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2 years ago

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Cold water (cp = 4.18 kJ/kg⋅°C) leading to a shower enters a well-insulated, thin-walled, double-pipe, counterflow heat exchange

r at 10°C at a rate of 0.95 kg/s and is heated to 70°C by hot water (cp = 4.19 kJ/kg⋅°C) that enters at 85°C at a rate of 1.6 kg/s. Determine (a) the rate of heat transfer and (b) the rate of entropy generation in the heat exchanger.

1 answer:

jeka57 [31]2 years ago

5 0

Answer:

(a) The rate of heat transfer is <em>238.26 kW.</em>

(b) Rate of entropy generation is <em>0.063 kW/K.</em>

Explanation:

(a) Taking the heat gained by the cold water into consideration,

<em> $Q_{in} =mc(T_{2} -T_{1})$ </em>

<em>Q = (0.95 kg/s)(4.18 kJ/kg.⁰C)(70-10) = 238.26 kW.</em>

(b) Entropy generation = entropy in cold water + entropy in hot water

$S_{generated} = (m_{c} c_{c} (ln(\frac{T_{c2}}{T_{c1}})) + (m_{h} c_{h} (ln(\frac{T_{h2}}{T_{h1}}))$

But $\frac{T_{h2}}{T_{h1}} = 1-\frac{Q_{h} }{m_{h}c_{h}T_{h1}}$

∴ $S_{generated}$ = (0.95)(4.18)(ln(343/283)) + (1.6)(4.19)(ln(1- (238.26/((1.6)(4.19)(358))))

$S_{generated}$ = 0.7636 + (- 0.7009) = 0.0627 kW/K.

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Answer:

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Part a)

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Since turn table is accelerating uniformly

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