LMTD for a heat exchanger is given by: (a) (frac{Delta t_{2}-Delta t_{1}}{log _{e} frac{Delta t_{2}}{Delta t_{1}}}) (b)

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LMTD for a heat exchanger is given by:

(a) $\frac{\Delta t_{2}-\Delta t_{1}}{\log _{e} \frac{\Delta t_{2}}{\Delta t_{1}}}$

(b) $\frac{\Delta t_{2}-\Delta t_{1}}{\log _{e} \frac{\Delta t_{1}}{\Delta t_{2}}}$

(c) Both

(d) None of these

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Basic Mechanical Engineering

ISBN: 9789386873293

2nd Edition

Authors: PRAVIN KUMAR

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Question Posted: Apr 09, 2024 02:14 AM

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LMTD for a heat exchanger is given by: (a) (frac{Delta t_{2}-Delta t_{1}}{log _{e} frac{Delta t_{2}}{Delta t_{1}}}) (b)

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Basic Mechanical Engineering

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