Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
Answer:
Machine 2 has a higher process capability index, it would be best considered for purchase.
Explanation:
Process capability index: Cpk= Min [(mean-L spec)/3sd; (U spec-mean)/3sd]
For machine 1, mean= 48mm and L spec= 46 and U spec= 50, Standard deviation sd= 0.7
Cpk= [0.952;0.952]= 0.952
For machine 2, mean= 47 and L spec= 46 and U spec= 50, Standard deviation sd= 0.3
Cpk= [1.111;3.333]= 1.111
It is clearly observed from the calculations above that the Cpk value of machine 2 is higher than that of machine 1.
Since machine 2 has a higher process capability index, it would be best considered for purchase.
Answer:
15625 moles of methane is present in this gas deposit
Explanation:
As we know,
PV = nRT
P = Pressure = 230 psia = 1585.79 kPA
V = Volume = 980 cuft = 27750.5 Liters
n = number of moles
R = ideal gas constant = 8.315
T = Temperature = 150°F = 338.706 Kelvin
Substituting the given values, we get -
1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin
n = (1585.79*27750.5)/(8.315 * 338.706) = 15625
I think that the answer would be B or C
