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
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
Answer:
The mechanical gauge would be the one for the job
Explanation:
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 = 

Answer:
Heat required (q) = 471.19kj/kg
Explanation:
Find attached below solution to problem