A simple Rankine cycle uses water as the working fluid. The boiler operates at 6000 kPa and the condenser at 50 kPa. At the entrance of the turbine the temperature is 450 deg C. The isentropic efficiency of the turbine is 94 percent, pressure and pump losses are negligible, and the water in the condenser is subcooled by 6.3 degC. The boiler is sized for a mass flow rate of 20 kg/s. Determine the rate at which heat is added in the boiler, the power required to operate the pumps, the net power produced by the cycle, and the thermal efficiency.
Answer:
the rate at which heat is added in the boiler = 59597.4 kW
the power required to operate the pumps = 122.57 kW
The net power produced by the cycle = 17925 kW.
The thermal efficiency = 30%.
Explanation:
The specific enthalpy of saturated liquid is equal to the enthalpy of the first point which is equal to 314 kJ/ kg.
The second enthalpy is calculated from the pump work. Therefore, the second enthalpy = first enthalpy point + specific volume of water [ the pressure of the boiler - the pressure of the condenser].
The second enthalpy = 314 + 0.00103 [ 6000 - 50 ] = 320.13 kJ/kg.
The specific enthalpy for the third point = 3300 kJ/kg.
Therefore, the rate at which heat is added in the boiler = 20 × [3300 - 320.13] = 59597.4 kW.
The rate at which heat is added in the boiler = 59597.4 kW.
Also, the power required to operate the pumps = 20 × 0.00103 [6000 - 50] = 122.57 kW.
The power produced by the turbine = 20 [ 300 - ( the fourth enthalpy value)].
The fourth enthalpy value = 3300 - 0.94 [ 3300 - 2340] = 2397.6 kJ/kg
Thus, the power produced by the turbine = 20 [ 300 - 2397.6] = 18048 kW.
The power produced by the turbine = 18048 kW.
The net power produced = 18048 + 122.57 = 17925 kW.
The thermal efficiency = [net power produced] / [the rate at which heat is added in the boiler].
The thermal efficiency = 17925/ 59597.4 = 30%.
A reservoir rock system located between a depth of 2153m and a depth of
2383m , as the pressure at these depths is 18.200 MPa , 19.643 MPa
respectively the thickness of oil zone 103m, if the density of water is 1060 kg/m3
Determine the oil and gas density. what is the pressure at the depth of 2200m ?
what is the depth at which the pressure is 1900 MPa? Determine the gas-oil and
oil- water contact depth.
(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are held constant at 300oC and 100oC, respectively.(b) Determine the heat loss per hour if the cross-sectional area of the sheet is 0.25 m2.(c) What will be the heat loss per hour if a sheet of soda-lime glass is used instead
Answer:
do the wam wam
Explanation:
The heat flux is =1038kW/m² , the heat lost per hour is =259.5 kW, the heat lost per hour using a sheet of soda- lime glass.
Calculation of heat fluxThe thickness of steel( t) = 10mm = 10× 10^-³m
The temperature difference on both sides = 300-100
∆T = 200°C
But the formula for heat flux = q = k∆T/t
Where K = thermal conductivity for steel = 51.9W/mK.
Substitute the variables into the formula for heat flux;
q = 51.9 × 200/10 × 10-³
q = 10380 × 10³/10
q = 10380000/10
q = 1038000 W/m² = 1038kW/m²
To calculate the heat lost per hour if the cross sectional area is = 0.25 m2 use the formula q × A
= 1038kW/m² × 0.25 m2
= 259.5 kW.
Learn more about heat flux here:
https://brainly.com/question/15217121
Solved this question??????????????????
I do not know!!!!!!!!
For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.6 MN/mm per bolt. The bolts are preloaded to 75 percent of proof strength. Assume the external load is equally distributed to all the bolts. The bolts are M6 × 1 class 5.8 with rolled threads. A fluctuating external load is applied to the entire joint with Pmax = 60 kN and Pmin = 20 kN.(a) Determine the yielding factor of safety.(b) Determine the overload factor of safety.(c) Determine the factor of safety based on joint separation.(d) Determine the fatigue factor of safety using the Goodman criterion.
Answer:
a) 0.978
b) 0.9191
c) 1.056
d) 0.849
Explanation:
Given data :
Stiffness of each bolt = 1.0 MN/mm
Stiffness of the members = 2.6 MN/mm per bolt
Bolts are preloaded to 75% of proof strength
The bolts are M6 × 1 class 5.8 with rolled threads
Pmax =60 kN, Pmin = 20kN
a) Determine the yielding factor of safety
[tex]n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }[/tex] ------ ( 1 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
Input the given values into the equation above
equation 1 becomes ( np ) = [tex]\frac{380*20.1}{0.277*7500*5728.5}[/tex] = 0.978
note : values above are derived values whose solution are not basically part of the required solution hence they are not included
b) Determine the overload factor of safety
[tex]n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}[/tex] ------- ( 2 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
input values into equation 2 above
hence : [tex]n_{L}[/tex] = 0.9191[tex]n_{L} = 0.9191[/tex]
C) Determine the factor of safety based on joint separation
[tex]n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }[/tex]
Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,
input values into equation above
Hence [tex]n_{0}[/tex] = 1.056
D) Determine the fatigue factor of safety using the Goodman criterion.
nf = 0.849
attached below is the detailed solution .
At steady state, the power input of a refrigeration cycle is 500 kW. The cycle operates between hot and cold reservoirs which are at 550 K and 300 K, respectively. a) If cycle's coefficient of performance is 1.6, determine the rate of energy removed from the cold reservoir, in kW. b) Determine the minimum theoretical power required, in kW, for any such cycle operating between 550 K and 300 K
Answer:
The answer is below
Explanation:
Given that:
Hot reservoir temperature ([tex]T_H[/tex]) = 550 K, Cold reservoir temperature ([tex]T_C[/tex]) = 300 K, power input ([tex]W_{cycle}=500 \ kW[/tex]), cycle's coefficient of performance([tex]\beta_{actual}[/tex]) = 1.6
a) The rate of energy removal in the cold reservoir ([tex]Q_C[/tex]) is given by the formula:
[tex]Q_C=\beta_{actual}* W_{cycle}\\\\Q_C=1.6*500\\\\Q_C=800\ kW[/tex]
b) The maximum cycle's coefficient of performance([tex]\beta_{max}[/tex]) is:
[tex]\beta_{max}=\frac{T_C}{T_H-T_C}=\frac{300}{550-300}=1.5\\\\For\ minimum\ theoretical\ power\ \beta_{max}=\beta_{actual}=1.5\\\\W_{cycle}=\frac{Q_C}{\beta_{actual}} =\frac{800}{1.5} \\\\W_{cycle}=533.3\ kW[/tex]
difference between theory and practice?
Answer:
There is a huge difference between theory vs. practice. Theory assumes an outcome, while practice allows you to test the theory and see if it is accurate.
Theory and Practice Explained
Practice is the observation of disparate concepts (or a phenomenon) that needs explanation. A theory is a proposed explanation of the relationship between two or more concepts, or an explanation for how/why a phenomenon occurs.
In order to properly construct this road you have determined that you need to bring in 135,000 CCY of fill material to construct the sub-base. The fill material that is available is dry earth, with a weight of 2800 lbs/BCY and a moisture content of 7%. The percent swell for this soil is 25%. The final compaction level will be 118 PCF dry density and the optimum moisture content is 6%. How many BCY of earth must be excavated to meet this requirement
Answer:
The volume of water needed to be removed = 4301100 lbs/ [ 62.43 × 27] = 2551.66.
Explanation:
The first thing to do right now is to determine the weight of soil solids. This can be gotten below as:
The weight of soil solids = 1 × 118 × 27 = 3186 lbs.
Also, there is the need to determine the value of the water that we have and the one that is needed. This is calculated below as:
The weight of water needed = 3186 × 6% = 191.16 lbs.
The water available = 3186 × 7% = 223.02%.
The next thing to determine is the value for the total BCY of available of soil.
Thus, the total BCY of available of soil = [ 3186 + 223.02 ]/ 2800 × 135000 = 3409.02/ 2800 = 1.217551 BCY × 135000 = 164,363.46 BCY.
The last thing to do here is to determine the value for the weight and volume of water to be removed.
The weight of water to be removed = [ 223.02 lbs - 191.16 lbs] × 135,000 = 4301100 lbs
The volume of water needed to be removed = 4301100 lbs/ [ 62.43 × 27] = 2551.66.
A car with tires pressurized to 270 kPa leaves
Los Angeles with the tire temperature at 30°C
Estimate the tire pressure
(gage) when the
car arrives in New York with a tire temperature of 65°C .
I think Charles law should work here
A 4-m-high and 6-m-long wall is constructed of twolarge 2-cm-thick steel plates (k 5 15 W/m·K) separated by1-cm-thick and 20-cm wide steel bars placed 99 cm apart. Theremaining space between the steel plates is filled with fiber-glass insulation (k 5 0.035 W/m·K). If the temperature dif-ference between the inner and the outer surfaces of the wallsis 22°C, determine the
Answer:
fart
Explanation:
A benefit to using the medium the author used in "Great Rock and Roll
Pauses" is that the audience:
Incomplete question. The options read;
A. can change the story's ending
B. listens to the dialogue
C. hears the rock songs
D. feels more connected to the text.
Answer:
D. feels more connected to the text.
Explanation:
Remember, the underlying motive of an author when writing any text is to make his readers or audience more connected to the material been read or heard.
Hence, in "Great Rock and Roll Pauses" we can conclude that the author's choice of medium was motivated by the same desire of making the audience feel more connected to the text.
Answer:
The audience can actually hear the music.
Explanation: dont listen to below average iq people :0