Transmission cleaners are used: A) Only in conjunction with fuel system cleanersB) Only in the colder monthsC) By themselvesD) I
2 answers:
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Solution :
<u>Sieve Size</u> (in) <u>Weight retain</u><u>(g)</u>
3 1.62
2 2.17
3.62
2.27
1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan <u> 6.3 </u> 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ()
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
x = 0.2634 mm
Effective size,
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
x = 1.6317 mm
Uniformity co-efficient =
Cu = 6.17
Now, fineness modulus =
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm
This question is not complete, the complete question is;
The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005
Answer:
the length of the pipe is 11583 in or 965.25 ft
Explanation:
Given the data in the question;
Static pressure ratio; p1/p2 = 10
friction coefficient f = 0.005
diameter of pipe, D =4 inch
first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so
4fL / D = 57.915
we substitute
(4×0.005×L) / 4 = 57.915
0.005L = 57.915
L = 57.915 / 0.005
L = 11583 in
Therefore, the length of the pipe is 11583 in or 965.25 ft
