Answer:
a) P = 1776.97 kJ/min
b) Source temperature, T = 640 K
Given:
[tex]\eta = 55%[/tex]
T = [tex]15^{\circ} = 288 K[/tex]
Q' = 800 kJ/min
Solution:
Refer to the given fig.:
a) We first calculate the value of Q from the eqn:
[tex]\eta = 1 -\frac{Q'}{Q}[/tex]
[tex]0.55 = 1 -\frac{800\times 1000}{Q}[/tex]
Solving for Q, we get:
0.55 Q = Q - 800000
Q = 1777.778 kJ/min
Now,
Output Work, W = Q - Q' = 1777.778 - 800 = 1776.97 kJ/min = power output, P
b) Temperature of source, T':
We know that:
T ∝ Q
Therefore,
[tex]\frac{T}{T'} = \frac{Q}{Q'}[/tex]
[tex]\frac{T}{288} = \frac{1777.78\times 1000}{800}[/tex]
T = 640 K
The power output of the engine is equal to 977.78 kJ/min.
Given the following data:
Thermal efficiency = 55 percent.
Temperature = 15°C to K = [tex]15+273[/tex] = 288 K.
Rate of rejection = 800 kJ/min.
First of all, we would calculate the quantity of heat input into the engine. Mathematically, the thermal efficiency of an engine is given by this formula:
[tex]\eta = 1-\frac{Q_{out}}{Q_{in}} \\\\0.55 = 1-\frac{800 \times 10^3}{Q_{in}} \\\\0.55 Q_{in} = Q_{in} - 800000\\\\Q_{in}-0.55Q_{in}=800000\\\\0.45Q_{in}=800000\\\\Q_{in}=\frac{800000}{0.45} \\\\[/tex]
Heat input = 1777.78 kJ/min.
For the power output, we have:
[tex]W=Q_{in}-Q_{out}\\\\W=1777.78-800[/tex]
W = 977.78 kJ/min.
[tex]T\;\alpha \;Q\\\\\frac{T_i}{T_o} =\frac{Q_{in}}{Q_{out}} \\\\\frac{T_i}{288} =\frac{1777.78 }{800} \\\\800T_i=1777.78\times 288\\\\800T_i=512000.64\\\\T_i=\frac{512000.64}{800}[/tex]
Ti = 640.0 Kelvin.
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