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Allisa [31]
3 years ago
6

Calculate the efficiency of a Carnot Engine working between temperature of 1200°C and 200°C.

Engineering
1 answer:
swat323 years ago
8 0

Answer:

efficiency = 0.678

Explanation:

First we have to change temperatures from °C to K (always in thermodynamics absolute temperature is used).

\text{hot body temperature,}T_H = 1200 \circC + 273.15 = 1473.15 K

\text{cold body temperature,} T_C = 200 \circC + 273.15 = 473.15 K

Efficiency \eta of a carnot engine can be calculated only with hot body and cold body temperature by

\eta = 1 - \frac{T_C}{T_H}

\eta = 1 - \frac{473.15 K}{1473.15 K}

\eta = 0.678

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marissa [1.9K]

Answer:

a. 130.73 atm

b. 102.62 atm

c. 87.1 atm

Explanation:

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6 0
3 years ago
HELP<br><br><br>the overall width of a part is dimensioned as 3.00 ± 0.02. what is the tolerance
MariettaO [177]

Answer:

Not knowing the units the tolerance is 0.02.  I would presume mm but hopefully your question has more detail.  

Explanation:

The tolerance is the portion after the main dimension (+/- 0.02).  In our case we have bilateral tolerance since there is tolerance in both directions (positive and negative).  If you were building a part the acceptable range would be 2.98 to 3.02 based on the tolerance provided.  

3 0
3 years ago
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Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
I really need help ASAP!!!
ValentinkaMS [17]

Explanation:

He would work on the thing like in the method you work on your question.

8 0
3 years ago
In the engineering design process, testing is the least important step.<br> True<br> False
Wittaler [7]

Answer:

false

Explanation:

7 0
3 years ago
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