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3 years ago

A brake technician typically visually inspects the blake units for all of the following

Engineering

1 answer:

-Dominant- [34]3 years ago

8 0

Explanation:

a gas in a rigid container has a pressure of 632 torrs and a temperature of 45 celsius. The pressure has increased to 842 torrs. What is the new temperature of the gas

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An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectivel

Tju [1.3M]

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

3 0

3 years ago

A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pul

Ghella [55]

Answer:

a) 1.16 m/s

b) 1/216000

c) (√15)/6480000

Explanation:

The parameters given are;

Length of boat prototype, lp = 30 m

Speed of boat prototype = 9 m/s

Length of boat model, lm= 0.5 m

a) lm/lp = 0.5/30 = 1/60 = ∝

(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)

vm = 9 × (1/60)^(1/2) = 1.16 m/s

b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;

Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³

Fm/Fp = (1/60)³ = 1/216000

c) The ratio of the model to prototype power pm/p $_p$ = (Fm/Fp) × (vm/vp) = ∝³×√∝

The ratio of the model to prototype power pm/p $_p$ = √(1/60) × (1/60)³

pm/p $_p$ = √(1/60) × (1/60)³ = (√15)/6480000

6 0

3 years ago

Air enters a tank through an area of 0.2 ft2 with a velocity of 15 ft/s and a density of 0.03 slug/ft3. Air leaves with a veloci

Mademuasel [1]

Answer:

please find attached.

Explanation:

4 0

3 years ago

Select the correct answer.

cricket20 [7]

Answer:

The power generated by a wind farm is not constant because of irregular wind patterns.

5 0

3 years ago

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

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago