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Ira Lisetskai [31]
3 years ago
12

A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g

eneration in the wire per unit volume, in Btu/hrft3 , and the heat flux on the outer surface of the wire, in Btu/hrft2 , as a result of this generation.
Engineering
1 answer:
SSSSS [86.1K]3 years ago
3 0

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

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Answer:

The minimum mass flow rate will be "330 kg/s".

Explanation:

Given:

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m_{s}=5.55 \ kg/s

\Delta h=2491 \ kg/kj

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\Delta T=10^{\circ}C

(Cp)_{w}=4.184 \ kJ/kg^{\circ}C

They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,

⇒  m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T

On putting the estimated values, we get

⇒  5.55\times 2491=M_{w}\times 4.184\times 10\\

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3 years ago
A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?
blagie [28]

Answer:

3A

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2 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and
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Answer:

A.) 0.3088

B.) 0.0017

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Explanation:

A.)

z1= \frac{\left(150-137\right)}{27.7}=0.4693

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P(0.4693

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3 years ago
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According to the
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Answer:

The part of the system that is considered the resistance force is;

B

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The effort, which is the input force at A gives the value of the tension at C and  D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

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C = B/2 = A

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The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

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Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

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