For H2O, determine the specific volume at the indicated state,
in m3/kg. (a) T = 400°C, p = 20 MPa. (b) T = 120°C, p = 20 MPa. (c)
T = 40°C, p = 2 MPa.
I have figured out how to do A, but can't get the correct answer
for B or C. All the examples I am finding are with numbers that are
listed already on the charts, and mine are not so I need to
approach the...
For H2O, determine the specific volume at the indicated state,
in m3/kg.
(a) T = 480°C, p = 20 MPa.
(b) T = 160°C, p = 20 MPa.
(c) T = 40°C, p = 2 MPa.
Problem 10.33
Complete the following table for an ideal gas:
P
V
n
T
2.10 atm
1.40 L
0.480 mol
? K
0.320 atm
0.250 L
? mol
22 ∘C
670 torr
? L
0.333 mol
360 K
? atm
595 mL
0.260 mol
295 K
A system contains H2O at P=150 kPa, v= 1.5 m^3 and a specific
volume of 0.5 m^3/kg (state 1). It is heated at constant volume
until the temperature reacues 160 celcius
What is the temperature at state 1 ?
Use the ideal gas law to complete the table:
P
V
n
T
1.06 atm
1.25 L
0.105 mol
___
115 torr
___
0.249 mol
309 K
___
28.2 mL
1.81×10−3 mol
26.0 ∘C
0.565 atm
0.440 L
___
257 K
Calculus dictates that
(∂U/∂V) T,Ni = T(∂S/∂V)T,Ni – p = T(∂p/∂T)V,Ni – p
(a) Calculate (∂U/∂V) T,N for an ideal gas [ for which p = nRT/V
]
(b) Calculate (∂U/∂V) T,N for a van der Waals gas
[ for which p = nRT/(V–nb) – a (n/V)2 ]
(c) Give a physical explanation for the difference between the
two.
(Note: Since the mole number n is just the particle number N
divided by Avogadro’s number, holding one constant is equivalent...
For H2O, determine the specified property at the
indicated state.
(a) T = 140°C, v = 0.5 m3/kg. Find
p, in bar.
(b) p = 30 MPa, T = 120°C. Find v, in
m3/kg.
(c) p = 10 MPa, T = 450°C. Find v, in
m3/kg.
(d) T = 80°C, x = 0.8. Find p, in bar,
and v, in m3/kg.
At the inlet of a steam turbine the following information are
given: p=1000 kPa, T=500 deg-C, u = 3124.3 kJ/kg, v = 0.35411, V=20
m/s, z=10 m, and mdot=10 kg/s.
determine the inlet area.
Determine the flow work at the inlet in MW
Determine the rate of transport of flow energy (Jdot).
If you neglect ke and pe in the expression for j, how would the
answer in part c change? in percentage