Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer and Explanation:
clear all; close all;
N=512;
t=(1:N)/N;
fs=1000;
f=(1:N)*fs/N;
x= sin(2*pi*200*t) + sin(2*pi*400*t);
y= sin(2*pi*200*t) + sin(2*pi*900*t);
for n = 1:20
a(n) = (2/N)*sum(x.*(cos(2*pi*n*t)))
b(n) = (2/N)*sum(x.*(sin(2*pi*n*t)))
c(n) = sqrt(a(n).^2+b(n).^2)
theta(n) =-(360/(2*pi))*atan(b(n)./a(n));
end
plot(f(1:20),c(1:20),'rd');
disp([a(1:4),b(1:4),c(1:4),theta(1:4)])
A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.
The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.
I wont give the answer but the steps
Your Welcome
Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...
Explanation:
Is the same as the example,
If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...
Then
(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)
Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...
The way to write this is
Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)
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