I think the answer is B. 10D
Answer:
MRR = 1.984
Explanation:
Given that
Depth of cut ,d=0.105 in
Diameter D= 1 in
Speed V= 105 sfpm
feed f= 0.015 ipr
Now the metal removal rate given as
MRR= 12 f V d
d= depth of cut
V= Speed
f=Feed
MRR= Metal removal rate
By putting the values
MRR= 12 f V d
MRR = 12 x 0.015 x 105 x 0.105
MRR = 1.984
Therefore answer is -
1.944
Answer: (a) 9.00 Mega Newtons or 9.00 * 10^6 N
(b) 17.1 m
Explanation: The length of wall under the surface can be given by
![b=25m/sin(60)\\=28.867](https://tex.z-dn.net/?f=b%3D25m%2Fsin%2860%29%5C%5C%3D28.867)
The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.
![F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N](https://tex.z-dn.net/?f=F%28resultant%29%20%3D%20Pavg%20%28%20A%29%20%3D%20%28Patm%20%2B%20%20%5Crho%20g%20h%20c%29%2AA%20%5C%5C%3D%20%5B100000%20N%2Fm%5E2%20%2B%20%281000%20kg%2Fm%5E3%20%2A%209.81%20m%2Fs%5E2%20%2A%2025m%2F2%29%5D%2A%20%28140%2A25m%2Fsin60%29%5C%5C%3D%208.997%2A10%5E8%20N%20%5C%5C%3D%209.0%2A10%5E8%20N)
Noting from the Bernoulli equation that
![Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\](https://tex.z-dn.net/?f=Po%2F%5Crho%20g%20sin60%20%3D%20100000%2F1000%20%2A%209.81%2A%20sin%2860%29%20%3D%2011.77%20m%20%5C%5C%20%5C%5C)
From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:
Substituting the values gives us the the distance of the surface to be equal to = 17.1 m
Answer:
Percentage change 5.75 %.
Explanation:Given ;
Given
Pressure of condenser =0.0738 bar
Surface temperature=20°C
Now from steam table
Properties of steam at 0.0738 bar
Saturation temperature corresponding to saturation pressure =40°C
So Δh=2573.5-167.5=2406 KJ/kg
Enthalpy of condensation=2406 KJ/kg
So total heat=Sensible heat of liquid+Enthalpy of condensation
![Total\ heat\ =C_p\Delta T+\Delta h](https://tex.z-dn.net/?f=Total%5C%20heat%5C%20%3DC_p%5CDelta%20T%2B%5CDelta%20h)
Total heat =4.2(40-20)+2406
Total heat=2,544 KJ/kg
Now film coefficient before inclusion of sensible heat
![h_1=\dfrac{\Delta h}{\Delta T}](https://tex.z-dn.net/?f=h_1%3D%5Cdfrac%7B%5CDelta%20h%7D%7B%5CDelta%20T%7D)
![h_1=\dfrac{2406}{20}](https://tex.z-dn.net/?f=h_1%3D%5Cdfrac%7B2406%7D%7B20%7D)
![h_1=120.3\frac{KJ}{kg-m^2K}](https://tex.z-dn.net/?f=h_1%3D120.3%5Cfrac%7BKJ%7D%7Bkg-m%5E2K%7D)
Now film coefficient after inclusion of sensible heat
![h_2=\dfrac{total\ heat}{\Delta T}](https://tex.z-dn.net/?f=h_2%3D%5Cdfrac%7Btotal%5C%20heat%7D%7B%5CDelta%20T%7D)
![h_2=\dfrac{2,544}{20}](https://tex.z-dn.net/?f=h_2%3D%5Cdfrac%7B2%2C544%7D%7B20%7D)
![h_2=127.2\frac{KJ}{kg-m^2K}](https://tex.z-dn.net/?f=h_2%3D127.2%5Cfrac%7BKJ%7D%7Bkg-m%5E2K%7D)
=5.75 %
So Percentage change 5.75 %.