Consider a normal shock wave in air. The upstream conditions are given by M1=3, p1 = 1 atm, and r1 = 1.23 kg/m3. Calculate the d
ownstream values of p2, T2, r2, M2, u2, p02 and T02. (Anderson 3.4)
1 answer:
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Solution :
Methods for selling pressure of a distillation column :
a). Set, the overhead stream using cooling water.
(minimum of approximate 45°C condenser temperature)
b). Set, of bottom product that avoids decomposition or reaction.
c). Set, not utility for reboiler.
Running the distillation column above the ambient pressure because :
The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.
Run the reactor at an evaluated temperature because :
a). The rate of reaction is taster. This results in a small reactor or high phase conversion.
b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.
Run the reaction at an evaluated pressure because :
The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.
The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.
Answer:
- <u>0.59</u>
Explanation:
The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.
The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.
The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.
The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.
By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.
The joint probability of all the samples are good is the product of each probability:
The result is: 0.41278 ≈ 0.41
The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.
Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.
Answer:
maximum isolator stiffness k =1764 kN-m
Explanation:
mean speed of rotation
=65.44 rad/sec
= 0.1*(65.44)^2
F_T =428.36 N
Transmission ratio
also
transmission ratio
SOLVING FOR Wn
Wn = 42 rad/sec
k = m*W^2_n
k = 1000*42^2 = 1764 kN-m
k =1764 kN-m