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3 years ago

6

For precipitation hardening, how many phases (in numeric form) exist (a) following the solution heat treatment, and (b) followin

g the precipitation heat treatment

1 answer:

Nitella [24]3 years ago

4 0

The correct answer is B

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a centrifugal pump delivers water at the rate of 50kg/is . the inlet and outlet pressure are 2 bar and 6.2 bar respectively. the

Monica [59]

Answer:

21.6 kw

Explanation:

Given data:

m = 50 kg/s

Inlet pressure (p1) = 2 bar

outlet pressure(p2) = 6.2 bar

suction ( h1 ) = -2.2 m

delivery ( h2 ) = 8.5 m

d1 = 200 mm = 0.2 m

d2 = 100 mm = 0.1 m

Vs of water = 0.001 m^3/kg

next we have to determine the Q value

Q = V*A

Q = 0.001 * 50 = 0.05 m^3/s

next we have to calculate the various V's

V1 = Q/A1 = $\frac{0.05*4}{\pi *0.2^2}$ = 1.59 m/s

V2 = Q/A2 = $\frac{0.05*4}{\pi *0.1^2}$ = 6.37 m/s

Determine the capacity of the electric motor

attached below is the detailed solution

7 0

3 years ago

You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?

Leviafan [203]

Answer:

18 Watts

Explanation:

For this problem, we simply need to understand the relationship of power to voltage and current. This relationship is derived from Ohm's law:

Power = Voltage * Current

Given this equation, we can say the following to find the power consumption of the motor:

Power = 12volts * 1.5amps

Power = 18 Watts

Hence, the motor is consuming 18 Watts of power.

Cheers.

3 0

3 years ago

Please read the short case at the end of Chapter 8, "Mary Barra of General Motors Values Culture". Use the following prompts to

Tom [10]

Answer: Incoherent question

Explanation: This is an act of plagiarism at subjecting the tutor to unnecessary stress at answering the purported question.

4 0

3 years ago

Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made

Zina [86]

Answer:

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

Explanation:

Since, the position of a car is random.Therefore, number ways in which cars can line up is given as:

<u>No. of ways = 20! = 2.43 x 10^18</u>

For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!)/20!

<u>Probability = 0.00001 = 0.001%</u>

4 0

3 years ago

Cooling water for a chemical plant must be pumped from a river 2500 ft from the plant site. Preliminary design cans for a flow o

Vesna [10]

Answer:

The power cost savings for the 8 inches pipe offsets the increased cost for the pipe therefore to save costs the 8-inch pipe should be used

Explanation:

The given parameters in the question are;

The distance of the river from the the site, d = 2,500 ft.

The planned flow rate = 600 gal/min

The diameter of the pipe, d = 6-in.

The pipe material = Steel

The cost of pumping = 3 cents per kilowatt-hour

The Bernoulli's equation is presented as follows;

$\dfrac{P_a}{\rho} + g\cdot Z_a + \dfrac{V^2_a}{2} + \eta\cdot W_p = \dfrac{P_b}{\rho} + g\cdot Z_b + \dfrac{V^2_b}{2} +h_f +W_m$

${P_a} = {P_b} = Atmospheric \ pressure$

$Z_a = Z_b$

Vₐ - 0 m/s (The river is taken as an infinite source)

$W_m$ = 0

The head loss in 6 inches steel pipe of at flow rate of 600 gal/min = 1.19 psi/100 ft

Therefore; $h_f$ = 1.19 × 2500/100 = 29.75 psi

$\eta\cdot W_p = \dfrac{V^2_b}{2} +h_f$

$V_b$ = Q/ $A_b$ = 600 gal/min/(π·(6 in.)²/4) = 6.80829479 ft./s

$V_b$ ≈ 6.81 ft./s

The pressure of the pump = P = 62.4 lb/ft³× (6.81 ft./s)²/2 + 29.75 psi ≈ 30.06 psi

The power of the pump = Q·P ≈ 30.06 psi × 600 gal/min = 7,845.50835 W

The power consumed per hour = 7,845.50835 × 60 × 60 W

The cost = 28,243.8301 kW × 3 = $847.31 per hour

Annual cost = $847.31 × 8766 = $7,427,519.46

Pipe cost = $15/ft × 2,500 ft = $37,500

Annual charges = 20% × Installed cost = 0.2 × $37,500 = $7,500

Total cost = $37,500 + $7,500 + $7,427,519.46 = $7,475,519.46

For the 8-in pipe, we have;

$V_b$ = Q/ $A_b$ = 600 gal/min/(π·(8 in.)²/4) = 3.83 ft./s

$h_f$ = 1.17 ft/100 feet

Total head loss = (2,500 ft/100 ft) × 1.17 ft. = 29.25 ft.

$h_p = \dfrac{V^2_b}{2 \cdot g} +h_f$

∴ $h_p$ = (3.83 ft./s)²/(2 × 32.1740 ft/s²) + 29.25 ft. ≈ 29.5 ft.

The power of the pump = ρ·g·h × Q

Power of pump = 62.4 lb/ft³ × 32.1740 ft/s² × 29.5 ft.× 600 gal/min = 3,363.8047 W

The cost power consumed per annum = 3,363.8047 W × 60 × 60 × 3 × 8766 = $3,184,608.1

The Cost of the pipe = $20/ft × 2,500 ft. = $50,000

The total cost plus charges = $3,184,608.1 + $50,000 + $10,000 = $3,244,608.1

Therefore it is more affordable to use the 8-in pipe which provides substantial savings in power costs

8 0

3 years ago