Answer:
Throat diameter
=28.60 mm
Explanation:
Bore diameter
⇒
Manometric deflection x=235 mm
Flow rate Q=240 Lt/min⇒ Q=.004
Coefficient of discharge
=0.8
We know that discharge through venturi meter


=13.6 for Hg and
=1 for water.

h=2.961 m
Now by putting the all value in



⇒
=28.60 mm
So throat diameter
=28.60 mm
Answer:
B. 180 million joules
Explanation:
Apply the formula for heat transfer given as;
Q=m*c*Δt where
Q = electrical energy consumed by the heater in joules
m= mass of air in the chamber in kg
c= specific heat of air in joules per kg degrees Celsius
Δt= change in temperatures in degrees Celsius
Given in the question;
m= 1200 kg
c= 1000 J/°C /kg
Δt = 180°-30°= 150° C
Substitute values in the equation to get Q as;
Q=m*c*Δt
Q= 1200 * 1000* 150
Q= 180000000 joules
Q = 180 million joules
<u>The correct answer option is B : 180 million joules.</u>
Answer & Explanation:
function Temprature
NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];
DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];
%AVERAGE CALCULATION AND ROUND TO NEAREST INT
avgNYC=round(mean(NYC));
avgDEN=round(mean(DEN));
fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);
fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);
%part B
count=1;
NNYC=0;
NDEN=0;
while count<=length(NYC)
if NYC(count)>avgNYC
NNYC=NNYC+1;
end
if DEN(count)>avgDEN
NDEN=NDEN+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);
fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);
%part C
count=1;
highDen=0;
while count<=length(NYC)
if NYC(count)>DEN(count)
highDen=highDen+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);
end
%output
check the attachment for additional Information
From the statement it is determined that the maximum limit is 100 and the lower limit is 0. At the same time there is an average of 85 that is closer to the upper limit than the lower limit, therefore the distribution will tend to occur to the left .
This will mean that people will begin to surround the average of 85 and that those who are 60, will be more than those who are below 50, and these 50 will be more than those who are by 40 and so on.