A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
1 answer:
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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
Answer:
a. V = 109.64 × 10⁵ ft/min
b. Mw = 654519.54 kg/hr
Explanation:
Given Parameters
mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s
inlet temperature of water, T1 = 84 F = 28.89 C
outlet temperature of water, T2 = 68 F = 20 C
specific heat capacity of water, c = 4.18kJ/kgK
rate of heat remover from water, Qw is given by
Qw = 6607.33[28.89 - 20] * 4.18
Qw = 245529.545kw
For air, inlet condition
DBT = 70 F hi = 43.43 kJ/kg
WBT = 60 F wi = 0.00874 kJ/kg
u1 = 0.8445 m/kg
oulet condition,
DBT = 70 F RH = 100.1
h1 = 83.504kJ/kg
Wo = 0.222kJ/kg
check the attached file for complete solution
