Consider R-134a as the substance. (a) Find the specific volume (ft3/lbm) and spe

Question

Consider R-134a as the substance.

(a) Find the specific volume (ft3/lbm) and specific entropy (Btu/lbm-R) when P = 90 psia and h = 125.695 Btu/lbm.

(b) Find the pressure (psia) and specific volume (ft3/lbm) when T = 60?F and s = 0.10305 Btu/lbm-R.

(c) Find the specific volume (ft3/lbm) and specific entropy (Btu/lbm-R) when P = 80 psia and T = 40?F.

(d) Locate states (a), (b), and (c) on P-v and T-s diagrams for R-134a. Label states, show property values, and indicate appropriate lines of constant temperature or pressure.

Explanation / Answer

(a) at P=90psia

Ts=72.8F

vf=0.0132, vg=0.5294

sf=0.0748, sg=0.2202

hf=35.7, hg=113.2

Cpl=0.3396, Cpg=0.2395

Since h=125.695>hg

so refrigerant is in superheated state

so specific volume=vg=0.5294 ft3/lb

let T2 be the temperature of refrigerant

so,

h=hg+Cpg*(T2-Ts)

125.695=113.2+0.2395(T2-72.8)

T2=124.96F

so entropy at this state

s2=sg+Cpg*ln(T2/Ts)=0.2202+0.2395*ln(124.96/72.8)=0.3496

(b) At T=60 F , P=72.167psia(saturation pressure)

sf=0.0666, sg=0.2208

vf=0.0129, vg=0.6622

since s=0.10305<sg

so refrigerant is inside vapour dome

let x be the quality of refrigerant

s= sf+x*(sg-sf)

0.10305=0.0666+x(0.2208-0.0666)

x=0.236

so specific volume v=vf+x(vg-vf)=0.0129+0.236(0.6622-0.0129)=0.1661 ft3/lb

(c)

at P=80psia, Ts=65.9F

sf=0.0704, sg=0.2205

vf=0.0130

Cpl=0.3359, Cpg=0.2342

since T=40 F<Ts

so refrigerant is in subcooled region

so, specific volume v=vf=0.0130 ft3/lb

specific entropy s= sf=0.0704

(d) Above states can be easily drawn as
I have mention where the points lies e.g. superheated zone, inside vapour zone and subcooled

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