Flip-flops are normally used for all of the following applications, except ________. logic gates data storage frequency division
1 answer:
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Explanation:
Specific cutting energy:
It the ratio of power required to cut the material to metal removal rate of material.If we take the force required to cut the material is F and velocity of cutting tool is V then cutting power will be the product of force and the cutting tool velocity.
Power P = F x V
Lets take the metal removal rate =MRR
Then the specific energy will be
If we consider that metal removal rate and cutting tool velocity is constant then when we increases the cutting force then specific energy will also increase.
Answer:
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Explanation:
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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