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LMTD

The hot gases...

Question

The hot gases enters a parallel flow heat exchanger at $425^{o} C$ and leaves at $260^{o} C$ , heating 25kg/s of water from $32^{o} C$ to $82^{o} C$ . For a surface area of $370 m^{2}$ , what is the overall heat transfer coefficient in $(W / m^{2} K)$

521.4

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13.03

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52.14

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26.07

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Solution

The correct option is C 52.14

Heat transfered by the heat exchanger = Heat gained by the cold water

$U A (L M T D) =_{c} C p_{c} (Δ T)_{e}$

$U = \frac{˙ m C_{p} (Δ T)}{A (L M T D)}$

$L M T D = \frac{(Δ T)_{i} - (Δ T)_{e}}{l n (\frac{(Δ T)_{i}}{(Δ T)_{e}})}$

From equation (1)

$U = \frac{25 \times 4200 (82 - 32)}{370 \times \frac{(393 - 178)}{l n (\frac{393}{178})}}$

$U = 52.27 W / m^{2} - K$

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The hot gases enters a parallel flow heat exchanger at 425oC and leaves at 260oC, heating 25kg/s of water from 32oC to 82oC. For a surface area of 370m2, what is the overall heat transfer coefficient in (W/m2K)

The correct option is C 52.14 Heat transfered by the heat exchanger = Heat gained by the cold water UA(LMTD)=˙mcCpc(ΔT)e U=˙mCp(ΔT)A(LMTD) LMTD=(ΔT)i−(ΔT)eln((ΔT)i(ΔT)e) From equation (1) U=25×4200(82−32)370×(393−178)ln(393178) U=52.27W/m2−K

The hot gases enters a parallel flow heat exchanger at $425^{o} C$ and leaves at $260^{o} C$ , heating 25kg/s of water from $32^{o} C$ to $82^{o} C$ . For a surface area of $370 m^{2}$ , what is the overall heat transfer coefficient in $(W / m^{2} K)$