5.20 Maintaining pipe wall temperature. Maintaining a constant pipe wall tempera

Question

5.20 Maintaining pipe wall temperature. Maintaining a constant pipe wall temperature in some hot process applications is critical. A new technique that utilizes bolt-on trace elements to maintain temperature was presented in the Journal of Heat Transfer (November 2000). Without bolt-on trace elements, the pipe wall temperature of a switch condenser used to produce plastic has a uniform distribution ranging from 260 to 290 F. When several bolt on trace clements are attached to the piping, the wall temperature is uniform from 278 to 285 F. a. Ideally, the pipe wall temperature should range between 2800 and 284°F. What is the probability that the temperature will fall in this ideal range when no bolt-on trace elements are used? When bolt-on trace elements are attached to the pipe? (0.1333, 0.5714) b. When the temperature is 268 F or lower, the hot liquid plastic hardens (or plates), causing a buildup in the piping. What is the probability ofplastic plating when no bolt-on trace elements are used? When bolt-on trace elements are attached to the pipe? (0.2667, 0)

Explanation / Answer

a)

WITHOUT BOLT ON:

Note that here,          
          
a = lower fence of the distribution =    260      
b = upper fence of the distribution =    290      
          
Thus, the area between the said numbers is          
          
c = lower number =    280      
d = higher number =    284      
          
Thus, the probability between these two values is          
          
P = (d - c)/(b - a) =    0.133333333   [ANSWER]  

**********************

WITH BOLT ON:

Note that here,          
          
a = lower fence of the distribution =    278      
b = upper fence of the distribution =    285      
          
  
          
Thus, the area between the said numbers is          
          
c = lower number =    280      
d = higher number =    284      
          
Thus, the probability between these two values is          
          
P = (d - c)/(b - a) =    0.571428571   [ANSWER]

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b)

WITHOUT BOLT ON:

Note that here,          
          
a = lower fence of the distribution =    260      
b = upper fence of the distribution =    290      
          
Note that P(x<c) = P(a<x<c) = (c-a)/(b-a). Thus, as          
          
c = critical value =    268      
          
Then          
          
P(x<c) =    0.266666667   [ANSWER]

********************

WITH BOLT ON:

As 268 is below the interval (278, 285), this is an impossible event, so

P(x<268) = 0 [ANSWER]  

  

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