Sample calculation

Efficiency for cold medium

ŋc=T c ,out−T c ,∈¿

T h ,∈¿−T c,∈¿×100 ¿¿¿

ŋc=30−2560−25

×100

ŋc=14.29%

Efficiency for hot medium

ŋh=T h ,∈¿−T h,out

T h ,∈¿−T c,∈¿×100 ¿¿¿

ŋh=60−4560−25

×100

ŋh=42.86%

Mean temperature efficiency

ŋh=ŋc+ŋh2

ŋh=14.29+42.86

2

ŋh=28.58%

Power emitted

Pem=∀̇h ρhC ph¿

Pem=(1.67×10−5 ) (983.3 )(4185)(60−45)

Pem=1028.78W

Power absorbed

Pab=∀̇ c ρcCpc¿

Pab=(3.33×10−5 ) (996 )(4178)(30−25)

Pab=693.55W

Power loss

Ploss=P em−Pab

Ploss=1028.78−693.55

Ploss=335.23W

Overall efficiency

ŋoverall=PabPem×100

ŋoverall=693.551028.78

×100

ŋoverall=67.41%

Logarithmic mean temperature difference

∆T m=∆T 1−∆T 2

ln(∆T 1∆T 2

)=¿¿

∆T m=(60−30 )−(45−25)

ln ((60−30)(45−25)

)

∆T m=24.66℃

Overall heat transfer coefficient

U=Power absorbedAs×Tm

U= 693.550.067×24.66

U=419.72W /m2℃