Question

one kg of steam at a pressure of 17 bar and dryness 0.95 is heated at a constant pressure, until it is completely dry . determine .bincrease in volume . quantity of heat added​

Answer 1

Answer:

Given: m = 1 kg, P = 17 bar, x = 0.95

Solution:

Now from steam tables,

h1=hf+xxhfg=871.8+0.95x1921.5=2697.32kJ/kg

h2=2793.4kJ/kg

Quantity of Heat added = m×(h2−h1)=96.08kJ

Now from steam tables

V1=xxVg=0.11664×0.95m3/kg

V2=Vg=0.11664m3/kg

Therefore, Increase in Volume = m×(V2−V1)=5.832×10−3m3

Answer 2

Answer:

Given: m = 1 kg, P = 17 bar, x = 0.95

Solution:

Now from steam tables,

h1=hf+xxhfg=871.8+0.95x1921.5=2697.32kJ/kg

h2=2793.4kJ/kg

Quantity of Heat added = m×(h2−h1)=96.08kJ

Now from steam tables

V1=xxVg=0.11664×0.95m3/kg

V2=Vg=0.11664m3/kg

Therefore, Increase in Volume = m×(V2−V1)=5.832×10−3m3.