Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:
![E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}](https://tex.z-dn.net/?f=E%20_%7Bpump%2Cu%7D%20%3D%20%5Cdfrac%7Bm%28P_2-P_1%29%7D%7B%5Crho%7D)
where;
m = mass flow rate = 120 kg/min
the pressure at the inlet
= 96 kPa
the pressure at the exit
= ???
the pressure
= 1000 kg/m³
∴
![0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}](https://tex.z-dn.net/?f=0.8%20%5Ctimes%2010%5E%7B3%7D%20%5C%20W%20%3D%20%5Cdfrac%7B%28120%20%5C%20kg%2Fmin%20%2A%201min%2F60%20s%29%28P_2-96000%29%7D%7B1000%7D)
![0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}](https://tex.z-dn.net/?f=0.8%20%5Ctimes%2010%5E%7B3%7D%5Ctimes%201000%20%3D%20%7B%28120%20%5C%20kg%2Fmin%20%2A%201min%2F60%20s%29%28P_2-96000%29%7D)
![800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}](https://tex.z-dn.net/?f=800000%20%3D%20%7B%28120%20%5C%20kg%2Fmin%20%2A%201min%2F60%20s%29%28P_2-96000%29%7D)
![\dfrac{800000}{2} = P_2-96000](https://tex.z-dn.net/?f=%5Cdfrac%7B800000%7D%7B2%7D%20%3D%20P_2-96000)
400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
Answer:
Explained below
Explanation:
The isohyetal method is one used in estimating Rainfall whereby the mean precipitation across an area is gotten by drawing lines that have equal precipitation. This is done by the use of topographic and other data to yield reliable estimates.
Whereas, the arithmetic method is used to calculate true precipitation by the way of getting the arithmetic mean of all the points or arial measurements that will be considered in the analysis.
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑
= mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:A. No one has ever beat Nancy.
Explanation:
The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.
The dormain of discourse is also known simply as universe, can also be said to be a set of entities o
upon which certain variables of interest in some formal treatment may range.
The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.
THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.
Answer:
beam with a span length of 10 ft, a width of 8 in, and an effective depth of 20 in. Normal weight concrete is used for the beam. This beam carries a total factored load of 9.4 kips. The beam is reinforced with tensile steel, which continues uninterrupted into the support. The concrete has a strength of 4000 psi, and the yield strength of the steel is 60,000 psi. Using No. 3 bars and 60,000 psi steel for stirrups, do the followings: