Use the following information to complete the concrete mixture designs for the following question.

The following materials are available for a concrete mixture design:
ASTM C 150 Type I/II Cement
Relative density of 3.15
Coarse Aggregate: well-graded 3⁄4 in. maximum-size aggregate (MSA), crushed limestone, angular
Oven-dry specific gravity: 2.6
Absorption: 0.72%
Oven-dry rodded bulk density: 102 lb/ft3 Moisture content of coarse stockpile: 0.6%
Fine Aggregate: well-graded natural sand Oven-dry specific gravity: 2.5 Absorption: 1.3%
Moisture content of fine stockpile: 3.6% Fineness Modulus: 2.2
An air entraining admixture will provide enough air for the design mixture at a dosage rate of 3 oz/ft3

QUESTION:
Concrete is required for an 8 in thick exterior concrete slab in central Texas. A specified compressive strength, f’c , of 5000 psi is required at 28 days using an ASTM C 150 Type II portland cement. The design calls for a minimum of 2 in. of concrete cover over the reinforcing steel. The minimum distance between reinforcing bars is 3 in. A slump of 3 inches should be the target. The sidewalk is located in an environment that does not require air entraining. The concrete is required to have low permeability when exposed to water and moderate sulfates. The concrete will be exposed water soluble sulfates in soil at 0.14%. No statistical data on previous mixes are available.

Determine:
a. The theoretical mixture proportions of all concrete constituents (fine aggregate, coarse aggregate, water, cement, chemical admixtures) based on 1 yd3 of concrete on an oven dry basis for aggregates.
b. The as-batched mixture proportions based on the moisture contents provided in the materials available section.
c. How much of each material would be required to make a 20 ft long by 15 ft slab from the concrete mixture described above?

Answer:

a. Heat Capacity = 1.756J/mol-K

b. Heat Capacity = 24.942J/mol-k

Explanation:

Given

Constant volume Cv = 0.81J/mol-k

T1 = 34K

Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K

First, The value of the temperature-independent constant.

Using Cv = AT³

Make A the subject of formula

A = Cv/T³

Substitute each values

A = 0.81/34³

A = 0.000020608589456543

A = 2.061 * 10^-5J/mol-k

The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature

Cv = AT³

So, The heat capacity when T = 44k is then calculated as

Cv = 2.061 * 10^-5 * 44³

Cv = 1.755522084266232

Cv = 1.756J/mol-K

(b) at 477 K.

Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:

Cv = 3R

Where R = universal gas constant

R = 8.314J/mol-k

Cv = 3 * 8.314

Cv = 24.942J/mol-k

Answer:

a) 1.75 J/mol-k

b) 24.94 J/mol-k

Explanation:

We are given:

Cv = 0.81J/mol-k

Debye temperature = 306k

Heat capacity varies with temperature and the relationship between them is given as:

Cv = AT³

To find temperature independent constant A, we have:

[tex] A = \frac {C_v}{T^3} [/tex]

[tex] A = \frac{0.81J/mol-k}{34^3} [/tex]

[tex] A = 2.06*10^-^5 [/tex]

a) for temperature at 44k

Cv = AT³

[tex] C_v = 2.06*10^-^5 * 44^3 [/tex]

Cv = 1.75 J/mol-k

b) for T at 477k

Since the temperature 447k is higher than Debye's temperature, the specific heat will be given as:

Cv = 3R

Where R = universal gas constant = 8.314 J/mol-k

Therefore,

Cv = 3 * 8.314 J/mol-k

Cv = 24.94 J/mol-k

Answer:

Check the explanation

Explanation:

The desired Cold working (which is any process or procedure of metalworking in which the metal body is formed at a lower temperature beyond its recrystallization, typically at the ambient temperature.) tensile strength in excess of 750 MPa and a ductility of at least 12 %EL can be seen in the attached image below.

Answer:

a) see attachment

b) A= m0m1+ m1m2+ m0m2

see attachment for K-map

c) see attachment

Explanation:

a) see attachment for truth table

b) see attachment for k-map

A= m0m1+ m1m2+ m0m2

c) see attachment for gate level circuit

Answer:

A student wants to determine experimentally, without disconnecting any wires in the circuit, the DC current moving through a copper wire. Which of the following items of laboratory equipment would be sufficient to make the necessary measurements for this determination?

(E) Voltmeter and current sensor

Explanation:

DC (direct current) is the unidirectional stream or development of electric charge transporters (which are generally electrons). The power of the current can shift with time, yet the general direction of development remains the equivalent consistently. As a descriptor, the term DC is utilized in reference to voltage whose extremity never turns around. In a DC circuit, electrons rise up out of the negative, or less, shaft and move towards the positive, or additionally, post. All things considered, physicists characterize DC as making a trip from in addition to short.

A voltmeter is an instrument utilized for estimating electrical potential contrast between two focuses in an electric circuit. Simple voltmeters move a pointer over a scale in relation to the voltage of the circuit; advanced voltmeters give a numerical presentation of voltage by utilization of a simple to computerized converter. A voltmeter in a circuit outline is spoken to by the letter V around. Voltmeters are made in a wide scope of styles. Instruments for all time mounted in a board are utilized to screen generators or other fixed mechanical assembly. Versatile instruments, typically prepared to likewise gauge flow and obstruction as a multimeter, are standard test instruments utilized in electrical and hardware work. Any estimation that can be changed over to a voltage can be shown on a meter that is reasonably adjusted; for instance, pressure, temperature, stream or level in a compound procedure plant.

A current sensor is a gadget that identifies electric current in a wire and creates a sign relative to that current. The produced sign could be simple voltage or current or even a computerized yield. The produced sign can be then used to show the deliberate current in an ammeter, or can be put away for additional investigation in an information securing framework, or can be utilized with the end goal of control. Current sensors are either open-or shut circle. Open-circle current sensors measure air conditioning and DC currents and give electrical confinement between the circuit being estimated and the yield of the sensor (the essential current is estimated without electrical contact with the essential circuit, giving galvanic detachment). More affordable than their shut circle cousins, open-circle current sensors are commonly favored in battery-controlled circuits given their low-working force prerequisites and little impression highlights.

Answer:

17060700 btu/h

Explanation:

Power output from power plant = 150 kW.

This power output is at 3% efficiency, this means power available in pond is 0.30P

Now, 150 kW = 0.03P

P = 150/0.03 = 5000 kW of power in the pond.

From basic conversion,

1 kW = 3412.14 btu/h

5000 kW = 5000 x 3412.14

average value of the required solar energy collection rate equal

17060700 btu/h

Solar pond power plant with 3% efficiency and 150 kW output needs ~17,060,000 Btu/h solar energy collection rate.

To find the average value of the required solar energy collection rate in Btu/h, we need to calculate the total solar energy collected per unit time and then convert it into Btu/h.

First, let's find the total solar energy collected per unit time:

Given:

- Net power output = 150 kW

- Efficiency = 3%

We know that efficiency is the ratio of output power to input power, so:

Efficiency = (Net Power Output / Solar Energy Input) * 100

Rearranging the equation to find the solar energy input:

Solar Energy Input = (Net Power Output / Efficiency) * 100

Now, let's substitute the given values:

Solar Energy Input = (150 kW / 3%) * 100

Solar Energy Input = (150,000 W / 0.03) * 100

Solar Energy Input = 5,000,000 W

Now, we convert this into Btu/h. To do this, we'll use the conversion factor:

1 watt = 3.412 Btu/h

So, the solar energy input in Btu/h will be:

Solar Energy Input (Btu/h) = (Solar Energy Input in watts) * (3.412 Btu/h per watt)

Solar Energy Input (Btu/h) = 5,000,000 W * 3.412 Btu/h per watt

Solar Energy Input (Btu/h) ≈ 17,060,000 Btu/h

Therefore, the average value of the required solar energy collection rate is approximately 17,060,000 Btu/h.

Answer:

Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s

Explanation:

du/u² = dt = ∫du/2.3183 = ∫dt

0.4313 u = t + c

(a) t = 0, u= 15m/s, c = 0.647

u = t+c/0.4313 = t + 0.647/0.4313

(a) when t= 1   u = 1+ 0.647/0.4313 = 3.8187m/s

(b) when t= 10   u = 10 + 0.647/0.4313 = 24.0858m/s

(c)when t= 1000  u = 1000 + 0.647/0.4313 = 3220.071m/s

Answer:

Explanation:

The solution code is written in Matlab.

Firstly, we initialize n to 10 and set n to f variable (Line 1-2).

Next, create a while loop that will continue to loop until n is equal or smaller than one. In the loop, decrement n by one and multiply n with current f_variable.

At last display value of f. The output is  3628800.

Answer:

See explaination

Explanation:

Looking at Reynolds number, we can go ahead and describe the Reynolds number as a dimensionless value that is used to determine whether the fluid is exhibiting laminar flow (R less than 2300) or turbulent flow (R greater than 4000). Laminar flow is when a fluid moves smoothly and is predictable.

Check attachments

Answer:

x =5

y =6

Explanation:

The reason behind that is that while using the reference P2 we have setted the values of X and Y both as 5 and 6 respectively in the main function.

After setting we are calling the setXY function which sets the value to 3 and 4 for X and Y respectively but does not implement in the value to the instance.

Apart from which setX and setY function are setting the values with respect to the instance variable using "this" keyword.

Hence, the values 5 and 6 are been displayed at the end of the main method.

Answer:

The answer which is a calculation can be found as an attached document

Explanation:

Answer:

Answer is -286.78 kJ/kg

Refer below.

Explanation:

Refer to the pictures for brief explanation.