Answer:
835,175.68W
Explanation:
Calculation to determine the required power input to the pump
First step is to calculate the power needed
Using this formula
P=V*p*g*h
Where,
P represent power
V represent Volume flow rate =0.3 m³/s
p represent brine density=1050 kg/m³
g represent gravity=9.81m/s²
h represent height=200m
Let plug in the formula
P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m
P=618,030 W
Now let calculate the required power input to the pump
Using this formula
Required power input=P/μ
Where,
P represent power=618,030 W
μ represent pump efficiency=74%
Let plug in the formula
Required power input=618,030W/0.74
Required power input=835,175.68W
Therefore the required power input to the pump will be 835,175.68W
Answer:
Divide one teeth count by the other.
Now that you know how many teeth are on each gear, you can find the gear ratio relatively simply. Divide the driven gear teeth by the drive gear teeth. Depending on your assignment, you may write your answer as a decimal, a fraction, or in ratio
Explanation:
Answer:
N_A=1.5*10^-8 kmol/s.m^2
Explanation:
<u>KNOWN: </u>
Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.
<u>FIND:</u>
Molar diffusion flux.
<u>ASSUMPTIONS:</u>
(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.
<u>ANALYSIS:</u> The molar flux may be obtained from
N_A=D_AB/L(C_A,1-C_A,2)
=10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3
N_A=1.5*10^-8 kmol/s.m^2
<u>COMMENTS:</u> The mass flux is:
n_A,x=M_a*N_A,x
n_A,x=6*10^-8 kg/s m^2
Answer:
maximum roll-off factor is 0.7
Explanation:
given data
rate of output data = 1,000,000 bits/sec
cutoff frequency fc = 850 kHz
solution
we know bandwidth frequency is same as that of bandwidth
we apply here fc formula that is
fc = ω ( 1 + α ) ..........1
here
ω = data rate ÷ 2 .............2
ω = 1000 ÷ 2
ω = 500000 Hz
so put now value in eq 1
fc = ω ( 1 + α )
850000= 500000 ( 1 + α )
solve it we get
α = 0.7
so, maximum roll-off factor is 0.7
Answer:
Power= Heat energy\ time
heat energy= mc∆T
specific heat of water = 4180
1.5*4180*(60-25)
=219450
time = 10× 60= 600 secs
power = 219450/600
Power= 365.75 Watt
=0.37 KW
Explanation:
the actual has to be bigger because the heater might be required to handle more ...also to accommodate extra energy.