A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of
60 percent. Determine the temperature of the source that supplies heat to this engine.
1 answer:
Answer:
The source temperature is 1248 R.
Explanation:
Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.
Given:
Temperature of the heat sink is 520 R.
Second law efficiency is 60%.
Actual thermal efficiency is 35%.
Calculation:
Step1
Reversible efficiency is calculated as follows:
Step2
Source temperature is calculated as follows:
T = 1248 R.
The heat engine is shown below:
Thus, the source temperature is 1248 R.
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Answer:
A.) 0.08 kJ/kg.K
B.) 207.8 KJ/Kg
C.) 0.808
Explanation:
From the question, the use of fluids mechanic table will be required. In order to get the compressor processes, the kinetic energy and the potential energy will be negligible while applying the ideal gas model.
Since the steam is a closed system, the carbon dioxide will be compressed adiabatically.
Please find the attached files for the solution and the remaining explanation.
Answer:
a) aA = - 13.33 mm/s²
aB = - 20 mm/s²
b) aD = - 13.33 mm/s²
c) vB = 70 mm/s
d) xB = 440 mm
Explanation:
Given
The initial speed of B is: v₀B = 150 mm/s
Distance moved by A is: xA = 240 mm
Velocity of A is: vA = 60 mm/s
Assuming:
Displacement of blocks are denoted by:
A = xA
B = xB
C = xC
D = xD
From the pic shown, the total length of the cable is:
xB + (xB - xA) + 2*(d - xA) = L
⇒ 2*xB - 3*xA = L - 2*d
where L - 2*d is constant. Differentiating the above equation with respect to time:
d(2*xB)/dt - d(3*xA)/dt = 0
⇒ 2*vB - 3*vA = 0 (i)
Substituting in equation (i)
2*(150 mm/s) - 3*vA = 0
⇒ v₀A = 100 mm/s (initial speed of A)
Then, we use the equation
vA² = v₀A² + 2*aA*xA
Substituting the values in above equation:
(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)
⇒ aA = - 13.33 mm/s²
If 2*vB - 3*vA = 0
Differentiating the above equation with respect to time:
d(2*vB)/dt - d(3*vA)/dt = 0
⇒ 2*aB - 3*aA = 0 (ii)
Substituting in equation (ii)
2*aB - 3*(- 13.33 mm/s²) = 0
⇒ aB = - 20 mm/s²
b) From the pic shown,
xD - xA = constant
If we apply
d(xD)/dt - d(xA)/dt = 0
⇒ vD - vA = 0
then
d(vD)/dt - d(vA)/dt = 0
⇒ aD - aA = 0
⇒ aD = aA = - 13.33 mm/s²
c) We use the formula
vB = v₀B + aB*t
Substituting the values in above equation:
vB = 150 mm/s + (- 20 mm/s²)*(4 s)
⇒ vB = 70 mm/s
d) We apply the equation
xB = v₀B*t + 0.5*aB*t²
Substituting the values in above equation:
xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²
⇒ xB = 440 mm
Answer:
a) 5.2 kPa
b) 49.3%
Explanation:
Given data:
Thermal efficiency ( л ) = 56.9% = 0.569
minimum pressure ( P1 ) = 100 kpa
<u>a) Determine the pressure at inlet to expansion process</u>
P2 = ?
r = 1.4
efficiency = 1 - [ 1 / (rp) ]
0.569 = 1 - [ 1 / (rp)^0.4/1.4
1 - 0.569 = 1 / (rp)^0.285
∴ (rp)^0.285 = 0.431
rp = 0.0522
note : rp = P2 / P1
therefore P2 = rp * P1 = 0.0522 * 100 kpa
= 5.2 kPa
b) Thermal efficiency
Л = 1 - [ 1 / ( 10.9 )^0.285 ]
= 0.493 = 49.3%