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a_sh-v [17]
3 years ago
10

Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water

at a rate of 800 per minute. The balls leave the oven at a uniform temperature of 900°C and are exposed to air at 25°C for a while before they are dropped into the water. If the temperature of the balls drops to 850°C prior to quenching, determine the rate of heat transfer from the balls to the air.
Engineering
2 answers:
icang [17]3 years ago
7 0

Answer:

rate of heat transfer from the balls to the air; Q' = 2.34 Kw

Explanation:

We are given;

Density of ball bearings; ρ = 8085 kg/m³

Initial Temperature; T_i = 900°C

Final Temperature; T_f= 850 °C

Rate at which stainless balls are quenched in water; N' = 800 per minute = 800/60 per sec = 40/3 per second

Diameter; d = 1.2cm = 0.012m

Specific heat capacity; cp = 0.480 kJ/kg·°C

The rate of heat transfer to the air is expressed as;

Q' = m'c(T_i - T_f)

mass/volume = density, thus;

Q' = ρV'c(T_i - T_f)

Volumetric rate;V' = VN'

Thus,

Q' = ρVN'c(T_i - T_f)

Where,

c is specific heat capacity

ρ is density

V is volume

Volume = πr³/3 or πd³/6. So volume = π(0.012³)/6 = 9.0478 x 10^(-7) m³

N' is numerical rate

T_i are T_f initial and final temperatures respectively

Thus,

Q' = 8085•9.0478 x 10^(-7)•(40/3)•(0.48)•(900 - 850)

Q' = 2.34 Kw

sveticcg [70]3 years ago
3 0

Answer:

\dot Q = -2.341\,kJ

Explanation:

The rate of heat transfer from the balls to the air is:

\dot Q = \left[(800\,\frac{1}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot(8085\,\frac{kg}{m^{3}})\cdot (\frac{4}{3}\pi )\cdot (6\times 10^{-3}\,m)^{3}\right]\cdot (0.480\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (850\,^{\textdegree}C-900\,^{\textdegree}C)\dot Q = -2.341\,kJ

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Answer: The electric field decreases because of the insertion of the Teflon.

Explanation:

If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.

However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.

In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and  increasing the capacitance proportionally to the dielectric constant of the Teflon.  

8 0
3 years ago
An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/
Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755  / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min  = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

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Answer:

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Blu ray can hold 25gb per layer

Dvd can hold 4.7GB on a single layer

Cd can hold around 737 mb

Also, dvds can go up to 2 layers

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3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

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= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

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