Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch ca
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Answer:
Explanation:
From the given information:
Water freezing temp.
Heat rejected temp
Recall that:
The coefficient of performance is:
Again:
The efficiency given by COP can be represented by:
Finally; the power input in an hour can be determined by using the formula:
In hp; since 1 kW = 1.34102 hp
6.86kW will be = (6.86 × 1.34102) hp
= 9.199 hp
Answer:
Explanation:
Availability:
It define as the probability of system which perform desired task before showing any failure .
The availability can be define as follows
Or we can say that
Availability can also be express as
Where MTBF is the mean time between two failure.
MTTR is the mean time to repair.
Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
Explanation:
from the T-S diagram, we get the overall pressure ratio of the cycle is 9
Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3
P5/P6 = P7/P8 = √9 =3
get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.
At temperature T1 =300K
Specific enthalpy of air h1 = 300.19 kJ/kg
Relative pressure pr1 = 1.3860
At temperature T5 = 1200 K
Specific enthalpy h5 = 1277.79 kJ/kg
Relative pressure pr5 = 238
Calculate the relative pressure at state 2
Pr2 = (P2/P1) Pr5
Pr2 =3 x 1.3860 = 4.158
get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure pr = 4.153
The corresponding specific enthalpy h = 411.12 kJ/kg
Relative pressure pr = 4.522
The corresponding specific enthalpy h = 421.26 kJ/kg
Find the specific enthalpy of state 2 by the method of interpolation
(h2 - 411.12) / ( 421.26 - 411.12) =
(4.158 - 4.153) / (4.522 - 4.153 )
h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))
h2 - 411.12 = 0.137
h2 = 411.257kJ/kg
Calculate the relative pressure at state 6.
Pr6 = (P6/P5) Pr5
Pr6 = 1/3 x 238 = 79.33
Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure Pr = 75.29
The corresponding specific enthalpy h = 932.93 kJ/kg
Relative pressure pr = 82.05
The corresponding specific enthalpy h = 955.38 kJ/kg
Find the specific enthalpy of state 6 by the method of interpolation.
(h6 - 932.93) / ( 955.38 - 932.93) =
(79.33 - 75.29) / ( 82.05 - 75.29 )
(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )
h6 - 932.93 = 13.427
h6 = 946.357 kJ/kg
Calculate the total work input of the first and second stage compressors
(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )
= 222.134 kJ/kg
Calculate the total work output of the first and second stage turbines.
(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )
= 662.866 kJ/kg
Calculate the net work done
Wnet = (Wturb)out - (Wcomp)in
= 662.866 - 222.134
= 440.732 kJ/kg
Calculate the minimum mass flow rate of air required to generate a power output of 105 MW
W = m × Wnet
(105 x 10³) kW = m(440.732 kJ/kg)
m = (105 x 10³) / 440.732
m = 238.2 kg/s
therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
