In an ideal Bryaton ccle, atmospheric air (ratio of specific heats, $c_{p} / c_{v} = 1.4$ specifci heat at constant pressure $= 1.005 k J / k g K$ ) at $1$ bar adn $300 K$ is compressed to $8$ bar. The maximum temperature in the cycle is limited to $1280 K$ . If the heat is suppied at hte rate of $80 M W$ , the mass flow rate (in $k g / s$ ) of air required in the cycle is
108.07

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Solution

The correct option is A 108.07

Given, $\frac{c_{p}}{c_{v}} = γ = 1.4$
$c_{p} = 1.005 k J / k g K$
$p_{1} = 1$ bar
$T_{1} = 330 K$
$p_{2} = 8$ bar
$T_{3} = T_{m a x} = 1280 K$
$Q_{s} = 80 M W = 80 \times 10^{3} k W$
For proccess $1 - 2$ ,
$\frac{T_{2}}{T_{1}} = {(\frac{p_{2}}{p_{1}})}^{\frac{γ - 1}{γ}} = 8^{\frac{0.4}{1.4}} = 1.8114$
or $T_{2} = 300 \times 1.8114 = 543.43 K$
$Q_{s} = m c_{p} (T_{3} - T_{2})$
$80 \times 10^{3} = m \times 1.005 (1280 - 543.43)$
or $m = 108.07 k / s$

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The correct option is A 108.07 Given, cpcv=γ=1.4 cp=1.005kJ/kgK p1=1 bar T1=330K p2=8 bar T3=Tmax=1280K Qs=80MW=80×103kW For proccess 1−2, T2T1=(p2p1)γ−1γ=80.41.4=1.8114 or T2=300×1.8114=543.43K Qs=mcp(T3−T2) 80×103=m×1.005(1280−543.43) or m=108.07k/s