ENGINEERING COLLEGE

For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration to 0.38 wt% at a point 2.3 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 5.3 mm position for an identical steel and at the same carburizing temperature.

Answers

Answer 1

Answer:

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

Related Questions

A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.
A furnace is shaped like a long equilateral triangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is ε1 = 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?
A chemical process converts molten iron (III) oxide into molten iron and carbon dioxide by using a reducing agent of carbon monoxide. The process allows 10.08 kg of iron to be produced from every 16.00 kg of iron (III) oxide in an excess of carbon monoxide. Calculate the percentage yield of iron produced in this process.
Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cooled water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T
Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate: a. the thermal efficiency of the cycle b. the back work ratio c. the net power developed, in kW d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.

A power plant burns natural gas to supply heat to a heat engine which rejects heat to the adjacent river. The power plant produces 800 MW of electrical power and has a thermal efficiency of 38%. Determine the heat transfer rates from the natural gas and to the river, in MW.

Answers

A. The heat transfer rate from natural gas is 2105.26 MW

B. The heat transfer rate to river is 1305.26 MW

Efficiency formula

Efficiency = (power output / power input) × 100

A. How to determine the heat transfer from natural gas

Efficiency = 38%
Power output = 800 MW

Power input =?

Power input = Power input / efficiency

Power input = 800 / 38%

Power input = 800 / 0.38

Power input = 2105.26 MW

Thus, the heat transfer from natural gas is 2105.26 MW

B. How to determine the heat transfer to the river

Total heat = 2105.26 MW
Heat used by plant = 800 MW
Heat to the river =?

Heat to the river = 2105.26 – 800

Heat to the river = 1305.26 MW

Learn more about efficiency:

brainly.com/question/2009210

Answer:

heat transfer from natural gas is 2105.26 MW

heat transfer to river is 1305.26 MW

Explanation:

given data

power output Wn = 800 MW

efficiency = 38%

solution

we know that efficiency is express as

$\eta = (Wn)/(Qin)$ ......................1

put here value we get

38% = $(800)/(Qin)$

Qin = 2105.26 MW

so heat supply is 2105.26

so we can say

Wn = Qin - Qout

800 = 2105.26 - Qout

Qout = 2105.26 - 800

Qout = 1305.26 MW

so heat transfer from natural gas is 2105.26 MW

and heat transfer to river is 1305.26 MW

An electric current of 237.0 mA flows for 8.0 minutes. Calculate the amount of electric charge transported. Be sure your answer has the correct unit symbol and the correct number of significant digits x10

Answers

Answer:

amount of electric charge transported = 1.13 × $10^(-2)$ C

Explanation:

given data

electric current = 237.0 mA = 0.237 A

time = 8 min = 8 × 60 sec = 480 sec

solution

we get here amount of electric charge transported that is express as

amount of electric charge transported = electric current × time ...........1

put here value and we get

amount of electric charge transported = 0.237 × 480

amount of electric charge transported = 113.76 C

amount of electric charge transported = 1.13 × $10^(-2)$ C

) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).

Answers

Answer:

Va / Vb = 0.5934

Explanation:

First step is to determine total head losses at each pipe

at Pipe A

For 1/4 open gate valve head loss = 17 *Va^2 / 2g

elbow loss = 0.75 Va^2 / 2g

at Pipe B

For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g

elbow loss = 0.75 * Vb^2 / 2g

Given that both pipes are parallel

17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g

∴ Va / Vb = 0.5934

In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.

Answers

The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

Vertical slope, = 2
Horizontal slope = 3
Clockwise moment = 330 ft-lb
Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

the A coordinate = (0, 5)
the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

$= (11 )/(25) * 5, \ \ (14)/(25) * 5\n\n= (2.2, \ 2.8)$

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2, 2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

Learn more about moment of force here: brainly.com/question/6278006

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provideA. expansion joints
B. isolation joints
C. control joints
D. construction joints

Answers

If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provide Control joints. The correct answer would be C.

Control joints are used to prevent random cracks from forming in large concrete slabs caused by shrinkage. These joints are placed at strategic locations in the slab to allow for the concrete to expand and contract without cracking. Expansion joints, on the other hand, are used to separate concrete from other structures or materials, and isolation joints are used to separate different sections of concrete.

Construction joints are used to connect two different pours of concrete. Therefore, the best option for preventing random cracks caused by shrinkage would be to use control joints.

Learn more about Control joints:

brainly.com/question/1992029

#SPJ11

The reservoir pressure of a supersonic wind tunnel is 5 atm. A static pressure probe is moved along the centerline of the nozzle, taking measurements at various stations. For these probe measurements, calculate the local Mach number and area ratio: a. 4 atm; b. 2.64 atm; c. 0.5 atm.