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3 years ago

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3

Engineering

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

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Problem 7.16 (GO Tutorial) A cylindrical bar of steel 10.4 mm (0.4094 in.) in diameter is to be deformed elastically by applicat

shepuryov [24]

Answer:

P = 18035.25 N

Explanation:

Given

D = 10.4 mm

ΔD = 3.2 ×10⁻³ mm

E = 207 GPa

ν = 0.30

σ = P/A

A = 0.25*π*D²

σ = E*εx

ν = - εz / εx

εz = ΔD / D

We can get εx as follows

εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴

Now we find εx

ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³

then

σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa

we have to obtain A:

A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²

Finally we apply the following equation in order o get P

σ = P/A ⇒ P = σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N

4 0

3 years ago

Evaluate (204 mm)(0.004 57 kg) / (34.6 N) to three

vesna_86 [32]

Answer:

the evaluation in SI unit will be $2.69\times 10^{-5}sec^{2}$

Explanation:

We have evaluate $\frac{(204mm\times 0.00457kg)}{34.6N}$

We know that 1 mm $=10^{-3}m$

So 240 mm $=204\times 10^{-3}m$

Newton can be written as $kgm/sec^2$

So $\frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}$

So the evaluation in SI unit will be $2.69\times 10^{-5}sec^{2}$

4 0

3 years ago

Write the work of an architectual engineer????

valentinak56 [21]

An architectural engineer is a person who is responsible for studying and evaluating construction blueprints and designs. This is to determine if the design is feasible and if it is feasible to construct. Building styles and aesthetics should not be prioritized over the structural integrity and safety of a proposed building. Plan, construct, and prototype buildings, and be able to identify, analyze, and resolve issues with them as they arise.

<em>Hope this helps :)</em>

3 0

2 years ago

Read 2 more answers

My computer has a mass of 0.031080997078386 slug the Earth's surface.

poizon [28]

Answer:

The answer is "0.187 lbm and 1 lbf".

Explanation:

The mass = $0.031080997078386\ slug$

Calculating mass on Mars:

$\to m=m_g\frac{g}{g_e}$

$=0.031080997078386 \times \frac{32.2}{5.35}\\\\=0.187 \ lbm$

$\to W=mg_e$

$=0.187 \times 5.35\\\\=1 \ lbf$

4 0

3 years ago

The device shown below contains 2 kg of water. The cylinder is allowed to fall 800 m during which the temperature of the water i

olganol [36]

Answer:

m_added = 2 kg

Explanation:

From the question, we are told that the cylinder is allowed to fall 800 m in height. Thus, the potential energy will be converted into heat energy which will increase the temperature of water .

Now, let the mass of the falling cylinder be denoted by "m1" and let h be the height of fall.

Thus;

Formula for potential energy = mgh

Thus, as said earlier it's converted to heat generated. So heat generated = m1gh

Now let's calculate the heat absorbed;

heat absorbed = (m2)cΔt

Where;

ΔT is change in temperature

c is specific heat of water .

m2 is mass of water

Heat absorbed = heat generated

Thus;

(m2)cΔt = m1gh

Δt = m1gh/(m2•c)

Now, in both cases of the water and cylinder, m1, g , h and c are constant

Thus, we have;

Δt = (m1gh/m2) × 1/c

Where;

(m1gh/m2) is denoted as a constant k.

Thus;

Δt = K/m

For the first experiment, we have;

m = 2 kg

Δt = 2.4

Thus;

2.4 = K/2

Multiply both sides by 2 to get;

K = 4.8

For the second experiment, we have;

Δt = 1.2

Also,we have seen that K = 4.8

Thus;

Δt = K/m

Thus;

1.2 = 4.8/m

m = 1.2

m = 4 kg

Thus,mass added is;

m_added = 4 - 2

m_added = 2 kg

6 0

3 years ago