# Chemical Engineering - Chemical Engineering Basics - Discussion

Discussion Forum : Chemical Engineering Basics - Section 6 (Q.No. 2)
2.
In a counter flow heat exchanger, hot fluid enters at 170°C & leaves at 150°C, while the cold fluid enters at 50°C & leaves at 70°C. The arithmetic mean temperature difference in this case is __________ °C.
20
60
120
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

TANMAY SHAH said:   3 months ago
Arithmetic Mean Temperature Difference is the average temperature of hot and cold fluids.

AMTD = ((HTs + HTe)/2) - ((CTs + CTe) /2)

Where HTs and HTe denote the start and end temperature of the Hot Stream. CTs and CTe indicates the start and end temperature of the Cool Stream.
So, here.

AMTD = ((170+150)/2) - ((50+70)/2).
AMTD = 100.

Nitesh Ghidode said:   1 year ago
Answer should be 100 for AMTD and Infinite for LMTD.

Nitesh ghidode said:   1 year ago
It's a log mean temperature difference.

Neeraj kumar said:   1 year ago
LMTD= ΔT1/ln(ΔT1/ΔT2)

ΔT1=170-70 = 100.
ΔT2=150-50 = 50.

Then LMTD = 100/Ln(100/100),
= 100/0.
= ~ infinite.

Options D is the right answer.

Nilesh Singh said:   2 years ago
T1 = 170 - 150 = 20,
T2 = 50 - 70 = -20.
AM=20+(-20)/2= infinity.

Raj Musale said:   4 years ago

Tejprakash said:   5 years ago
Yes, I agree with you @Ashok.

Sk jha said:   6 years ago
@Ashok

I don't think so.

AMTD = (170+150)/2 - (70+50)/2,
= 320/2 - 120/2,
= 160 - 60 = 100 Ans.

Romil kikani said:   6 years ago
I think it is logarithmic mean, not an arithmetic mean.

Ashok kumar said:   7 years ago
= (170 - 70) / (150 - 50),

ln (100 / 100) is equal to infinity.