Physics, 25.01.2020 06:31 mr1martin

Acarnot heat engine receives 3,200 kj/s of heat from a high temperature source at 825 ℃ and rejects heat to a cold temperature sink at 15 ℃. solve following questions:

a. what is the thermal efficiency of this engine?
b. what is the power delivered by the engine in watts?
c. at what rate is heat rejected to the cold temperature sink?
d. what is the entropy change of the sink?

Answers: 3

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Answers

Answer from: payshencec21

Explanation:

Given:

source temperature T1= 825° C = 1099 K

Sink temperature T2= 15°C = 288 K

Heat Supplied to the engine Qs= 3200 KW

a) Efficiency of a Carnot engine is

$\eta =1-\frac{T_1}{T_2}$

$=1-\frac{1099}{288}$

=0.7379

=73.79%

$\eta= \frac{Power delivered}{power recieved}$

let W be the powe delivered

$0.7379= \frac{W}{3200}$

W= 2361.408 KW

c) Heat rejected to the cold temperature Qr= Qs-W

= 3200-2361.408=838.60 KW

d) Entropy change in the sink

$\Delta S= \frac{Qr}{T_{sink}}$

=838.60/288= 2.911 W/K

Answer from: love7973

a. $\eta_{th}=0.7377=73.77\%$

b. $P=2360655.7377\ W$

c. $Q_o=839344.2622\ W$

d. $\Delta s=2914.3897\ J.K^{-1}$

Explanation:

Given:

temperature of the source, $T_H=825+273=1098\ K$ heat received from the source, $Q_i=3200000\ J.s^{-1}$ temperature of the sink, $T_L=15+273=288\ K$

Since the engine is a Carnot engine:

$\eta_{th}=1-\frac{T_L}{T_H}$

$\eta_{th}=1-\frac{288}{1098}$

$\eta_{th}=0.7377=73.77\%$

Now the power delivered:

since here we are given with the rate of heat transfer

$P=\eta_{th}\times Q_i$

$P=0.7377\times 3200000$

$P=2360655.7377\ W$

rate of heat rejection to the sink:

Q_o=Q_i-P

Q_o=3200000-2360655.73770491808

$Q_o=839344.2622\ W$

$\Delta s=\frac{Q_o}{T_L}$

$\Delta s=\frac{839344.2622}{288}$

$\Delta s=2914.3897\ J.K^{-1}$

Answer from: Quest

i think it is non rem sleep

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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Acarnot heat engine receives 3,200 kj/s of heat from a high temperature source at 825 ℃ and rejects...

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