2 dT Figure 8-13 8-28 Figure 8-13, which is similar to Fig. 8-9(b), shows a Carn

Question

2 dT Figure 8-13 8-28 Figure 8-13, which is similar to Fig. 8-9(b), shows a Carnot cycle in the liquid vapor region. The working substance is 1 kg of water, and T 453 K, T. 313 K Steam tables list values of T, P, u, s, and h at points on the saturation lines and these are tabulated below, in MKS units, for points a, b, e, and f. e wish to make a complete analysis of the cycle. Point TOK) PON m 2) uo kg 1) s kg 1 K-1) 180 453 7.82 x 105 10 x 105 7.60 x 105 2140 180 453 27.7 x 103 10 x 105 25.8 x 105 6590 40 313 074 x 105 1.67 x 105 1.67 x 105 572 400 313 074 x 105 24.3 x 105 25.6 x 105 8220

Explanation / Answer

(e) Mass fraction or vapour quality is,

x = havg-hf / hfg

For point c,

x2 = havg - he / hf - he

Also, TdS = dh - VdP

and if pressure is assumed constant, TdS = dh

So, x2 = Sb-Se / Sf-Se

Similarly, at point d,

x1 = Sa-Se / Sf-Se

(f) Now,

  x2 = Sb-Se / Sf-Se

At constant volume, dh = TdS = dU

So, x2 = Ub-Ue / Uf -Ue

   Internal energy is same at constant pressure, so Ub = Uc

      x2 = Uc-Ue / Uf -Ue

or Uc = Ue+x2(Uf-Ue)

In terms of specific heat,

hc = he+x2(hf-he)

From the quality at point c, we have

x1 = Sa-Se / Sf-Se

Similarly as above, we get

Ud = Ue+x1(Uf-Ue)

and   hd = he+x1(hf-he)

(g)

Work done in expansion,

W = -nRdT

= -1*8.314*(453-313)

= - 1163.96 J

(h)

Work done in comression,

W = nRdT

= 1163.96 J

So the ratio of exapansion work to compression work is unity.

(i)

Net work done,

W = Wexpansion + Wcompression

   = -1163.96 + 1163.96

= 0 J

so, no work is done throughout the cycle.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.