GATE 2024 BT Q.54: solution and answer

GATE 2024 · BTQ.54 · 2 marks · NAT

Fermentation medium is cooled from 121 °C to 30 °C in a double pipe heat exchanger. If cold water is flowing in the counter-current direction and is heated from 10 °C to 70 °C, then the Log-Mean Temperature Difference (LMTD) is ________ °C (rounded off to the nearest integer).

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Answer and solution

ANSWER

In counter-current flow the end differences are $Δ T_{1} = 121 - 70 = 51^{\circ} C$ and $Δ T_{2} = 30 - 10 = 20^{\circ} C$ . This gives $Δ T_{l m} = 31/ ln 2.55 = 33.12 \approx 33^{\circ} C$ .

Official answer: 33 to 35Source: Official answer key, GATE 2024 BT (IISc Bengaluru)

Write down the four end temperatures. Hot fermentation medium: $T_{h, in} = 121^{\circ} C$ , $T_{h, out} = 30^{\circ} C$ . Cold water: $T_{c, in} = 10^{\circ} C$ , $T_{c, out} = 70^{\circ} C$ .
In counter-current flow the two streams enter at opposite ends of the exchanger. So the hot inlet faces the cold outlet at one end, and the hot outlet faces the cold inlet at the other end.
Temperature difference at the hot-inlet end: $Δ T_{1} = T_{h, in} - T_{c, out} = 121 - 70 = 51^{\circ} C$ .
Temperature difference at the hot-outlet end: $Δ T_{2} = T_{h, out} - T_{c, in} = 30 - 10 = 20^{\circ} C$ . Both differences are positive, so the hot stream stays hotter than the cold stream all along the exchanger and the duty is feasible.
Apply the log-mean temperature difference definition: $Δ T_{l m} = \frac{Δ T _{1} - Δ T _{2}}{ln ( \frac{Δ T _{1}}{Δ T _{2}} )} = \frac{51 - 20}{ln ( \frac{51}{20} )} = \frac{31}{ln 2.55}$
Use the natural logarithm: $ln 2.55 = 0.9361$ , so $Δ T_{l m} = \frac{31}{0.9361} = 33.12^{\circ} C$ .
Rounded to the nearest integer, $Δ T_{l m} \approx 33^{\circ} C$ .

KEY IDEA

The LMTD is the log mean of the temperature differences at the two ends of the exchanger, $Δ T_{l m} = (Δ T_{1} - Δ T_{2}) / ln (Δ T_{1} /Δ T_{2})$ . The flow direction decides which terminal temperatures are paired. In counter-current flow, each stream's inlet sits at the same end as the other stream's outlet.

WATCH OUT FOR

(1) Pairing the ends as if the flow were co-current: $121 - 10 = 111$ and $30 - 70 = - 40$ . The negative difference shows that co-current flow cannot deliver this duty at all, because the water leaves hotter than the medium. (2) Using $lo g_{10}$ in place of $ln$ , which gives $31/0.4065 \approx 76^{\circ} C$ . (3) Using the arithmetic mean $(51 + 20) /2 = 35.5$ , which rounds to 36 and falls outside the key.

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