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5)a heat engine is operating on a carnot cycle and has a thermal efficiency of 55 percent. the waste heat from this engine is rejected to a nearby lake at 60 f at a rate of 800 btu/min. determine (1) the power output of the engine and (b) the temperature of the source.



Answer :

(1) the determined power output of the heat engine is  1075.66kJ/min.

(b) the temperature of the source 676.891 K.

η = 55%

T = 15 degree = 288K

Q' = 800 btu/ mint

To calculate Q

η = 1 - Q'/Q

0.55 = 1 - (800 x1.05506)/Q

(∵ 1 btu = 1.05506 kJ )

btu = British thermal unit

then

0.55 = 1 - (800 x 1.05506)/Q

Q = 1875.66kJ/min

To calculate output work

W = Q - Q' = 1075.66kJ/min.

b) Temperature of the source is given as:

T ∝ Q

T/T' = Q/Q'

T'  = 60F = 288.705556 K

Then

T/288.705556 K = 1875.66 /800

T = (1875.66 x 288.705556) /800

T = 676.891 K

(1) the determined power output of the heat engine is 1075.66kJ/min.

(b) the temperature of the source 676.891 K.

To know more about temperature

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