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2 years ago

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The flow of a liquid in a 2 inch nominal diameter steel pipe produces a pressure drop due to friction of 78.86 kPa. The length o

f pipe is 40 m and the mean velocity is 3 m/s. if the density of the liquid is 1000 kg/m^3, then a) determine the reynolds number b) determind if the flow is laminar or tubulent c) compute viscosity of the liquid d) compute the mass flow rate (assume ?, equivalent roughness factor, for Steel pipe to be 45.7 x 10-6 m)

1 answer:

Anastasy [175]2 years ago

5 0

Answer:

Explanation:

The detailed steps and careful analysis is as shown in the attached file.

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Write cout statements with stream manipulators that perform the following:

Semenov [28]

Answer:

A)cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B)cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C)cout<<fixed<<5.789E12;

D)cout<<left<<setw(7)<<67;

Explanation:

Stream Manipulators are functions specifically designed to be used in conjunction with the insertion (<<) and extraction (>>) operators on stream objects in C++ programming while the 'cout' statement is used to display the output of a C++to the standard output device.

setw: used to specify the minimum number of character positions on the output field

setprecision: Sets the decimal precision to be used to format floating-point values on output operations.

fixed: is used to set the floatfield format flag for the specified str stream.

left: adjust output to the left.

A) To display the number 34.789 in a field of eight spaces with two decimal places of precision. cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B) To display the number 7.0 in a field of six spaces with three decimal places of precision. cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C) To print out the number 5.789e+12 in fixed-point notation. cout<<fixed<<5.789E12;

(D) To display the number 67 left-justified in a field of six spaces. cout<<left<<setw(7)<<67;

7 0

2 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

2 years ago

A pair of spur gears with 20 degree pressure angle, full-depth, involute teeth transmits 65 hp. The pinion is mounted on a shaft

vfiekz [6]

Answer:

See explaination

Explanation:

a) The pitch circle diameter of pinion in inches is given by

Dp=Np/P

Where

Np= No. of teeth in pinion = 26

P =diametral pitch= 6

Hence

Dp= 26/6 = 4.333 in

Pinion angular speed\omega _{p} =1250 rpm = 130.9 rad/s

Therefore pitch line speed

V=Dp/2\omega _{p} = 4.333/2x130.9

= 283.62 in/s

V= 23.63 ft/s

V=1418 ft/min

b) The pitch circle diameter og gear is given by

Dg= Ng/P= 48/6 = 8 in

The center distance is given by

C=(Dp+Dg)/2

= (4.333+8)/2

C= 6.167 in

c) The torque on the pinion is given by

Tp= P/\omega _{p}

Where

P = transmitted power, =65 hp = 65x550= 35750 lt-lb/s

Tp= 35750/130.9

= 273.1 ft-lb

d) Speed ratio is given as

R=Ng/Np= 48/26 = 1.8461

Hence speed of gear is

\omega _{g}=\frac{\omega _{p}}{R}

= 130.9/1.8461

= 70.9 rad/s

Therefore torque on gear is given as

Tg= P/\omega _{g} = 35750/70.9

= 504.2 ft-lb

e) Assuming transmission eficiency of 100%

Output hp=input hp= 65 hp

f) Tangential force on gear teeth is given by

Fgt= Tg/(Dg/2)

= 504.2x2/8

= 126.05 lb

g) Radial force on ger teeth is given as

Fgr= Fgt tan\phi

Where

\phi is pressure angle = 200

Hence

Fgr= 126.05tan200

= 45.88 lb

h) The normal force on gear teeth is given as

F=Fgt/cos\phi

= 126.05/cos200

= 134.14 lb

4 0

2 years ago

Air in tankBis at 200 kPa, 280 K and mass 1 kg. It is connected to an empty piston/cylinder with a float pressure of 400 kPa sim

Romashka [77]

Answer:

Explanation: see attachment

7 0

2 years ago

Outline an algorithm in **pseudo code** for checking whether an array H[1..n] is a heap and determine its time efficiency.

svlad2 [7]

Answer:

Condition to break: $H[j] \geq max {H[2j] , H[2j+1]}$

Efficiency: O(n).

Explanation:

Previous concepts

Heap algorithm is used to create all the possible permutations with K possible objects. Was created by B. R Heap in 1963.

Parental dominance condition represent a condition that is satisfied when the parent element is greater than his children.

Solution to the problem

We assume that we have an array H of size n for the algorithm.

It's important on this case analyze the parental dominance condition in order to the algorithm can work and construc a heap.

For this case we can set a counter j =1,2,... [n/2] (We just check until n/2 since in order to create a heap we need to satisfy minimum n/2 possible comparisions $and we need to check this:Break condition: [tex]H[j] \geq max {H[2j] , H[2j+1]}$

And we just need to check on the array the last condition and if is not satisfied for any value of the counter j we need to stop the algorithm and the array would not a heap. Otherwise if we satisfy the condition for each $j =1,2,.....,[n/2]p$ then we will have a heap.

On this case this algorithm needs to compare 2*(n/2) times the values and the efficiency is given by O(n).

3 0

2 years ago