Answer:
temperature T2 = 826.9°C
= -1142.7 kJ/kg
Explanation:
given data
initial state temperature = 120°C
final state pressure p1 = 1 bar
pressure p2 = 100 bar
solution
we use here super heated water table A6 that is
specific internal energy u1 = 2537.3 kJ/kg
specific entropy s1 = 7.4668 kJ/kg.K
and
here specific entropy stage 1 = stage 2
so for specific entropy and pressure 100 bar
specific internal energy u2 = 3680 kJ/kg
and temperature T2 = 826.9°C
so here now we get specific work of steam is
ΔU = -W ...........1
m ( u1 - u2) = W
= u1 - u2
= 2537.3 - 3680
= -1142.7 kJ/kg
C smaller than that of the larger-diameter cylinder
Answer:
<u>note:</u>
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
Answer:
19063.6051 g
Explanation:
Pressure = Atmospheric pressure + Gauge Pressure
Atmospheric pressure = 97 kPa
Gauge pressure = 500 kPa
Total pressure = 500 + 97 kPa = 597 kPa
Also, P (kPa) = 1/101.325 P(atm)
Pressure = 5.89193 atm
Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)
Temperature = 28 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K
⇒n = 595.76 moles
Molar mass of oxygen gas = 31.9988 g/mol
Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g