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3 years ago

13

Consider coaxial, parallel, black disks separated a distance of 0.20 m. The lower disk of diameter 0.40 m is maintained at 500 K

and the surroundings are at 300 K. What temperature will the upper disk of diameter 0.20 m achieve if electrical power of 17.5 W is supplied to the heater on the back side of the disk?

1 answer:

ANTONII [103]3 years ago

4 0

Answer:

In this picture is the answer

Explanation:

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The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long hori

Dima020 [189]

Answer:

The maximum discharge rate of water is 4.6 L/s

Explanation:

Given data:

d=diameter=8 m

h=height=3 m

The mathematical expression for the theoritical velocity is:

$v_{th} =\sqrt{2gh} =\sqrt{2*9.8*3} =7.6681m/s$

The maximum discharge can be calculate by:

$Q=C_{d} Av_{th}$

Here

Cd=coefficient of discharge=0.855

$Q=0.855*\frac{\pi }{4} *d^{2} *vx_{th} =0.855*\frac{\pi }{4} *0.03^{2} *7.6681=0.0046m^{3} /s=4.6L/s$

8 0

3 years ago

A pump is used to deliver water from a lake to an elevated storage tank. The pipe network consists of 1,800 ft (equivalent lengt

Nataly_w [17]

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

7 0

3 years ago

12.28 LAB: Output values in a list below a user defined amount - functions Write a program that first gets a list of integers fr

Anastaziya [24]

Answer:

def output_ints_less_than_or_equal_to_threshold(user_values, upper_threshold):

for value in user_values:

if value < upper_threshold:

print(value)

def get_user_values():

n = int(input())

lst = []

for i in range(n):

lst.append(int(input()))

return lst

if __name__ == '__main__':

userValues = get_user_values()

upperThreshold = int(input())

output_ints_less_than_or_equal_to_threshold(userValues, upperThreshold)

8 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine h

IrinaVladis [17]

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

7 0

3 years ago