1.

Δtmax − Δtmin
T
LMTD = Δt
ln( max )
t hi = 500ᵒF Δtmin

t ho = 400ᵒF
Δtmax = 400 – 120 = 280ᵒF
Δtmin = 500 – 310 = 190ᵒF
t co = 310ᵒF
(280−190) ᵒF
t ci = 120ᵒF LMTD = 280ᵒF = 232.1ᵒF
ln( )
190ᵒF
L

2.

Δtmax + Δtmin
AMTD =
T 2
t hi = 145ᵒC
Δtmax = 145 – 138 = 107ᵒC
t ho = 45ᵒC Δtmin = 45 – 30 = 15ᵒC
(107 + 15) ᵒC
t co = 38ᵒC AMTD = = 61ᵒC
2
t ci = 30ᵒC

L
 3. Pabs = Patm − Pvac
Pabs = 760 – 702 = 58 mm Hg = 0.007732 MPa
T
t hi = 40.586ᵒC t ho = 40.586ᵒC From steam table: Table 2 (using interpolation)
t sat = thi = tho = 40.856ᵒC
t co = 37.5ᵒC Δtmax − Δtmin
LMTD = Δt
ln( max )
Δtmin
t ci = 29.5ᵒC
Δtmax = 40.856 – 29.5 = 11.356ᵒC
L
Δtmin = 40.856 – 37.5 = 3.356ᵒC
(11.356 − 3.356) ᵒC
LMTD = 11.356ᵒC = 6.563ᵒC
ln( )
3.356ᵒC

4.

Δtmax − Δtmin
T t hi = 40ᵒC t ho = 40ᵒC LMTD = Δt
ln( max )
Δtmin

t co = 35ᵒC
Δtmax = 40 – 20 = 20ᵒC
Δtmin = 40 – 35 = 5ᵒC
t ci = 20ᵒC (20 − 5) ᵒC
LMTD = 20ᵒC = 10.82ᵒC
L ln( )
5ᵒC
 5. Q absorbed by the oil = Q released from the gas

moil Cp oil (TH oil − TL oil) = mgas Cp gas (TH gas − TL gas )
moil Cp oil (TH oil − TL oil )
TL gas = TH gas –
T mgas Cp gas
t hi = 200ᵒC
kg kJ
(100 ) (2.5 ) (65 K)
min kg−K
t ho = 135ᵒC TL gas = 473 K – kg kJ
(250 )(1 )
min kg−K

t co = 100ᵒC TL gas = 408 K – 273 = 135ᵒC = tho

t ci = 35ᵒC
kg 1 min kJ
Q = (100 x ) (2.5 ) (65 K) = 270833 W
L min 60 sec kg−K

Δtmax + Δtmin
AMTD = 2

Δtmax = 200 – 100 = 100ᵒC

Δtmin = 135 – 35 = 100ᵒC
(100 + 100) ᵒC
AMTD = = 100ᵒC
2

Q = A U AMTD
Q 270833 W
A = U AMTD = W = 36.11 𝐦𝟐
(75 )(100ᵒC)
m2 −C