m (evaluated at pump’s maximum height of 1.22 m)

input height, z_i = 0 m

hydraulic power of the pump,P =? kW

Initially, the volumetric flow (Q) needs to be determined

Q = \frac{\dot{m}}{\rho}

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}

v_o = \frac{0.0606 m^3/s}{0.069 m^2}

v_o = 0.88 m/s

v_i = \frac{Q}{A_i}

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

6 0
3 months ago
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