Air is contained inside a vertical piston-cylinder assembly by a piston of mass 105 kg and face area of 0.05 m2 . The mass of air inside the cylinder is 20 grams, and the air initially occupies a volume of 50 liters. The atmosphere exerts a pressure of 101.3 kPa on the top of the piston. The volume of the air slowly decreases to 0.005 m3 , while maintaining constant pressure, as the specific internal energy of the air decreases by 175 kJ/kg. Neglecting friction between the piston and the cylinder wall, determine the heat between the surroundings and the air inside the piston-cylinder assembly in kJ

Answers

Answer 1

Answer:

ΔQ = 1.06 KJ

Explanation:

The amount of heat transfer between the piston-cylinder system and the surrounding can easily be found by using the First Law of Thermodynamics. The first law of thermodynamics can be written as follows:

ΔQ = ΔU + W

ΔQ = mΔu + PΔV

where,

ΔQ = Heat transfer between system and surrounding = ?

Δu = specific change in internal energy of the system = - 175 KJ/kg

m = mass of air = 20 g = 0.02 kg

P = Constant Pressure = 101.3 KPa

ΔV = Change in Volume = 0.05 m³ - 0.005 m³ = 0.045 m³

Therefore,

ΔQ = (0.02 kg)(-175 KJ/kg) + (101.3 KPa)(0.045 m³)

ΔQ = -3.5 KJ + 4.56 KJ

ΔQ = 1.06 KJ


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Required:
Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperatureis maintained at Tt < 85°C for the unfinned and finned packages.

Answers

Answer:

For unfined package = 1.30 × 10^-3 W.

Dor fined package = 8.64 × 10^-3 W.

Explanation:

STEP ONE: The first thing to do is to determine the cross sectional area and input the value into the convectional resistance formula/equation in order to determine the heat rate for the unfined package.

Thus, Area = (dimension of the electronic device) ^2. = (10 × 10^-6)^2 m.

Convectional Resistance = (100 × 10^-9)÷ (10 × 10^-6)^2 × 490 = 2.04 K/W.

Heat transfer coefficient = 1/ (10 × 10^-7)^2 × 10^5. = 1 × 10^5 K/W.

Thus, the heat rate for the unfined package = 2{(85 - 20) ÷ (10^5 + 2.04)} = 1.30 × 10^-3 W.

STEP TWO: Determine the surface area of each fin and the prime area.

Surface area of each fin = {300 + 15/4} × 10^-9 × 10^-9 × π × 15 = 1.43 × 10^-14 m^2.

Prime area = (10^-6 × 10)^2 - (200 × 200) × π × [ ( 15 × 10^-9)^2 ÷ 4] = 9.29 × 10^-11 m^2.

STEP THREE: Determine the efficiency

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Efficiency= tanh ( 7.09 × 10^-2) ÷ 7.09 × 10^-2 = 0.998.

Thus, the coefficient = 1 - (1.43 × 10^-14 × (200 × 200) ÷ (6.65 × 10^-10) × (1 - 0.998) = 0.99.

STEP FOUR: Determine the heat rate for the finned package.

Thermal resistance = 1/ (10^5 × 0.999 × 10^-10 × 6.65 = 1.50 × 10^4 K/W.

The heat rate for the finned package = 2 { 85 - 20} ÷ { 1.50 × 10^4 + 2.04} = 8.64 × 10^-3 W

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A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°C and convection heat transfer coefficient of 220 W/m^2•K. The 10-cm thick brass plate (rho = 8530 kg/m^3, cp = 380 J/kg•K, k = 110 W/m•K, and α = 33.9×10^–6 m^2/s) has a uniform initial temperature of 900°C, and the bottom surface of the plate is insulated.

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Answers

Answer:

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Explanation:

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STEP THREE: This is the last step for the question, here we will be calculating the temperature of the center plane of the brass plate after 3 minutes.

Thus, the temperature of the center plane of the brass plane after 3 minutes = (1.00705) (0.89199) (900- 15) + 15.

= > the temperature of the center plane of the brass plane after 3 minutes = 809.98°C.

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Directions and Analysis
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About the building/work
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