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2 years ago

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A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k = 70 W/m·K) with a uniform cl

earance of 0.6 mm filled with lubricating oil (μ = 0.03 N·s/m2 and k = 0.14 W/m·K). The bearing is cooled externally by a liquid, and its outer surface is maintained at 40°C. Disregarding heat conduction through the shaft and assuming one-dimensional heat transfer, determine (a) the rate of heat transfer to the coolant, (b) the surface temperature of the shaft, and (c) the mechanical power wasted by the viscous dissipation in oil.

1 answer:

-BARSIC- [3]2 years ago

5 0

Answer:

(a) the rate of heat transfer to the coolant is Q = 139.71W

(b) the surface temperature of the shaft T = 40.97°C

(c) the mechanical power wasted by the viscous dissipation in oil 22.2kW

Explanation:

See explanation in the attached files

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The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h

diamong [38]

Answer:

$M=0.0411 kg/h or 4.1*10^{-2} kg/h$

Explanation:

We have to combine the following formula to find the mass yield:

$M=JAt$

$M=-DAt(ΔC/Δx)$

The diffusion coefficient : $D=6.0*10^{-8} m/s^{2}$

The area : $A=0.25 m^{2}$

Time : $t=3600 s/h$

ΔC: $(0.64-3.0)kg/m^{3}$

Δx: $3.1*10^{-3}m$

Now substitute the values

$M=-DAt(ΔC/Δx)$

$M=-(6.0*10^{-8} m/s^{2})(0.25 m^{2})(3600 s/h)[(0.64-3.0kg/m^{3})(3.1*10^{-3}m)]$

$M=0.0411 kg/h or 4.1*10^{-2} kg/h$

8 0

2 years ago

Python Homework

DerKrebs [107]

Answer:

TAX_RATE = 0.20
STANDARD_DEDUCTION = 10000.0
DEPENDENT_DEDUCTION = 3000.0
# Request the inputs
grossIncome = float(input("Enter the gross income: "))
numDependents = int(input("Enter the number of dependents: "))
# Compute the income tax
taxableIncome = grossIncome - STANDARD_DEDUCTION - \
DEPENDENT_DEDUCTION * numDependents
incomeTax = taxableIncome * TAX_RATE
# Display the income tax
print("The income tax is $" + str(round(incomeTax,2)))

Explanation:

We can use round function to enable the program to output number with two digits of precision.

The round function will take two inputs, which is the value intended to be rounded and the number of digits of precision. If we set 2 as second input, the round function will round the incomeTax to two decimal places. The round function has to be enclosed within the str function so that the rounded value will be converted to a string and joined with the another string to display the complete a sentence of income tax info.

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2 years ago

A cylindrical drum (2 ft. dia ,3 ft height) is filled with a fluid whose density is 40 lb/ft^3. Determine (a. the total volume o

Ksivusya [100]

Answer:

a) $V=9.42\ ft^3$

b)Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c) $v=0.025\ ft^3/lb$

d) $w=1276 \ lb/ft.s^2$

Explanation:

Given that

d= 2 ft

r= 1 ft

h= 3 ft

Density

$\rho = 40\ lb/ft^3$

a)

We know that volume V given as

$V=\pi r^2 h$

$V=\pi \times 1^2\times 3$

$V=9.42\ ft^3$

b)

Mass = Density x volume

$mass =40\times 9.42\ lb$

mass= 376.8 lb

We know that

1 lb = 0.031 slug

So 376.8 lb= 11.71 slug

Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)

we know that specific volume(v) is the inverse of density.

$v=\dfrac{1}{\rho}\ ft^3/lb$

$v=\dfrac{1}{40}\ ft^3/lb$

$v=0.025\ ft^3/lb$

d)

Specific weight(w) is the product of density and the gravity(g).

w= ρ X g

w = 40 x 31.9

$w=1276 \ lb/ft.s^2$

8 0

2 years ago

A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes

ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f = 0.77 for 232 ksi

because for Se 0.5 Su at $10^{6}$ cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = $\frac{(f*Su)^2}{Se}$

a = $\frac{(0.77*1600)^2}{700}$

a = 2168.3 Mpa

so

coefficient value (b) will be

a = - $\frac{1}{3}$ log $\frac{(f*Su)}{Se}$

b = - $\frac{1}{3}$ log $\frac{(0.77*1600)}{700}$

b = -0.0818

so no of cycle N is

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value

N = $(\frac{ 900}{2168.3})^{1/-0.0818}$

N = 46400

the life (N) of the specimen is 46400 cycles

5 0

2 years ago

For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active

Sonja [21]

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

6 0

2 years ago