Hi plz delete this question i had to edit it cuz it was wrong question

Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur

ruslelena [56]

Answer:

$\frac{\delta p }{l} = 30.4 lb/ft^3$

Explanation:

Given data:

flow rate = 10 gallon per minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q = VA

solve for velocity V = \frac{Q}{A}[/tex]

$V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}$

V = 7.27 ft/sec

we know that Reynold number

$Re = \frac{VD}{\nu}$

$Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}$

$Re = 3.76 \times 10^4$

calculate the $\frac{\epsilon }{D}$ ratio to determine the fanning friction f

$\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008$

from moody diagram f value corresonding to Re and $\frac{\epsilon }{D}$ is 0.037

for horizontal pipe

$\delta p = \frac{f l \rho v^2}{2D}$

$\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}$

where 1.94 slug/ft^3is density of water

$\frac{\delta p }{l} = 30.4 lb/ft^3$

3 0

3 years ago

A natural-draft cooling tower receives 250,000 ft3/min of air at standard atmospheric pressure, 70oF, and 45 percent relative hu

notsponge [240]

Find the attachment for complete solution

5 0

3 years ago

Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

3 years ago

What are the benefits of using the engineering design process

Shalnov [3]

Answer:

Some of the benefits are tangible for they are visible in the design and production process, while the other benefits are intangible which may not be visible directly but result in improvement in the quality of product, better control over designing and production process, reduction of stress on the designers etc.

7 0

3 years ago

ILL GIVE BRAINLIEST!!!

Sedbober [7]

Get the app socratic I saw the answer to your question on the app but I ran out of screen time to show you

6 0

2 years ago