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Professor: Euiwon Bae

Lecture Hours : MWF 1:30-2:20 in ME1130

Office Hours : MWF 2:30-3:30 in ME1091

Email : ebae@purdue.edu

Phone : 765-494-6849

Course website : http://engineering.purdue.edu/ME200

ME200 : Thermodynamics I

Lecture 23 : Temperature scales &

maximum performance measures

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Outline

The Kelvin and temperature scales

Maximum performance measures for cycles operating

Between Two thermal reservoirs

Examples

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Kelvin scales

From the Carnot corollary:

Hot reservoir

System A

Cold reservoir

Q

Q

W

𝜂𝐴

Hot reservoir

System B

Cold reservoir

Q

Q

W

𝜂𝐵

TH TH

TC TC

𝜼𝑨 𝜼𝑩 =

=

=

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Kelvin scales

η = 1 −𝑄𝑜𝑢𝑡𝑄𝑖𝑛

Can argue that efficiency for the reversible cycle is the function of (TH, TC)

From the Carnot corollary:

All reversible power cycles with the same two thermal reservoirs

have the same 𝜼

(Regardless of substance making up the system (air, steam etc))

-> Temperature is the important factor

𝑄𝑐

𝑄𝑕𝑟𝑒𝑣 = 𝑓(TH, TC)

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Kelvin scales

In Kevlin scale, the ratio of temperature is equal to the amount of heat transfer for an reversible cycle.

Define a Kelvin scales as :

𝑄𝑐

𝑄𝑕𝑟𝑒𝑣 =

𝑇𝐶𝑇𝐻

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Maximum performance measures

For power cycles

η𝑚𝑎𝑥 =𝑊𝑐𝑦𝑐𝑙𝑒

𝑄𝑖𝑛= 1 −

𝑄𝑜𝑢𝑡

𝑄𝑖𝑛 = 1 −

𝑇𝐶

𝑇𝐻

Maximum achievable efficiency between two thermal reservoir : Carnot efficiency

Wcycle

= Qin- Qout

TH

(Hot)

Tc

(Cold)

W=Qin-Qout

Qin System

𝑄𝑐

𝑄𝑕𝑟𝑒𝑣 =

𝑇𝐶𝑇𝐻

Qout

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Maximum performance measures

For power cycles with Tc=298 K and

TH

η

η𝑚𝑎𝑥= 1 −𝑇𝐶

𝑇𝐻

Rate of efficiency increase Is high at tower TH

Decreasing Tc below certain

level is impractical

Real power plant 40% compared to theoretical max.

η𝑚𝑎𝑥= 1 −𝑇𝐶

𝑇𝐻

= 1 −298

745= 0.6 (60%)

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Maximum performance measures

For power cycles with TH=298 K and

TC

η

TC=TH

η𝑚𝑎𝑥= 1 −𝑇𝐶

𝑇𝐻

Rate of efficiency decreases as TC increases

Tc=Th, efficiency is zero

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Maximum performance measures

For Refrigeration and heat pump cycles

Maximum achievable efficiency between two thermal reservoir

𝑄𝑐

𝑄𝑕𝑟𝑒𝑣 =

𝑇𝐶𝑇𝐻

Wcycle =

Qout- Qin>0

TH Tc

W=Qout-Qin

Qout Qin System

β𝑚𝑎𝑥 =𝑄𝑖𝑛

(𝑄𝑜𝑢𝑡;𝑄𝑖𝑛)=

𝑇𝑐

𝑇𝐻;𝑇𝑐; Refrigeration cycle

γ𝑚𝑎𝑥 =𝑄𝑜𝑢𝑡

(𝑄𝑜𝑢𝑡;𝑄𝑖𝑛)=

𝑇𝐻𝑇𝐻;𝑇𝐶

; Heat pump cycle

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Maximum performance measures

For Refrigeration cycles with TH=298 K and

TC

𝛽 𝛽𝑚𝑎𝑥 =

1

𝑇𝐻𝑇𝑐

− 1

COP will gradually increase until Tc gets closer to the Th

TH

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Maximum performance measures

For Refrigeration cycles with TL=298 K and

TH

𝛽 𝛽𝑚𝑎𝑥 =

1

𝑇𝐻𝑇𝑐

− 1

COP will gradually increase when TH is closer to the TL

TL

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Example 1 Power cycle

5.30 An inventor claims to have developed a power cycle operating between TH=1000 K and Tc=250 K that develops net work equal to a multiple of the amount of energy Qc rejected to the cold reservoir – that is Wcycle=NQc, where all quantities are positive. What is the theoretical maximum N?

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Example 1 Power cycle

Given : TH, TC, W=N Qc

Find : maximum N

System :

TH

(Hot)

Tc

(Cold)

W=Qin-Qout

Qin System

Qout

Sol: 𝜂 = 𝑊𝑐𝑦𝑐𝑙𝑒

𝑄𝐻= 1 −

𝑄𝑐

𝑄𝐻= 1 −

𝑇𝑐

𝑇𝐻

𝑊𝑐𝑦𝑐𝑙𝑒= 𝑁𝑄𝑐

𝑁𝑄𝑐

𝑄𝐻≤ 1 −

𝑄𝐶

𝑄𝐻⇒ 𝑁 0.25 ≤ 1 − 0.25 ⇒ 𝑁 ≤ 3

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Example 1 Power cycle

5.38 At steady state, a 750 MV power plant receives energy by heat transfer from the combustion of fuel at an average T of 317 C. The plant discharges energy by heat transfer to a river whose mass flow rate is 1.65x105 kg/s. Upstream of the power plant the T is 17 C. Determine the increase in the T of the river if the thermal efficiency of the power plant is 40%

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Example 1 Power cycle

Given : 750 MW power plant, TH=317 m=1.65x10^5 kg/s

Tup=17, efficiency 0.4

Find : ΔT

System :

TH

(Hot)

Tc

(Cold)

W=750 MW

Qin System

Qout

T1=17 C

T2=?

M=1.65e5 kg/s

Assumption : steady state, incompressible flow with constant specific heat

Sol: 𝜂 =

𝑊𝑐𝑦𝑐𝑙𝑒

𝑄𝐻= 1 −

𝑄𝑐𝑄𝐻

; 𝑄𝐻 =750

0.4𝑀𝑊 = 1875 𝑀𝑊

𝜂 = 1 −𝑄𝑐𝑄𝐻

; 𝑄𝐶 = 0.6 𝑄𝐻 = 1125 𝑀𝑊

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Example 1 Power cycle

𝑄𝑐 = 𝑚𝐶 𝑇2 − 𝑇1 ; 𝑤𝑕𝑒𝑟𝑒 𝐶 = 4.2𝑘𝐽

𝑘𝑔 𝑘 𝑓𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 𝐴 − 19

Qout

T1=17 C

T2=?

m=1.65e5 kg/s Tc

(Cold)

Heat exchanger; Qcv=m (h2-h1)

𝑇2 =𝑄𝑐𝑚𝐶

+ 𝑇1 =1125 MW

1.65x105kgs 4.2

𝑘𝐽𝑘𝑔 𝑘

+ 290 𝐾 = 291.62 𝐾

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Example 2 Refrigeration cycle

5.60 The refrigerator operates at steady state with a COP of 4.5 and a power input of 0.8 kW. Energy is rejected from the refrigerator to the surroundings at 20C by heat transfer from metal coils whose average surface temperature is 28 C. Determine (a) 𝑄 in KW (b) Lowest theoretical T inside the refrigerator in K (c) Maximum theoretical power in kW that could be developed by a power cycle operating between coils and the surroundings. Would you recommend making use of this opportunity for developing power?

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Example 2

Given : COP 4.5 Win=0.8 kW TH=20 C Tcoil=28 C

Find : Q, Tc, max theoretical power

System :

TH=20C Tc

W=0.8 kW

Qout Qin System

COP=4.5

Tcoil=28C

Sol β𝑚𝑎𝑥 =

𝑄𝑖𝑛𝑊𝑖𝑛

=𝑄𝑖𝑛

(𝑄𝑜𝑢𝑡−𝑄𝑖𝑛); 𝑄𝑖𝑛 = 3.6 𝑘𝑊

𝑄𝑜𝑢𝑡 =β − 1

β𝑄𝑖𝑛 = 4.4 𝑘𝑊

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Example 2

TH=20C Tc

W=0.8 kW

Qout Qin System

COP=4.5

Tcoil=28C

β𝑚𝑎𝑥 =𝑄𝑖𝑛

(𝑄𝑜𝑢𝑡−𝑄𝑖𝑛)=

𝑇𝑐

𝑇𝐻 − 𝑇𝑐;

𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑇𝑐 =𝑇𝐻1

β𝑚𝑎𝑥

:1= 239.72 𝐾 = −33 𝐶

𝜂 = 1 −𝑇𝐶𝑇𝐻

= 1 −𝑇𝐻𝑇𝑐𝑜𝑖𝑙

= 1 −293

301= 0.0266 = 2.6%

𝜂 =𝑊𝑐𝑦𝑐𝑙𝑒

𝑄𝐻≤ 1 −

𝑇𝐶𝑇𝐻

⇒ 𝑊 ≤ 0.12 𝑘𝑊

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Example 3 Heat pump cycle

5.66 A heat pump with a COP of 3.8 provides energy at an average rate of 75,000 kJ/h to maintain a building at 21 C on a day when the outside temperature is 0 C. If electricity cost is 8 cents per kWh (a) Actual operating cost and the minimum theoretical operating

cost in $/day (b) Compare this with the cost of electrical resistance heating

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Example 3 Heat pump cycle

Given : heat pump to maintain temperature Find : theoretical and actual cost/ cost for using electrical resistance System : W

Qin Qout

γ=3.8

Qout=75000 kJ/h

T=0 C

Tin=21 C

Assumption : electricity cost 8 cent KWH Sol :

Since COP is given 𝑊 =𝑄 𝑜𝑢𝑡

𝛾=

75000𝑘𝐽

ℎ

3.8= 19737

𝑘𝐽

ℎ

Since 1 KWH=3600 kJ => $0.08 per 3600 kJ Therefore for 24 h work input 19737 kJ/h*24h=473688 kJ => 473688 kJ/3600 kJ*0.08=$10.52

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Example 3 Heat pump cycle

If we need the minimum cost -> when the COP is maximum Using two temperature only

𝛾𝑚𝑎𝑥 =𝑇𝐻

𝑇𝐻 − 𝑇𝐶=

1

1 −𝑇𝐶𝑇𝐻

= 14

Since COP is given 𝑊 =𝑄 𝑜𝑢𝑡

𝛾=

75000𝑘𝐽

ℎ

14= 5357

𝑘𝐽

ℎ

Since 1KWH=3600 kJ => $0.08 per 3600 kJ Therefore for 24 h work input 5357 kJ/h*24h=128568kJ => 128568 kJ/3600 kJ*0.08=$2.85

For resistance heating 75000kJ/h/3600*24*0.08=$40/day