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AE 231 Thermodynamics

Recitation 11

Instructor: Sinan Eyi

Question: A cyclic machine, shown in the figure, receives 325 kJ

from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K

energy reservoir and the cycle produces 200 kJ of work as

output. Is this cycle reversible, irreversible, or impossible?

Solution:

hCarnot = 1 – TL/TH

hCarnot = 1 - 400/1000 = 0.

heng = W/QH = 200/325 = 0.615 > hCarnot

This is impossible.

Solution:

C.V. House. For constant 20°C the heat pump must provide

C.V. Heat pump. Definition of the coefficient of performance and the fact that

the maximum is for a Carnot heat pump.

Substitute into the first equation to get

0.6 (TH - TL ) = [ 0.5 * TH / (TH - TL ) ] 1 =>

(TH - TL )^2 = (0.5 / 0.6) TH * 1 = 0.5 / 0.6 *293.15 = 244.

TH - TL = 15.63 => TL = 20 - 15.63 = 4.4 °C

QH = Qleak = 0.6( TH − TL (^) )= W

 = QH / W = QH / (^) ( QH − QL )= 0.5 carnot = 0.5*T / TH ( (^) H −TL)

Question: Consider a Carnot cycle heat engine

operating in outer space. Heat can be rejected from

this engine only by thermal radiation, which is

proportional to the radiator area and the fourth power

of absolute temperature, Q rad

~ KAT

4

. Show that for

given engine work output and given T H

, the radiator

area will be minimum when the ratio T L

/T H

= 3/4.

4 4 4 1

net H L L L H L L

H L H H L H H

W T T AT AT T T T

A

KT T T T T T T

−       ^    − = (^)   = (^)  − (^)  = (^)   ^   − 

      ^     

3 4

4

net L L

H H H

W T T

A const KT T T

3 4 4

net

H

W

A x x const

KT

= ^ − = = 

L

H

T

where x T

A 3

x x



−

( )

( ) (^ )

( )

( )

( )

( )

2 3 2 3 2 3

3 4 2 3 4 3 4 3 4

dA x^ x^ x^ x^ x^ x A dx (^) x x x x x x x x

− −^ −^ −

( )

( )

2 3 4

dA x Ax dx (^) x x

x =

3

4

L

H

T

T

=

Q

q m

= qH = 250 kJ /kg TH = 600 K TL = 300 K P 3 = 75 kPa

Cv =0.717 kJ /kgK R =0.287 kJ /kgK

T 1 = T 2 = TH = 600 K^3

T = T = TL = 300 K

3 3 3 3

RT

v m kg P

Solution:

From the first law of thermodynamics for a closed system

q = du +w

Between the states 2 and 3 we have an adiabatic process

q= 0 du =Cv 0 dT^ w^ =Pdv

from the equation of state for an ideal gas

P =RT /v

dv w RT v

v 0

dv q C dT RT v

 = + v 0

dT dv C R T v

^ q =^0  = −

2

1

H H^ ln

v q RT v

2

1

ln (^) H / (^) H 250 / 0.287 *600 1.

v q RT v

v 1 = v 2 / exp(1.4518) = 0.2034 / exp(1.4518) =0.

in Carnot Cycle

3 4

2 1

v (^) v

v v

= v 4 = v 1 * v 3 / v 2 = 0.04763*1.148 / 0.2034 =0.

P 1 = RT 1 / v 1 = 0.287 *600 / 0.04763 = 3615 kPa

P 2 = RT 2 / v 2 = 0.287 *600 / 0.2034 =846.6kPa

P 4 = RT 4 / v 4 = 0.287 *300 / 0.2688 =846.6kPa