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ludmilkaskok [199]
2 years ago
9

For a turbulent flow of a fluid in 0.6 m diameter pipe, the velocity 0.15 m from the wall is 2.7 m/s. Estimate the wall shear st

ress using the 1/7th expression for the velocity profile.
Engineering
1 answer:
MAVERICK [17]2 years ago
3 0

Answer:

If the turbulent velocity profile in a pipe of diameter 0.6 m may be approximated by u/U=(y/R)^(1/7), where u is in m/s and y is in m and 0.15 m from the pipe.

Explanation:

hope it helps

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Write a C++ program to display yearly calendar. You need to use the array defined below in your program. // the first number is
ddd [48]

Answer:

//Annual calendar

#include <iostream>

#include <string>

#include <iomanip>

void month(int numDays, int day)

{

int i;

string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

// Header print

      cout << "\n----------------------\n";

      for(i=0; i<7; i++)

{

cout << left << setw(1) << weekDays[i];

cout << left << setw(1) << "|";

}

cout << left << setw(1) << "|";

      cout << "\n----------------------\n";

      int firstDay = day-1;

      //Space print

      for(int i=1; i< firstDay; i++)

          cout << left << setw(1) << "|" << setw(2) << " ";

      int cellCnt = 0;

      // Iteration of days

      for(int i=1; i<=numDays; i++)

      {

          //Output days

          cout << left << setw(1) << "|" << setw(2) << i;

          cellCnt += 1;

          // New line

          if ((i + firstDay-1) % 7 == 0)

          {

              cout << left << setw(1) << "|";

              cout << "\n----------------------\n";

              cellCnt = 0;

          }

      }

      // Empty cell print

      if (cellCnt != 0)

      {

          // For printing spaces

          for(int i=1; i<7-cellCnt+2; i++)

              cout << left << setw(1) << "|" << setw(2) << " ";

          cout << "\n----------------------\n";

      }

}

int main()

{

int i, day=1;

int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};

string months[] = {"January",

"February",

"March",

"April",

"May",

"June",

"July",

"August",

"September",

"October",

"November",

"December"};

for(i=0; i<12; i++)

{

//Monthly printing

cout << "\n Month: " << months[i] << "\n";

month(yearly[i][1], day);

if(day==7)

{

day = 1;

}

else

{

day = day + 1;

}

cout << "\n";

}

return 0;

}

//end

3 0
3 years ago
A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1.  Steady operating conditions exist.

2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

PP_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The  isentropic  efficiency, n_{T} = \frac{h_{3}-h_{4}  }{h_{3}- h_{4s} }

==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
3 years ago
Read 2 more answers
The purpose of pasteurizing milk is to A. Kill pathogens B. Break down milk fat C. Add vitamins and minerals D. Prevent spoilage
motikmotik
Answer: A Kill pathogens
7 0
3 years ago
I wish to have a computer whose machine-level instructions are all 32 bits each. If I want to have all instructions of the form
melisa1 [442]

Answer:

Maximum number that can be represented by 13 bits  is 8192 Instructions

Explanation:

number of instructions = 1000

number of bits = log(1000) x number of register

                          = 6 bits

Since the complete instruction must have 32 bits, then

remaining number of bits = 32 - 6 = 236

number of registers in instruction = 2

number of bits per register = 26/2 = 13

Maximum number that can be represented by 13 bits = 2^{n}

                       = 2¹³ = 8192

4 0
3 years ago
What does the line strung between the batter boards represent?
soldi70 [24.7K]

Batter boards (or battre boards, Sometimes mispronounced as "battle boads") are temporary frames, set beyond the corners of a planned foundation at precise elevations. These batter boards are then used to hold layout lines (construction twine) to indicate the limits (edges and corners) of the foundation.

7 0
2 years ago
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