1-cm-diameter opening and also exits with a 1.5-cm-diameter opening. Recognize the velocity the the water at the inlet and outlet when the mass circulation rate v the pump is 0.5 kg/s. Just how much work-related is required to run the pump?






First, we space going to need the water details volume at 15ºC: v=0.001001

. The density "
" of the water is the inverse of the specific volume:

First, think about the mass flow, i m sorry is pertained to the volumetric flow (density and velocity) and the area:


The area of every cross-section is:

(in square meters). Here, the radius was no used but the diameter, which way a department by 4 (2 squared).

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From mass circulation isolate the velocity and also calculate it:




The occupational of the pump is calculate considering an energy balance top top the pump:


Considering the isentropic procedure may give us the relation:


Applying the to the pump,


Multiplying the by the fixed flow:


The work-related is an unfavorable because that is entering come the system, yet the forced is positive. (It is just a standard rule)

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Arrange the species according come the oxidation state the nitrogen in each. Note that highest possible refers come the most positive oxidatio
castortr0y <4>


A. NO₃⁻ ⇒ +5

B. NO₄⁺ ⇒ +9

C. N₂ ⇒ 0

D. NH₂OH ⇒ -1

E. NO₂⁻ ⇒ +3


To calculation the oxidation number of an facet in a compound, we have to know the oxidation variety of the other elements. Then, we have actually to think about that the sum of the oxidation number of each atom multiply by the subscripts is same to the net fee of the compound.

A. NO₃⁻

This is the ion nitrate. Oxygen atoms (O) has actually an oxidation variety of -2 due to the fact that it derives from an oxide. In this case, the net fee of the ion is -1. Thus, we calculate the oxidation variety of N together follows:

N + (3 x (-2)) = -1

N - 6 = -1 ⇒ N= -1 + 6 = +5

B. NO₄⁺

In this case, the amount of the oxidation numbers of O and N multiplied by the subscripts is equal to +1:

N + (4 x (-2)) = +1

N - 8 = +1 ⇒ N = +1 +8 = +9

C. N₂

The oxidation number of N is 0 due to the fact that N₂ is an element substance.


The oxidation variety of H is +1 and -2 for O. The net fee of the molecule is 0.

N + (H x 3) + (O) = 0

N + (+1 x 3) + (-2) = 0

N + 3 -2 = 0 ⇒ N = 2 - 3 = -1

E. NO₂

The net fee of the ion nitrite is -1.

N + (2 x O) = -1

N + (2 x (-2)) = -1 ⇒ N = -1 + 4 = +3


Highest oxidation number = B. NO₄⁺ (+9)

Lowest oxidation number = D. NH₂OH (-1)


B. CCl₄


Solubility is figured out by the principle , "like disappear like" .

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i.e. , if a compound is polar then it will certainly dissolve in a polar compound only , and

if a compound is no - polar then it will certainly dissolve in a non - polar compound just .

Hence , native the question ,

Benzene is a no - polar link , i.e. Go not separate out into poles , and hence , it will certainly get liquified into the CCl₄ , as it is also non - polar in nature , according to the above theory of solubility.