ENGINEERING COLLEGE

An interest buydown program offers to reduce interest rates by 4% from the base rate. Suppose the base rate for a loan of $8000 is 8% for 10 years. What is the monthly payment before and after the buydown? In this case, use monthly compounding, that is, the term is 120 payment periods, and the interest per month is 0.667% before and 0.333% after the buydown.

Answers

Answer 1

Answer:

Answer: The monthly payment before the buydown is $71.3

The monthly payment after the buydown is $68.9

Explanation: The payment is compounding so we use compound interest;

A= P[1+(r/n)^nt]

Where;

A= Compounded amount

P = principal

r= interest rate per payment

n= number of payment per period

t= number of period.

NOTE: from our questions, the period is yearly and the payment is monthly. Therefore;

number of payment per period (n) is 12

number of payment period (t) is 10

P=$8000, r= 0.667% or 0.333%

FIND MONTHLY PAYMENT BEFORE BUYDOWN:

Step 1: find the Compounded amount to pay.

A= $8000[1+(0.00667÷12)^(12×10)]=

$8551.64 this is the total amount he has to pay for a period of 10years

Step 2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months

$8551.64÷120= $71.3 monthly

FIND MONTHLY PAYMENT AFTER BUYDOWN:

Step 1: find the compounded amount to pay.

A= 8000[1+(0.00333÷12)^(12×10)=

$8270.85 this is the total amount he has to pay for a period of 10years

Step2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months;

$8270.85÷120= $68.9 monthly

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Assignment 1: Structural Design of Rectangular Reinforced Concrete Beams for Bending Perform structural design of a rectangular reinforced concrete beam for bending. The beam is simply supported and has a span L=20 feet. In addition to its own weight the beam should support a superimposed dead load of 0.50 k/ft and a live load of 0.65 k/ft. Use a beam width of 12 inches. The depth of the beam should satisfy the ACI stipulations for minimum depth and be proportioned for economy. Concrete compressive strength f’c = 4,000 psi and yield stress of reinforcing bars fy = 60,000 psi. Size of stirrups should be chosen based on the size of the reinforcing bars. The beam is neither exposed to weather nor in contact with the ground, meaning it is subjected to interior exposure.
• Use the reference on "Practical Considerations for Rectangular Reinforced Concrete Beams"
• Include references to ACI code – see slides from second class
• Include references to Tables from Appendix A
• Draw a sketch of the reinforced concrete beam showing all dimensions, number and size of rebars, including stirrups.

Answers

Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL

Thus;

W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"

Thus;

d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²

Hence;

ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.

Consider the products you use and the activities you perform on a daily basis. Describe three examples that use both SI units and customary units for measurement.

Answers

Three activities that I can do on a daily basis that involve both metric units (SI units) and customary units are: measuring the length of a door with a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that calls for one teaspoon (customary unit) of baking soda, which can also be converted to four grams (SI unit); and weighing myself on a weighing scale, which can be measured in pound and kilogram (metric unit).

Answer: Three examples of activities that I can perform on a daily basis that involves both metric units (SI units) and customary units include: measuring the length of a door using a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that requires one teaspoon (customary unit) of baking soda, which could also be converted into four grams (SI unit); weighing myself on a weighing scale, which can be measured by pounds (customary unit) or kilograms (metric unit).

Explanation:I big brain :) (Not Really I Just Wanted To Help) I hope this helped! ;)

An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Answers

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .

A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa

Answers

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto (1)/(V)$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 150 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v L

$V_2$ = final volume of gas = $(v)/(3)L$

$150* v=P_2* (v)/(3)$

$P_2=450kPa$

Therefore, the new pressure of the gas will be 450 kPa.

Internal and external flow boiling regimes?

Answers

Answer:

The flow boiling is also classified as either external and internal flow boiling depending on whether the fluid is forced to flow over a heated surface or inside a heated channel. The two-phase flow in a tube exhibits different flow boiling regimes, depending on the relative amounts of the liquid and the vapor phases.

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C?

Answers

Answer:

$\eta _(max) = 0.2413 = 24.13%$

$\eta' _(max) = 0.5061 = 50.61%$

Given:

$T_(1max) = 100^(\circ) = 273 + 100 = 373 K$

operating temperature of heat engine, $T_(2) = 10^(\circ) = 273 + 10 = 283 K$

$T_(3max) = 300^(\circ) = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _(max)$ is given by:

$\eta _(max) = 1 - (T_(2))/(T_(1max))$

$\eta _(max) = 1 - (283)/(373) = 0.24$

$\eta _(max) = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_(3max)$ changes to $300^(\circ)$ , so, the new maximum efficiency, $\eta' _(max)$ is given by:

$\eta' _(max) = 1 - (T_(2))/(T_(3max))$

$\eta _(max) = 1 - (283)/(573) = 0.5061$

$\eta _(max) = 0.5061 = 50.61%$

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