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2 answers:
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Answer:
a)exit velocity of the steam, V2 = 2016.8 ft/s
b) the amount of entropy produced is 0.006 Btu/Ibm.R
Explanation:
Given:
P1 = 100 psi
V1 = 100 ft./sec
T1 = 500f
P2 = 40 psi
n = 95% = 0.95
a) for nozzle:
Let's apply steady gas equation.
h1 and h2 = inlet and exit enthalpy respectively.
At T1 = 500f and P1 = 100 psi,
h1 = 1278.8 Btu/Ibm
s1 = 1.708 Btu/Ibm.R
At P2 = 40psi and s1 = 1.708 Btu/Ibm.R
1193.5 Btu/Ibm
Let's find the actual h2 using the formula :
solving for h2, we have
Take Btu/Ibm = 25037 ft²/s²
Using the first equation, exit velocity of the steam =
Solving for V2, we have
V2 = 2016.8 ft/s
b) The amount of entropy produced in BTU/ lbm R will be calculated using :
Δs = s2 - s1
Where s1 = 1.708 Btu/Ibm.R
At h2 = 1197.77 Btu/Ibm and P2 =40 psi,
S2 = 1.714 Btu/Ibm.R
Therefore, amount of entropy produced will be:
Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R
= 0.006 Btu/Ibm.R
Answer:21.3%
Explanation:
Given
80 % reduction in tool life
According to Taylor's tool life
=c
where V is cutting velocity
T=tool life of tool
80 % tool life reduction i.e. New tool Life is 0.2T
Thus
=1.213V
Thus a change of 21.3 %(increment) is required to reduce tool life by 80%
Answer:
Explanation:
Given
Required
Plot a steam and leaf display for the given data
Start by categorizing the data by their tenth values:
The 0.3's is will be plotted as thus:
The 0.4's is as follows:
The 0.5's is as follows:
The 0.6's is as thus:
Lastly, the 0.7's is as thus:
The combined steam and leaf plot is: