Question

1. Consider a cascade system that uses vapor compression refrigeration for both high and low temperature cycles. Saturated vapor ammonia enters the compressor in the low temperature cycle at 20 psi and exits at 90 psi as it enters the heat exchanger. R134a enters the compressor in the high temperature cycle at 20^oF and a quality of 0.84, and exits at 120 psi as it enters the high temperature condenser. Given a refrigeration capacity of 10 tons, determine:

1. Ratio of the mass flow rate of ammonia to R134a
2. Power input of each compressor (hp)
3. Coefficient of performance

The answer are:

a)0.083

b)8.38 hp and 10.5 hp

c)2.5

I just need to know the process to get the answers

Answer 1

7 At state & (P= 20 psi, a = 1) hi - 605.76 bullb = hal popsi S] 1.3687 bolur Si- zopsi 2. on A+ 8 A state 2 (p=gopsi, isi=S2) from superheated table h = 691.98 btuleb At states (p=gopsi ,x=0) that gopsi Now, Erat =holl = 98.05 btolkb At ma ha + no ho = ma ho et mehr > Macha-ha) = MiB (hs - hs) state y hz = hy = 98.05 btulb At hs-hs h₂ - hz states (p=20°F, &=0.54) hes from propesfic R-134 a table hal h48/20°F 90.7 ~ 40.9) + ax ㄷ hr 98-05 6914.98 0.0838 A = 17.74 + o. suxl 8f. 87) 90.71 brullb ax Ssa SH2001 S2/0°F WI MA Chhi) = 0.0393 + 0.84% (0.2205-0.6393) 0.1915 btulub R At from Nowo, E = 10x3.55 ng x Chihu) =) MA = 0.0 689 kots WI = 0.0659% (691.98 – 605.76) 5.543 kw - Ft hp 8.08 Ma el ho-hs) 100.07 El 0.0689 x (4282 - 90.71) wsit Wn 10.87 hp At state of (p=ropsi, a=0) has hy = 40.91 btullb 7. 695 rewa [llw= 1.36 hp] Af state 8 haths = 40.91 brolly 10x3.5 At stafe 6: Solp=ropsi 0.2165 btulbr al cop = 5.94347.695 Witwi - 2.56 SocSg Hence, Doint 6 lies in wet region

Ss 256 = 001915 tulko-R =) SH + ax Stel 12opsi =oolgis repsi 10.0839 + ax (0.2165-0.0839) -0.1915 =) x = 0.811 hoa hall propsi + ax host proin = 4o.glotoistIN7 2.91 I 100.07 Brullb

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