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Hot and cold fluids enter a parallel flow double tube heat exchanger at 100oC and 15oC, respectively. The heat capacity rates of hot and cold fluids are Ch = 2000 W/K and Cc = 1200 W/K, respectively. If the outlet temperature of the cold fluid is 45oC, the log mean temperature difference (LMTD) of the heat exchanger is ________ K (round off to two decimal places).
    Correct answer is '57.71°'. Can you explain this answer?
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    Hot and cold fluids enter a parallel flow double tube heat exchanger a...

    Using the energy balance equation for the heat exchanger


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    Hot and cold fluids enter a parallel flow double tube heat exchanger a...
    Given data:
    Hot fluid inlet temperature (Th1) = 100°C
    Cold fluid inlet temperature (Tc1) = 15°C
    Outlet temperature of cold fluid (Tc2) = 45°C
    Heat capacity rate of hot fluid (Ch) = 2000 W/K
    Heat capacity rate of cold fluid (Cc) = 1200 W/K

    To find:
    Log mean temperature difference (LMTD)

    Solution:
    Step 1: Calculate the temperature difference between the hot and cold fluids at the outlet:
    ΔT1 = Th1 - Tc2
    ΔT1 = 100°C - 45°C
    ΔT1 = 55°C

    Step 2: Calculate the temperature difference between the hot and cold fluids at the inlet:
    ΔT2 = Th2 - Tc1
    Since Th2 is not given, we need to find it.

    Step 3: Calculate the outlet temperature of the hot fluid (Th2) using the heat capacity rates:
    Q = Ch * ΔT1 = Cc * ΔT2
    2000 * 55 = 1200 * ΔT2
    ΔT2 = (2000 * 55) / 1200
    ΔT2 = 91.67°C

    Step 4: Calculate the temperature difference between the hot and cold fluids at the inlet:
    ΔT2 = Th2 - Tc1
    91.67°C = Th2 - 15°C
    Th2 = 91.67°C + 15°C
    Th2 = 106.67°C

    Step 5: Calculate the log mean temperature difference (LMTD):
    LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
    LMTD = (55°C - 91.67°C) / ln(55°C / 91.67°C)
    LMTD = -36.67°C / ln(0.6)
    LMTD = -36.67°C / -0.5108
    LMTD = 71.74°C

    Step 6: Round off the LMTD to two decimal places:
    LMTD = 71.74°C ≈ 71.74°C

    The correct answer is 71.74°C, rounded to two decimal places, which is approximately 71.74°C.
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    Hot and cold fluids enter a parallel flow double tube heat exchanger at 100oC and 15oC, respectively. The heat capacity rates of hot and cold fluids are Ch = 2000 W/K and Cc = 1200 W/K, respectively. If the outlet temperature of the cold fluid is 45oC, the log mean temperature difference (LMTD) of the heat exchanger is ________ K (round off to two decimal places).Correct answer is '57.71°'. Can you explain this answer?
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    Hot and cold fluids enter a parallel flow double tube heat exchanger at 100oC and 15oC, respectively. The heat capacity rates of hot and cold fluids are Ch = 2000 W/K and Cc = 1200 W/K, respectively. If the outlet temperature of the cold fluid is 45oC, the log mean temperature difference (LMTD) of the heat exchanger is ________ K (round off to two decimal places).Correct answer is '57.71°'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Hot and cold fluids enter a parallel flow double tube heat exchanger at 100oC and 15oC, respectively. The heat capacity rates of hot and cold fluids are Ch = 2000 W/K and Cc = 1200 W/K, respectively. If the outlet temperature of the cold fluid is 45oC, the log mean temperature difference (LMTD) of the heat exchanger is ________ K (round off to two decimal places).Correct answer is '57.71°'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hot and cold fluids enter a parallel flow double tube heat exchanger at 100oC and 15oC, respectively. The heat capacity rates of hot and cold fluids are Ch = 2000 W/K and Cc = 1200 W/K, respectively. If the outlet temperature of the cold fluid is 45oC, the log mean temperature difference (LMTD) of the heat exchanger is ________ K (round off to two decimal places).Correct answer is '57.71°'. Can you explain this answer?.
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