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bekas [8.4K]
4 years ago
12

Consider the following two-dimensional velocity field V = (u,v)

Engineering
1 answer:
elena-14-01-66 [18.8K]4 years ago
3 0

Answer:

a) C1 = 3C2

b) C1 = 1 , C2 = -3  

c) w = \frac{-x^2}{2} + \frac{y^2}{2} + 3xy + C

d) (v.v)T = 0

Explanation:

u = 3x + C1y

v = x + C2y

A) determining all  stagnation points

At The stagnation points : u = 0, v = 0

for all  values of C1 and C2 , C1 = 3C2

B) The coefficients of C1 and C2 so that the flow is potential

C1 = 1 , C2 = -3  

C) Determine the expression of the stream function

w = \frac{-x^2}{2} +\frac{y^2}{2} +3xy+ C

D) The value of (v.v)T at the point (x,y) = (1,2)

(v.v)T = 0

Attached is the detailed solution

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Seawater containing 3.50 wt% salt passes through a series of 11 evaporators. Roughly equal quantities of water are vaporized in
statuscvo [17]

Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

Explanation:

F, W and B are the fresh feed, brine and total water obtained

w = 2 x 10^4 L/h

we know that

F = W + B

we substitute

F = 2 x 10^4 + B

F = 20000 + B .................EQUA 1

solute

0.035F = 0.05B

B = 0.035F/0.05

B = 0.7F

now we substitute value of B in equation 1

F = 20000 + 0.7F

0.3F = 20000

F = 20000/0.3

F = 66666.67 kg/hr

B = 0.7F

B = 0.7 * F

B = 0.7 * 66666.67

B = 46,666.669 kg/hr

the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

8 0
4 years ago
Q2: The average water height of an ocean area is 2.5 m high and each wave lasts for an average period of 7 s. Determine (a) the
navik [9.2K]

Answer:

(a) 561.12 W/ m² (b) 196.39 MW

Explanation:

Solution

(a) Determine the energy and power of the wave per unit area

The energy per unit are of the wave is defined as:

E = 1 /16ρgH²

= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²

=3927. 83 J/m²

Thus,

The power of the wave per unit area is,

P = E/ t

= 3927. 83 J/m² / 7 s = 561.12 W/ m²

(b) The average and work power output of a wave power plant

W = E * л * A

= 3927. 83 J/m² * 0.35 * 1 *10^6 m²

= 1374.74 MJ

Then,

The power produced by the wave for one km²

P = P * л * A

= 5612.12 W/m² * 0.35 * 1* 10^6 m²

=196.39 MW

8 0
3 years ago
What type of oil pressure gauge should be used when
liq [111]

Answer:

The mechanical gauge would be the one for the job

Explanation:

6 0
3 years ago
The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort < 0. Fort 2 0 they areV =75 ~75e-10
masya89 [10]

Answer:

maximum value of the power delivered to the circuit =3.75W

energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

Explanation:

V =75 - 75e-1000t V

l = 50e -IOOOt mA

power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)

=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)

=

maximum value of the power delivered to the circuit =3.75W

the total energy delivered to the element = \int\limits^t_0  {3.75(e^{ -IOOOt} - e ^{-2OOOt} )} , dx \\\\

3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

5 0
3 years ago
Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 440°C.
lawyer [7]

Answer:

Heat required (q) = 471.19kj/kg

Explanation:

Find attached below solution to problem

7 0
3 years ago
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