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ladessa [460]
2 years ago
7

Q5. A hypothetical metal alloy has a grain diameter of 2.4 x 10-2 mm. After a heat treatment at 575°C for 500 min, the grain dia

meter has increased to 7.3 x 10-2 mm. Compute the time required for a specimen of this same material (i.e., d0 = 2.4 x 10-2 mm) to achieve a grain diameter of 5.5 x 10-2 mm while being heated at 575°C. Assume the n grain diameter exponent has a value of 2.2. (10 points)
Engineering
1 answer:
alexdok [17]2 years ago
8 0

Time is 244.89 minutes

<u>Explanation:</u>

<u />

Given:

Hypothetical diameter, d₀ = 2.4 X 10⁻² mm

Increase in diameter, d = 7.3 X 10⁻² mm

d₀ = 2.4 X 10⁻² mm

Time, t = 500 min

To solve K:

K = \frac{d^n - d_o^n}{t}

On substituting the value:

K = \frac{(7.3 X 10^-^2)^2^.^2 - (2.4 X 10^-^2)^2^.^2}{500} \\\\\\K = 5.8 X 10^-^6 mm^2^.^2/min

From the value of K, t can be calculated as:

t = \frac{d^2^.^2 - d_o^2^.^2}{K} \\\\t = \frac{(5.5 X 10^-^2)^2^.^2 - (2.4 X 10^-^2)^2^.^2}{5.8 X 10^-^6} \\\\t = 244.89 min

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simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The
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Answer:

A. ) 591.7 v

B.) 991.7v

C.) 59.7%

D.) 47.9 Kw

E.) 247925 W

F.) 59.7 %

Explanation:

Given that a simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The transmission line is 100 km copper wire of 3 cm diameter. If the voltage at the load side is 400 V,

Let first calculate the resistance in the wire.

The resistivity (rho) of a copper wire is 1.673×10^-8 ohm metres

Resistance R =( L× rho)/A

Where Area = πr^2 = π × 0.015^2

Area = 0.00071 m^2

R = (100000 × 1.673×10^-8) / 0.00071

Resistance in wire = 2.367 ohms

Then let calculate the resistance in the load.

Also, since Power P = V^2 /R

Make R the subject of formula

R = V^2/ P

R = 400^2/100000

Resistance in load = 1.6 Ohms

Current l = V / R

I = 400/1.6 = 250 Ampere

a.) Voltage drop across the line V line will be achieved by using Ohms law.

V = I R

V = 250 × 2.367

V = 591.7 v

B.) Voltage at the source side Vsource will be

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V = 400 + 591.7

V = 991.7 v

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591.7/991.7 × 100 = 59.7%

D.) Line losses

P = I V

P = 250 × 591.7

P = 147925 W

Power loss = 147925 - 100000

Power loss = 47,925 W

Power loss = 47.9 Kw

E.) Power delivered by the source

P = IV

P = 250 × 991.7 = 247925 W

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Efficiency = 59.7 %

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Answer:

The Net Present Worth at i 5% of Super Tools new test equipment is -$33,992.45

Explanation:

I calculated the net present based on cash flow projections that $300000 in years 1 and 2 ,declining by $50000 every year and it terminated year 8 with zero cash inflow.

I then discounted the cash flows using discounting factor calculated as (1+i)^n where i is the interest rate of 5% given as the n is the relevant year of cash flow.

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2 years ago
1. Consider the steady flow in a water pipe joint shown in the diagram. The areas are:
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Answer:

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Explanation:

Assuming steady and incompressible flow and uniform properties at each section

V_1A_1+V_2A_2+A_3V_3+Q_4=0

Here V is velocity of flow and A is area, Q is flow rate out of the leak, subscript 1-4 represent different sections

At the surface,  is negative hence the equation above will be

-V_1A_1+V_2A_2+A_3V_3+Q_4=0

Making  the subject of the formula then

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Substituting the given values then

V_2=\frac {(5\times 0.2)-(12\times 0.15)-0.1}{0.2}=-4.5 m/s\\V_2=-4.5m/s

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

Otto cycle T-S diagram

T₂ = 288.706*6.25^{0.393} = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}}  \right )^{K-1}

T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

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