Answer:
pneumatic power system
Explanation:
pneumatic power can be used to quietly operate power windows, door locks, power mirrors, and much much more, also negative pressure pneumatics (vacuum) is used to control many engine and fuel systems
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)])
Question:
The question is not complete. See the complete question and the answer below.
A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.
Answer:
T = 0.11029m²/sec
Radius of influence = 93.304m
expected drawdown = 3.9336m
Explanation:
See the attached file for the explanation.
Answer:
14.52 minutes
<u>OR</u>
14 minutes and 31 seconds
Explanation:
Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.
Specific heat at constant volume at 27°C = 0.718 kJ/kg*K
Initial temperature of room (in kelvin) = 283.15 K
Final temperature (required) of room = 293.15 K
Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg
Heat required at constant volume: 0.718 * (change in temp) * (mass of air)
Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ
Time taken for temperature rise: heat required / (rate of heat change)
Where rate of heat change = 10000 - 5000 = 5000 kJ/hr
Time taken = 1210.26 / 5000 = 0.24205 hours
Converted to minutes = 0.24205 * 60 = 14.52 minutes