Answer:
![(M_t)_{rated}=61.11lb-in](https://tex.z-dn.net/?f=%28M_t%29_%7Brated%7D%3D61.11lb-in)
Explanation:
speed of motor (N)=1500 rpm
power=4 hp =
=2.9828 KW
service factor(k)= 2.75
now,
![KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }](https://tex.z-dn.net/?f=KW%3D%5Cfrac%7B2%5Cpi%20n%20M_t%7D%7B60%20%5Ctimes%2010%5E6%7D%20%5C%5C2.9828%3D%5Cfrac%7B2%5Cpi%20%5Ctimes%201500%20M_t%7D%7B60%20%5Ctimes%2010%5E6%7D%5C%5CM_t%3D%5Cfrac%7B2.9828%5Ctimes%2060%20%5Ctimes%2010%5E6%7D%7B2%5Cpi%20%5Ctimes%201500%20%7D)
![M_t= 18,989.09 \ N-mm= 168.06 lb-in](https://tex.z-dn.net/?f=M_t%3D%2018%2C989.09%20%5C%20N-mm%3D%20168.06%20lb-in)
torque rating
![(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in](https://tex.z-dn.net/?f=%28M_t%29_%7Bdesign%7D%3Dk_s%5Ctimes%20%28M_t%29_%7Brated%7D%5C%5C168.06%3D%202.75%5Ctimes%20%28M_t%29_%7Brated%7D%5C%5C%28M_t%29_%7Brated%7D%3D%5Cfrac%7B168.06%7D%7B2.75%7D%20%5C%5C%28M_t%29_%7Brated%7D%3D61.11lb-in)
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
Answer:
All 4 could be justified.
Explanation:
They all represent ultimate improvement.
Answer:
Option E
Explanation:
All the given statements are true except the velocity gradients normal to the flow direction are small since these are not normally small. It's true that viscous effects are present only inside the boundary layer and the fluid velocity equals the free stream velocity at the edge of the boundary layer. Moreover, Reynolds number is greater than unity and the fluid velocity is zero at the surface of the object.
The weight of the specimen in SSD condition is 373.3 cc
<u>Explanation</u>:
a) Apparent specific gravity = ![\frac{A}{A-C}](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7BA-C%7D)
Where,
A = mass of oven dried test sample in air = 1034 g
B = saturated surface test sample in air = 1048.9 g
C = apparent mass of saturated test sample in water = 975.6 g
apparent specific gravity =
= ![\frac{1034}{1034-675 \cdot 6}](https://tex.z-dn.net/?f=%5Cfrac%7B1034%7D%7B1034-675%20%5Ccdot%206%7D)
Apparent specific gravity = 2.88
b) Bulk specific gravity ![G_{B}^{O D}=\frac{A}{B-C}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BO%20D%7D%3D%5Cfrac%7BA%7D%7BB-C%7D)
![G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BO%20D%7D%3D%5Cfrac%7B1034%7D%7B1048.9-675%20%5Ccdot%206%7D)
= 2.76
c) Bulk specific gravity (SSD):
![G_{B}^{S S D}=\frac{B}{B-C}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BS%20S%20D%7D%3D%5Cfrac%7BB%7D%7BB-C%7D)
![=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209%7D%7B1048%20%5Ccdot%209-675%20%5Ccdot%206%7D)
= 2.80
d) Absorption% :
![=\frac{B-A}{A} \times 100 \%](https://tex.z-dn.net/?f=%3D%5Cfrac%7BB-A%7D%7BA%7D%20%5Ctimes%20100%20%5C%25)
![=\frac{1048 \cdot 9-1034}{1034} \times 100](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209-1034%7D%7B1034%7D%20%5Ctimes%20100)
Absorption = 1.44 %
e) Bulk Volume :
![v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}](https://tex.z-dn.net/?f=v_%7Bb%7D%3D%5Cfrac%7B%5Ctext%20%7B%20weight%20of%20dispaced%20water%20%7D%7D%7BP%20%5Comega%20t%7D)
![=\frac{1048 \cdot 9-675 \cdot 6}{1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209-675%20%5Ccdot%206%7D%7B1%7D)
= ![373.3 cc](https://tex.z-dn.net/?f=373.3%20cc)