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Juliette [100K]

2 years ago

10

Which material has the highest cp value?

1 answer:

agasfer [191]2 years ago

6 0

The answer is copper. With 10 XP weighing one pound

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The Cv factor for a valve is 48. Compute the head loss when 30 GPM of water passes through the valve.

dlinn [17]

Answer:

The head loss in Psi is 0.390625 psi.

Explanation:

Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.

Given:

Factor cv is 48.

Flow rate of water is 30 GPM.

GPM means gallon per minute.

Calculation:

Step1

Expression for head loss for the water is given as follows:

$c_{v}=\frac{Q}{\sqrt{h}}$

Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.

Step2

Substitute 48 for cv and 30 for Q in above equation as follows:

$48=\frac{30}{\sqrt{h}}$

${\sqrt{h}}=0.625$

h = 0.390625 psi.

Thus, the head loss in Psi is 0.390625 psi.

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3 years ago

An isothermal CSTR with a first order irreversible reaction A â> B (and rA[mol/(ft3*min)] = - 0.5 CA) has a constant flow rat

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What type of spring is mounted on a mcpherson strut suspension system?

AysviL [449]

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Explanation:

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2 years ago

Why are funeral home adding dining facilities to the business

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A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

3 years ago