A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is th
2 answers:
Answer:
98.13kJ
Explanation:
Given that;
The rigid 10-L vessel initially contains a mixture of liquid water & vapor at
We are to calculate the heat transfer required during the process by obtaining our data from the steam tables.
In order to do that, let start with our Energy Balance
So, Energy Balance for closed rigid tank system is given as:
Since the K.E and P.E are insignificant;
∴ K.E = P.E = 0
Where;
m = mass flow rate of the mixture
= corresponding change in the internal energy at state point 2 and 1
However, since we are informed that the vessel is rigid, then there is no work done in the system, then W turn out to be equal to zero .i,e
W = 0
we have our above equation re-written as:
We were told to obtain our data from the steam table, so were going to do just that
∴ At inlet temperature , the given quality of mixture of liquid water and vapor
= 123% = 0.123
Using the equation:
where;
= specific volume at state 1
= specific volume of the liquid
= specific volume of the liquid vapor mixture
The above data from the steam table is given as;
= 0.001043 m³/kg
= 1.6720 m³/kg
so; we have
At = 100°C and
;
the following steam data from the tables were still obtained for the internal energy; which is given as:
Internal Energy at the state 1
where;
specific internal energy of the liquid = 419.06 kJ/kg
The specific internal energy of the liquid vapor mixture = 2087.0 kJ/kg
∴ since ;
= 419.06 + (0.123 × 2087.0)
= 675.761 kJ/kg
As the tank is rigid, so as the volume which is kept constant:
Now, let take a look at when from the data in the steam tables
Specific volume of the liquid = 0.00113 m³/kg
specific volume of the liquid vapor mixture = 0.19384 m³/kg
The quality of the mixture at the final state 2 can be determined by using the equation shown below:
= 1.0657
From our usual steam table; we still obtained data for the Internal Energy when
Specific internal energy of the liquid = 761.92 kJ/kg
Specific internal energy of the liquid vapor mixture = 1820.88 kJ/kg
Calculating the internal energy at finsl state point 2 ; we have:
= 761.92 + (1.0657 × 1820.88)
= 761.92 + 1940.511816
= 2702.431816
≅ 2702.43 kJ/kg
Furthermore, let us calculate the mass in the system; we have:
where V₁ = the volume 10 - L given by the system and v₁ = specific volume at state 1 as 0.2065
V₁ = the volume 10 - L = 10 × ( 0.001 m³/L)
v₁ = 0.2065
∴
mass (m) =
= 0.04842 kg
Now, we gotten all we nee do calculate for the heat transfer that is required during the process:
Therefore, the heat transfer that is required during the process = 98.13 kJ
There you have it!, I hope this really helps alot!
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Answer:
minimum electric power consumption of the fan motor is 0.1437 Btu/s
Explanation:
given data
area = 3 ft by 3 ft
air density = 0.075 lbm/ft³
to find out
minimum electric power consumption of the fan motor
solution
we know that energy balance equation that is express as
E in - E out = ......................1
and at steady state = 0
so we can say from equation 1
E in = E out
so
minimum power required is
E in = W = m =
put here value
E in =
E in =
E in = 0.1437 Btu/s
minimum electric power consumption of the fan motor is 0.1437 Btu/s
Answer:
0.31
126.23 kg/s
Explanation:
Given:-
- Fluid: Water
- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%
- Pump: Isentropic
- Net cycle-work output, Wnet = 100 MW
Find:-
- The thermal efficiency of the cycle
- The mass flow rate of steam
Solution:-
- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.
First process: Isentropic compression by pump
P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )
h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )
s1 = s-P1 = 0.6492 KJ/kg.K
v1 = v-P1 = 0.001010 m^3 / kg
P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )
s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )
- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:
- From the following relation we can determine ( h2 ) as follows:
h2 = h1 + wp
h2 = 191.81 + 8.0699
h2 = 199.88 KJ/kg
Second Process: Boiler supplies heat to the fluid and vaporize
- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).
- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).
P3 = 8 MPa
T3 = ? ( assume fluid exist in the saturated vapor phase )
h3 = hg-P3 = 2758.7 KJ/kg
s3 = sg-P3 = 5.7450 KJ/kg.K
- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:
Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).
- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.
- Under the isentropic conditions the steam exits the turbine at the following conditions:
P4 = 10 KPa
s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )
- Compute the quality of the mixture at condenser inlet by the following relation:
- Determine the isentropic ( h4s ) at this state as follows:
- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:
- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.
- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.
Answer: The mass flow rate of the steam would be 126.23 kg/s
- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):
Answer: The thermal efficiency of the cycle is 0.31