In order to impress your neighbors and improve your vision in traffic jams, you decide to mount a cylindrical periscope 2.0 m hi
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Answer:
hello your question is incomplete attached below is the complete question
A) overall mean = 5.535, standard deviation ≈ 0.3239
B ) upper limit = 5.85, lower limit = 5.0
C) Not all the samples meet the contract specifications
D) fluctuation ( unstable Asphalt content )
Explanation:
B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday
The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85
The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0
attached below is the required plot
C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :
15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20
D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed
Answer:
a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr
Explanation:
Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2
Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr
1W = 3.41 BTU/hr
Given parameters:
thickness, t = 7.5mm = 7.5/1000 = 0.0075m
Temperatures 150 C = 150 + 273 = 423 K
50 C = 50 + 273 = 323 K
Temperature difference, T = 423 - 323 = 100 K
We are assuming steady heat flow;
a.) Heat flux, Q" = kT/t
K= thermal conductivity of the material
The thermal conductivity of brass, k = 109.0 W/m.K
Heat flux,
b.) Area of sheet, A = 0.5m2
Heat loss, Q = kAT/t
Heat loss,
Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr
c.) Material is now given as soda lime glass.
Thermal conductivity of soda lime glass, k is approximately 1W/m.K
Heat loss,
Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr
d.) Thickness, t is given as 15mm = 15/1000 = 0.015m
Heat loss,
Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr