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nasty-shy [4]
2 years ago
8

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

e gradient, ?P/L, along the pipe.
Engineering
1 answer:
ruslelena [56]2 years ago
3 0

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

You might be interested in
6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right
dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0
2 years ago
A 227 pound compressor is supported by four legs that contact the floor of a machine shop. At the bottom of each leg there is a
Ganezh [65]

Answer:

1.312 in

Explanation:

Data provided in the question:

Weight of the compressor, W = 227 pound

Number of legs = 4

Maximum pressure = 42 psi

Now,

Let F be the force taken by the legs

Therefore,

W = 4F

or

227 pound = 4F

or

F = 56.75 pounds

Also,

Force = Pressure × Area

or

56.75 pounds = 42 psi × πr²                      [ r is the diameter of one leg]

or

r² = 0.4301

or

r = 0.656

therefore,

diameter = 2r = 2 × 0.656

= 1.312 in

6 0
2 years ago
A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses
yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature T_1 = 25° C

T{\infty} =600^0C

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}

Bi = \dfrac{2.5}{231}

Bi = 0.0108

The time constant value \tau_t is :

\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}

\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}

\tau_t = \dfrac{2702* 0.025*1033}{100}

\tau_t = 697.79

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]

where;

Q = -\Delta E _{st} which correlates with the change in the internal energy of the solid.

So;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}

The maximum value for the change in the internal energy of the solid  is :

(pVc)\theta_1 = -\Delta E _{st}max

By equating the two previous equation together ; we have:

\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{  (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

0.75=  [1-e^{\dfrac {-t}{ \tau_1}}]}

So;

0.75=  [1-e^{\dfrac {-t}{ 697.79}}]}

1-0.75=  [e^{\dfrac {-t}{ 697.79}}]}

0.25 =  e^{\dfrac {-t}{ 697.79}}

In(0.25) =  {\dfrac {-t}{ 697.79}}

-1.386294361= \dfrac{-t}{697.79}

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}

\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}

\dfrac{T - 600}{25-600}= 0.25

\dfrac{T - 600}{-575}= 0.25

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0
2 years ago
(a) If 15 kW of power from a heat reservoir at 500 K is input into a heat engine with an efficiency of 37%, what is the power ou
julsineya [31]

Answer:

(A) Power output will be 5.55 KW (b) lower temperature will be 315 K

Explanation:

We have given efficiency of heat engine \eta =37% = 0.37

Input power = 15 KW

Temperature of heat reservoir T_H=500K

(A) We know that \eta =\frac{output}{input}

So  [text]0.37=\frac{output}{15}[text]

Output = 5.55 KW

(B) We also know that [text]\eta =1-\frac{T_L}{T_H}0.37=\frac{output}{15}[text], here T_L  is lower temperature and T_H is higher temperature

So 0.37=1-\frac{T_L}{T_H}

0.37=1-\frac{T_L}{500}

T_L=315K

5 0
2 years ago
It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if King Hiero’s cr
arlik [135]

Answer: The crown is not made of pure gold.

Explanation:

Archimedes discovered that any solid, of any shape, when submerged in a liquid receives an upward force, equal to the weight of the volume of the liquid removed by the solid, which is equal to the solid volume.

So, if any body is weighed in air, the normal force will be equal to the gravity force (which we call weight) which can be expressed as follows:

Fg = m g = δ V g = 34.7 N

When submerged in water, the normal force is equal to the difference between the actual weight, and the upward force due to Archimedes' principle, called buoyant force, as follows:

Fn = Fg - Ep = δx. V. g - δH20 . V. g = 31.5 N

Dividing Fg between Fn, and simplifying common terms, we have:

δx / (δx - δh20) = 34.7 / 31.5 = 1.10

Solving for δx, we get the following value:

δx = 11,000 Kg/m3, less dense than pure gold, so we can conclude that the crown was not made of pure gold.

3 0
2 years ago
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