Q1: Hot water (0.01 m³ /min) enters the tube side of a counter current shell and tube heat exchanger at 80°C and leaves at 50°C. Cold oil (0.05 m³/min) of density 800 kg/m³ and specific heat of 2 kJ/kg.K enters at 20°C. The log mean temperature difference in °C is approximately

ANS:A - 32

To calculate the log mean temperature difference (LMTD) for a counter-current heat exchanger, we can use the formula: LMTD=Δ�1−Δ�2ln⁡(Δ�1/Δ�2)LMTD=ln(ΔT1​/ΔT2​)ΔT1​−ΔT2​​ Where:

• Δ�1ΔT1​ is the temperature difference at one end of the exchanger.
• Δ�2ΔT2​ is the temperature difference at the other end of the exchanger.

The formula for Δ�ΔT for a given fluid is: Δ�=�inlet−�outletΔT=Tinlet​−Toutlet​ Given:

• Hot water flow rate = 0.01 m³/min
• Hot water inlet temperature = 80°C
• Hot water outlet temperature = 50°C

For hot water: Δ�1=80°�−50°�=30°�ΔT1​=80°C−50°C=30°C Δ�2=20°�−50°�=−30°�ΔT2​=20°C−50°C=−30°C For cold oil: Δ�1=80°�−20°�=60°�ΔT1​=80°C−20°C=60°C Δ�2=50°�−20°�=30°�ΔT2​=50°C−20°C=30°C Now, let's substitute these values into the LMTD formula: LMTD=(30°−(−30°))ln⁡(30°−30°)=60°ln⁡(−1)LMTD=ln(−30°C30°C​)(30°C−(−30°C))​=ln(−1)60°C​ LMTD=60°ln⁡(1)LMTD=ln(1)60°C​ LMTD=60°0LMTD=060°C​ As the natural logarithm of a negative number is undefined, this indicates an arithmetic error in our calculations. To solve this issue, we need to recalculate the LMTD taking the absolute values of the temperature differences: LMTD=Δ�1−Δ2ln⁡(Δ1/Δ2)LMTD=ln(ΔT1​/ΔT2​)ΔT1​−ΔT2​​ LMTD=60°−30°�ln⁡(60°30°)LMTD=ln(30°C60°C​)60°C−30°C​ LMTD=30°ln⁡(2)LMTD=ln(2)30°C​ Using a calculator, we can approximate ln⁡(2)ln(2) to be about 0.693: LMTD≈30°0.693≈43.24°LMTD≈0.69330°C​≈43.24°C So, the approximate LMTD is 43.24°C. Therefore, among the provided options, the closest value to the calculated LMTD is 45. Hence, the answer is 45°C.