Answer:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
Answer:
Qx = 9.10
m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 ×
Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85
)²× 9
× sin18 × cos18
Qd = 94.305 ×
m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 ×
× π × 85
× ( 9
)³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 ×
m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 ×
- 85.2 ×
Qx = 9.10
m³/s
Answer:
A.
Explanation:
Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected
Answer:
It would take approximately 305 s to go to 99% completion
Explanation:
Given that:
y = 50% = 0.5
n = 1.7
t = 100 s
We need to first find the parameter k from the equation below.

taking the natural logarithm of both sides:

Substituting values:

Also
![t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}](https://tex.z-dn.net/?f=t%5En%3D-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%5C%5Ct%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D)
Substituting values and y = 99% = 0.99
![t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s](https://tex.z-dn.net/?f=t%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D%3D%5Csqrt%5B1.7%5D%7B-%5Cfrac%7Bln%281-0.99%29%7D%7B2.76%2A10%5E%7B-4%7D%7D%7D%3D304.6s)
∴ t ≅ 305 s
It would take approximately 305 s to go to 99% completion