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wariber [46]
3 years ago
8

(8 points) Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight

are given as 1:2:3 for cement, sand, and gravel with w/c of 0.50. The specific gravities of sand and gravel are 2.60 and 2.70 respectively. Entrained air content is 7.5%. How many pounds of cement, water, sand, and gravel are needed for the driveway?
Engineering
1 answer:
FromTheMoon [43]3 years ago
8 0

Answer:

Weight of cement = 10909.08 pounds

Weight of sand = 18008.64 lb

Weight of gravel = 28051.92 lb

Weight of water = 5454.54 lb

Explanation:

given data

length = 40 feet

width = 12 feet

thickness = 6 in = 0.5 feet

mix proportion = 1:2:3

specific gravity of cement = 3.15

specific gravity of sand = 2.60

specific gravity of gravel = 2.70

water to cement ratio = 0.50

solution

first we get here total volume that is

volume = length × width × thickness    ...........1

volume = 40  × 12 × 0.5

volume = 240 feet³

and

now we get here dry volume of concrete that will be

Dry volume =  wet volume ×  1.54    .......................2

Dry volume = 240 × 1.54

Dry volume =  360 feet³

and

we have given air content entrained is = 7.5 %

so required volume will be as

required volume = Dry volume × ( 1 - 7% )    .............3

required volume = 360 × ( 1 - 7.5 % )

required volume = 333 feet³  

and

total ratio is

total ratio = 1+ 2+ 3

total ratio = 6

so volume of sand cement and gravel will be

volume of  cement = \frac{1}{6} \times 333  

volume of  cement = 55.5 feet³

volume of  sand  = \frac{2}{6} \times 333

volume of  sand  = 111 feet³

volume of  gravel = \frac{3}{6} \times 333  

volume of  gravel = 166.5 feet³

and

now we get here weight of sand cement and gravel as

weight = volume ×  specific gravity × density of water   .............4

and now put here value and we get

Weight of cement = 55.5 × 3.15 × 62.4

Weight of cement = 10909.08 pounds

and

Weight of sand = 111 × 2.60 × 62.4

Weight of sand = 18008.64 lb

and

Weight of gravel = 166.5 × 2.7 × 62.4

Weight of gravel = 28051.92 lb

and

Weight of water = 0.5 of weight of cement

Weight of water  = 0.5 ×  1090.9.08

Weight of water = 5454.54 lb

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vovangra [49]

Answer:

The correct answer is "O (n\times Log n)". A further explanation is given below.

Explanation:

  • Throughout all the three instances (worst, average as well as best), the time complexity including its Merge sort seems to be O (n\times Log n) as the merge form often splits the array into two halves together tends to linear time to combine multiple halves.
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3 years ago
A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg

The quality at state 4 is determined from the condition  s_{4} =s_{3} and the entropies of the components at the condenser pressure taken from table:

 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

The enthalpy at state 2 is determined from an energy balance on the pump:

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

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Answer:

F=710KHZ

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