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AlexFokin [52]
3 years ago
6

A liquid with a specific gravity of 2.6 and a viscosity of 2.0 cP flows through a smooth pipe of unknown diameter, resulting in

a pressure drop of 0.183 lb/in? for 1.73 mi. What is a pipe diameter in inches if the mass rate of flow is 7000 lb/h?
Engineering
1 answer:
sasho [114]3 years ago
6 0
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You might be interested in
The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%
Aleonysh [2.5K]

Answer:

4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C  having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

4 0
3 years ago
At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut
professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ \frac{1}{\mu(1-\frac{\lambda}{\mu})}

thus,

on substituting the respective values, we get

Average time spent by the vehicle = \frac{1}{60(1-\frac{30}{60})}

or

Average time spent by the vehicle = \frac{1}{60(1-0.5)}

or

Average time spent by the vehicle = \frac{1}{60(0.5)}

or

Average time spent by the vehicle = \frac{1}{30} hr/veh

or

Average time spent by the vehicle = \frac{1}{30}\times60 min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0
3 years ago
An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60
belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F              hi = 43.43 kJ/kg

WBT = 60 F             wi = 0.00874 kJ/kg

                                u1 = 0.8445 m/kg

oulet condition,

DBT = 70 F        RH = 100.1

h1 = 83.504kJ/kg

Wo = 0.222kJ/kg

check the attached file for complete solution

3 0
3 years ago
Multiply. Write the answer in simplest form. 1 3/10×1/8
kicyunya [14]

9514 1404 393

Answer:

  13/80

Explanation:

The product is ...

  (1 3/10)×(1/8) = (13/10)×(1/8) = (13×1)/(10×8) = 13/80

4 0
2 years ago
A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th
Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0
3 years ago
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