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3 years ago

7

Tech A says that a cylinder leakage test is performed on a cylinder with low compression to determine the severity of the leak a

nd where it is located. Tech B says that most manufacturers consider up to 50% cylinder leakage acceptable. Who is correct?

1 answer:

Karolina [17]3 years ago

4 0

Tech A djjdjdndnndndbdbx

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. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C

MrRissso [65]

Answer:

The inside temperature, $T_{in}$ is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×( $T_s$ - $T_{\infty$ ) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

$P = \dfrac{K \times A \times \Delta T}{L}$

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

$2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}$

$T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C$

The inside temperature, $T_{in}$ = 247.99 °C ≈ 248 °C.

3 0

4 years ago

Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 90-gallon hot

Rudik [331]

Answer:

Q=36444.11 Btu

Explanation:

Given that

Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

Volume of water = 90 gallon

Mass = Volume x density

$1\ gallon = 0.13ft^3$

Mass ,m= 90 x 0.13 x 62.36 lbm

m=729.62 lbm

We know that sensible heat given as

Q= m Cp ΔT

Now by putting the values

Q= 729.62 x 0.999 x (110-60) Btu

Q=36444.11 Btu

5 0

3 years ago

Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at

makkiz [27]

Answer:

$\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}$

Explanation:

The Young's module is:

$E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }$

$E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}$

Let assume that both specimens have the same geometry and load rate. Then:

$E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}$

The displacement rate for steel is:

$\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}$

$\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )$

$\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}$

7 0

3 years ago

Read 2 more answers

If you need to write a function that will compute the cost of some candy, where each piece costs 25 cents, which would be an app

masya89 [10]

The best answer would be

D. Int calculateCost(int count);

6 0

3 years ago

The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th

Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

$R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}$

Given that t= 200 days

$R(200)=e^{-\left(\dfrac{200}{300}\right)^2}$

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

$0.95=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=67.74 days

When R=90 %

$0.90=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=97.2 days

7 0

3 years ago