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3 years ago

Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters

Engineering

1 answer:

KonstantinChe [14]3 years ago

4 0

Answer:

Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37

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Identify the measurement shown in figure 7 and state in centimeters

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Answer:

1.3cm

Explanation:

the arrow is 3 lines past the 1 so it is 1.3cm

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3 years ago

Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and

lina2011 [118]

Answer:

y[n] = x[n] x[n-1] x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is $\frac{1}{x[n] . x[n-1] x[n+1]}$

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

for x[n] = 2 $\pi$ , y[n] = cos(2 $\pi$ ) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

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2 years ago

A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as

Ulleksa [173]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ = 254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa

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3 years ago

What are the 5 major forest types?

Nataly [62]

Answer:

1. Equatorial Evergreen or Rainforest

2. Tropical forest

3. Mediterranean forest

4. Temperate broad-leaved forest

5. Warm temperate forest

Explanation:

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3 years ago

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Answer and Explanation:

The answer is attached below

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3 years ago