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Answer:
Total wight =640.7927 KN
Explanation:
Given that
do= 61 cm
L =120
t= 0.9 cm
That is why inner diameter of the pipe
di= 61 - 2 x 0.9 cm
di=59.2 cm
Water density ,ρ = 1 kg/L = 1000 kg/m³
Weight of the pipe ,wt = 2500 N/m
wt = 2500 x 120 N = 300,000 N
The wight of the water
wt ' = ρ V g
wt'=340792.47 N
That is why total wight
Total wight = wt + wt'
Total wight =300,000+ 340792.47 N
Total wight =640,792.47 N
Total wight =640.7927 KN
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:
Explanation:
k_max = 26.9 w/mk
k_min = 22.33 w/mk
Explanation:
a) the maximum thermal conductivity is given as
K_MAX = k_m v_m + k_p v_p
where k_m is thermal conductvitiy of metal
k_p is thermal conductvitiy of carbide
v_m = proportion of metal in the cement = 0.15
v_p = proportion of carbide in the cement = 0.85
= 66*0.15 + 20*0.85
k_max = 26.9 w/mk
b) the minimum thermal conductivity is given as
= \frac{20*66}{20*0.15 +66*0.85}
k_min = 22.33 w/mk
Answer:
1.8 mm
Explanation:
given data
thick = 2.5 mm
flux = 1 × kg/m²
high pressure surface is 2 kg/m³
solution
we use fick first law for steady state diffusion
J = D × ..........1
we take here Ca to point which concentration of nitrogen is 2 kg/m³
so we solve Xb
Xb = Xa + D ×
assume Xa = 0 at surface
Xb = 0 + ( 12 × ) ×
Xb = 1.8 ×
Xb = 1.8 mm