Elementary Principles 257 - 7.26 H H 2 O l , 24 C, 10 bar =...

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7- 12 7.26 ± ,. Hl H O C, 10 bar kJ kg 2 bg ch 24 100 6 °= (Table B.5 for saturated liquid at 24 o C; assume ± H independent of P ). ± . H 10 bar, 2776 2 sat'd steam kJ kg = (Table B.6) Δ ± .. . H =− = 2776 2 100 6 26756 kJ kg 22 o 3 [kg H O(l)/h] [kg H O(v)/h] 24 C, 10 bar 15,000 m /h @10 bar (sat'd) (kW) mm Q ²² ² ² . . m == × A 15000 01943 772 10 4 m kg h m kg h 3 3 Table 8.6 Energy balance ΔΔ Δ ² , ² : ²²² EW H E Q ps k =+ = 0 di ² ² EE E kk k final initial final E k initial = 0 Δ ² ² . E mu AD k f × = A 2 2 3 2 2 3 2 11 1 4 015 4 2 1 2 7.72 10 kg 15,000 m h h J h m 3600 s kg m / s 43 23 2 2 π ² ² ± ² QmH E k = × + × × 596 10 57973 5 45 kg 2675.6 kJ 1 h h kg 3600 s J1 k J s 10 J kJ s .80 10 kW 3 4 7.27 (a) 228 g/min 228 g/min 25 o C T ( o C) ² ( Q kW) ² , ² , ² E x E p W s Energy balance: = 0 ² QH Q =⇒ = Δ W 228 g min J min 60 s g 1( ±± ) HH out in ⇒= ± . ² HQ W out Jg b g 0 263 T W ° = C b g 25 26 4 27 8 29 0 32 4 0 263
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