Answer:
[tex]$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$[/tex]
The entropy change of the carbon dioxide is -0.1104 kJ/kg.K
Explanation:
We are given that carbon dioxide undergoes a process in a closed system.
We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.
The entropy change of the carbon dioxide is given by
[tex]$ S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1}) $[/tex]
Where Cp is the specific heat constant
Cp = 0.846 kJ/kg.K
R is the universal gas constant
R = 0.1889 kJ/kg.K
T₁ and T₂ is the initial and final temperature of carbon dioxide.
P₁ and P₂ is the initial and final pressure of carbon dioxide.
[tex]$ S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50}) $[/tex]
[tex]$ S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888) $[/tex]
[tex]$ S_2 - S_1 = 0.5864 - 0.6968 $[/tex]
[tex]$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$[/tex]
Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K
How old are you? answer this question plz lol I will mark someone as brainliest
Answer:
100000000000000000000000
A steel alloy is known to contain 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C. Assume that there are no alterations in the positions of other phase boundaries with the addition of Ni. (a) What is the approximate eutectoid temperature of this alloy
Answer:
650°C or 1,200°F
Explanation:
Data provided in the question
Steel alloy contains 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C
Plus we also assume that there are no changes in the boundaries of postions who have other phases but there is an addition of Ni.
Based on the above information, the approximate eutectoid temperature of this alloy for 6.1 wt% is 650°C or 1,200°F
In real world, sampling and quantization is performed in an analog to digital converter (ADC) and reconstruction is performed in a digital to analog converter (DAC). Which of the following statements hold true (fs denotes the sampling frequency)?
a. the reconstruction filter can be found in the DAQ
b. the antialiasing filter removes all frequencies of the continuous-time analog input signal that are above fs/2
c. the DAC needs to know the sampling frequency of the ADC to correctly reconstruct the signal.
d. the reconstructed continuous-time signal only contains frequencies up to fs/2
Answer:
b
Explanation:
a) ADC is located on DAQ filter but not the reconstruction filter
b) to remove aliasing, the sampling rate must be greater than or equal ot twice the highest frequency component in the input signal. In other words, all frequencies in input sgnal are less than fs/2. Therefore, frequencies greater than fs/2 are removed by anti-aliasing filter
c) the DAC can have different sampling rate from ADC
A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum allowable normal stress of 150 MPa . If the inner diameter of the tank is 3 m , what is the minimum thickness, t, of the wall
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress [tex]\sigma[/tex] = 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress [tex]\sigma = \dfrac{pd}{2t}[/tex]
Making thickness t the subject; we have
[tex]t = \dfrac{pd}{2* \sigma}[/tex]
[tex]t = \dfrac{560000*3}{2*150000000}[/tex]
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
[tex]\sigma = \dfrac{pd}{4t}[/tex]
[tex]t= \dfrac{pd}{4*\sigma }[/tex]
[tex]t = \dfrac{560000*3}{4*150000000}[/tex]
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in.2 and a temperature of 60F at the beginning of compression. The maximum temperature in the cycle is 5200R. Based on this model, calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
Answer:
the net work per cycle [tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, W = 88.0144746 hp
Explanation:
the information given includes;
diameter of the four-cylinder bore = 3.7 in
length of the stroke = 3.4 in
The clearance volume = 16% = 0.16
The cylindrical volume [tex]V_2 = 0.16 V_1[/tex]
the crankshaft N rotates at a speed of 2400 RPM.
At the beginning of the compression , temperature [tex]T_1[/tex] = 60 F = 519.67 R
and;
Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²
= 2088 lb/ft²
The maximum temperature in the cycle is 5200 R
From the given information; the change in volume is:
[tex]V_1-V_2 = \dfrac{\pi}{4}D^2L[/tex]
[tex]V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)[/tex]
[tex]V_1-0.16V_1= 36.55714291[/tex]
[tex]0.84 V_1 =36.55714291[/tex]
[tex]V_1 =\dfrac{36.55714291}{0.84 }[/tex]
[tex]V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3[/tex]
[tex]V_1 = 0.02518 \ ft^3[/tex]
the mass in air ( lb) can be determined by using the formula:
[tex]m = \dfrac{P_1V_1}{RT}[/tex]
where;
R = 53.3533 ft.lbf/lb.R°
[tex]m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}[/tex]
m = 0.0018962 lb
From the tables of ideal gas properties at Temperature 519.67 R
[tex]v_{r1} =158.58[/tex]
[tex]u_1 = 88.62 Btu/lb[/tex]
At state of volume 2; the relative volume can be determined as:
[tex]v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}[/tex]
[tex]v_{r2} = 158.58 \times 0.16[/tex]
[tex]v_{r2} = 25.3728[/tex]
The specific energy [tex]u_2[/tex] at [tex]v_{r2} = 25.3728[/tex] is 184.7 Btu/lb
From the tables of ideal gas properties at maximum Temperature T = 5200 R
[tex]v_{r3} = 0.1828[/tex]
[tex]u_3 = 1098 \ Btu/lb[/tex]
To determine the relative volume at state 4; we have:
[tex]v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}[/tex]
[tex]v_{r4} =0.1828 \times \dfrac{1}{0.16}[/tex]
[tex]v_{r4} =1.1425[/tex]
The specific energy [tex]u_4[/tex] at [tex]v_{r4} =1.1425[/tex] is 591.84 Btu/lb
Now; the net work per cycle can now be calculated as by using the following formula:
[tex]W_{net} = Heat \ supplied - Heat \ rejected[/tex]
[tex]W_{net} = m(u_3-u_2)-m(u_4 - u_1)[/tex]
[tex]W_{net} = m(u_3-u_2- u_4 + u_1)[/tex]
[tex]W_{net} = m(1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (410.08)[/tex]
[tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, in horsepower. can be calculated as follows;
In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:
[tex]W = 4 \times N' \times W_{net[/tex]
where ;
[tex]N' = \dfrac{2400}{2}[/tex]
N' = 1200 cycles/min
N' = 1200 cycles/60 seconds
N' = 20 cycles/sec
W = 4 × 20 cycles/sec × 0.777593696
W = 62.20749568 Btu/s
W = 88.0144746 hp
The net work per cycle and the power developed by this combustion engine are 0.7792 Btu and 88.20 hp.
Given the following data:
Diameter of bore = 3.7 inStroke length = 3.4 inClearance volume = 16% = 0.16Speed of 2400 RPM.Initial temperature = 60 F to R = 519.67 R. Initial pressure = 14.5 [tex]lbf/in^2[/tex] to [tex]lbf/ft^2[/tex] = 2088 [tex]lbf/ft^2[/tex] Maximum temperature = 5200 R.Note: The cylindrical volume is equal to [tex]0.16V_1[/tex]
How to calculate the net work per cycle.First of all, we would determine the volume, mass and specific energy as follows:
[tex]V_1-V_2=\frac{\pi D^2L}{4} \\\\V_1-0.16V_1=\frac{3.142 \times 3.7^2 \times 3.4}{4}\\\\0.84V_1=36.56\\\\V_1=\frac{36.56}{0.84} \\\\V_1=43.52\;in^3 \;to \;ft^3 = 0.0252\;ft^3[/tex]
For the mass:
[tex]M=\frac{PV}{RT} \\\\M=\frac{2088 \times 0.0252}{53.3533 \times 519.67} \\\\M=\frac{52.6176}{27726.109411}[/tex]
M = 0.0019 lb.
At a temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r1}=158.58\\\\u_1 = 88.62\;Btu/lb[/tex]
For the relative volume at the second state, we have:
[tex]v_{r2}=v_{r1}\times \frac{V_2}{V_1} \\\\v_{r2}=158.58\times 0.16\\\\v_{r2}=25.3728[/tex]
Note: At 25.3728, specific energy ([tex]u_2[/tex]) is 184.7 Btu/lb.
At a maximum temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r3}=0.1828\\\\u_3 = 1098\;Btu/lb[/tex]
For the relative volume at state 4, we have:
[tex]v_{r4}=v_{r3}\times \frac{V_1}{V_3} \\\\v_{r4}=0.1828\times \frac{1}{0.16}\\\\v_{r4}=1.1425[/tex]
Note: At 1.1425, specific energy ([tex]u_4[/tex]) is 591.84 Btu/lb.
Now, we can calculate the net work per cycle by using this following formula:
[tex]W=Heat\;supplied -Heat\rejected\\\\W=m(u_3-u_2)-m(u_4-u_1)\\\\W=0.0019(1098-184.7)-0.0019(591.84-88.62)\\\\W=1.73527-0.956118[/tex]
W = 0.7792 Btu.
How to calculate the power developed.In a four-cylinder, four-stroke internal combustion engine, power is given by this formula:
[tex]W=4N'W_{net}[/tex]
But;
[tex]N'=\frac{N}{2 \times 60} \\\\N'=\frac{2400}{120} \\\\N'=20\;cycle/sec[/tex]
Substituting the given parameters into the formula, we have;
[tex]W=4 \times 20 \times 0.7792[/tex]
W = 62.336 Btu/sec.
In horsepower:
W = 88.20 hp.
Read more on net work here: https://brainly.com/question/10119215
A 3-phase, 50 Hz, 110 KV overhead line has conductors placed in a horizontal plane 3 m apart. Conductor diameter is 2.5 cm. If the line length is 220 km, determine the charging current per phase assuming complete transposition. (6 Marks)
Answer:
A 3-phase, 50 Hz, 110 KV overhead line has conductors
Explanation:
hope it will helps you
Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 50°C at a rate of 0.02 kg/s and leaves at 750 kPa subcooled by 3°C. The refrigerant enters the compressor at 200 kPa superheated by 4°C. Determine the isentropic efficiency of the compressor, the rate of heat supplied to the room, COP of the Heat Pump and the rate of heat supplied to this room if the heat pump operated on an ideal vapor compression cycle between pressure limits of 200 and 800 kpa
Explanation:
The value of enthalpy and entropy at state 1 will be determined according to the given pressure and temperature as follows using interpolation from A-13 is as follows.
[tex]h_{1}[/tex] = 247.88 kJ/kg, [tex]S_{1}[/tex] = 0.9579 kJ/kg K
At state 2, isentropic enthalpy will be determined from the condition [tex]S_{2} = S_{1}[/tex] and given pressure at 2 with data from A-13 using interpolation is:
[tex]h_{2s}[/tex] = 279.45 kJ/kg
We will calculate actual enthalpy at state 2 using given pressure and temperature from A-13 as follows.
[tex]h_{2}[/tex] = 286.71 kJ/kg
Hence, isentropic compressor efficiency will be calculated using standard relation as:
[tex]\eta_{c} = \frac{h_{2s} - h_{1}}{h_{2} - h_{1}}[/tex]
= [tex]\frac{279.45 - 247.88}{286.71 - 247.88}[/tex]
= 0.813
Now, at state 3 enthaply is determined by temperature at state 3, that is, [tex]26^{o}C[/tex] for given pressure as per saturated liquid approximation and data from A-11.
[tex]h_{3}[/tex] = 87.83 kJ/Kg
Using energy balance in 2-3, the rate of heat supplied to the heated room is as follows.
[tex]Q_{H} = m(h_{2} - h_{3})[/tex]
= 0.022 (286.71 - 87.83) kW
= 4.38 kW
Now, COP will be calculated using power that is expressed through energy balance in 1-2 as follows.
COP = [tex]\frac{Q_{H}}{W}[/tex]
= [tex]\frac{Q_{H}}{m(h_{2} - h_{1})}[/tex]
= [tex]\frac{4.38}{0.022 (286.71 - 246.88)}[/tex]
= 5.13
In an ideal vapour-compression cycle, the enthalpy and entropy at state 1 will be obtained from given pressure and state with data from A-12:
[tex]h_{1}[/tex] = 244.5 kJ/kg
[tex]S_{1}[/tex] = 0.93788 kJ/kg K
[tex]h_{2}[/tex] = 273.71 kJ/kg
At state 3, enthalpy will be determined from given pressure and state with data from A-12 as follows.
[tex]h_{3}[/tex] = 95.48 kJ/kg
Hence, using energy balance in 2-3 the rate of heat supplied will be calculated as follows.
[tex]Q_{H} = m(h_{2} - h_{3})[/tex]
= 0.022 (273.31 - 95.48) kW
= 3.91 kW
The power input which is expressed through energy balance in 1-2 will be used to determine COP as follows.
COP = [tex]\frac{Q_{H}}{W}[/tex]
= [tex]\frac{Q_{H}}{m (h_{2} - h_{1})}[/tex]
= [tex]\frac{3.91}{0.022(273.31 - 244.5)}[/tex]
= 6.17
For the following peak or rms values of some important sine waves, calculate the corresponding other value:
(a) 117 V rms, a household-power voltage in North America
(b) 33.9 V peak, a somewhat common peak voltage in rectifier circuits
(c) 220 V rms, a household-power voltage in parts of Europe
(d) 220 kV rms, a high-voltage transmission-line voltage in North America
Answer:
A) V_peak ≈ 165 V
B) V_rms ≈ 24 V
C) V_peak ≈ 311 V
D) V_peak ≈ 311 KV
Explanation:
Formula for RMS value is given as;
V_rms = V_peak/√2
Formula for peak value is given as;
V_peak = V_rms x √2
A) At RMS value of 117 V, peak value would be;
V_peak = 117 x √2
V_peak = 165.46 V
V_peak ≈ 165 V
B) At peak value of 33.9 V, RMS value would be;
V_rms = 33.9/√2
V_rms = 23.97 V
V_rms ≈ 24 V
C) At RMS value of 220 V, peak value is;
V_peak = 220 × √2
V_peak = 311.13 V
V_peak ≈ 311 V
D) At RMS value of 220 KV, peak value is;
V_peak = 220 × √2
V_peak = 311.13 KV
V_peak ≈ 311 KV
If the resistance reading on a DMM'S meter face is to 22.5 ohms in the range selector switch is set to R X 100 range, what is the actual measure resistance of the circuit?
Answer:
The answer is 2.25 kΩ
Explanation:
Solution
Given that:
The resistance reading on a DMM'S meter face = 22.5 ohms
The range selector switch = R * 100 range,
We now have to find the actual measure resistance of the circuit which is given below:
The actual measured resistance of the circuit is=R * 100
= 22.5 * 100
=2.25 kΩ
Hence the measured resistance of the circuit is 2.25 kΩ
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Answer:
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify.... ... has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Explanation:
At steady state, a refrigerator whose coefficient of performance is 3 removes energy by heat transfer from a freezer compartment at 0 degrees C at the rate of 6000 kJ/hr and discharges energy by heat transfer to the surroundings, which are at 20 degrees C. a) Determine the power input to the refrigerator and compare with the power input required by a reversible refrigeration cycle operating between reservoirs at these two temperatures. b) If electricity costs 8 cents per kW-hr, determine the actual and minimum theoretical operating costs, each in $/day
Answer:
(A)0.122 kW (B) Actual cost = 1.056 $/day, Theoretical cost = 0.234 $/day
Explanation:
Solution
Given that:
The coefficient of performance is =3
Heat transfer = 6000kJ/hr
Temperature = 20°C
Cost of electricity = 8 cents per kW-hr
Now
The next step is to find the power input to the refrigerator and compare with the power input considered by a reversed refrigeration cycle operating between reservoirs at the two temperatures.
Thus
(A)The coefficient of performance is given below:
COP = Heat transfer from freezer/Power input
3 =6000/P
P =6000/3
P= 2000
P = 2000 kJ/hr = 2000/(60*60) kW
= 2000 (3600)kW
= 0.55 kW
Thus
The ideal coefficient of performance = T_low/(T_high - T_low)
= (0+273)/(20-0)
= 13.65
So,
P ideal = 6000/13.65 = 439.6 kJ/hr
= 439.6/(60*60) kW
= 0.122 kW
(B)For the actual cost we have the following:
Actual cost = 0.55 kW* 0.08 $/kW-hr = $ 0.044 per hour
= 0.044*24 $/day
= 1.056 $/day
For the theoretical cost we have the following:
Theoretical cost = 0.122 kW* 0.08 $/kW-hr = $ 0.00976 per hour
= 0.00976*24 $/day
= 0.234 $/day
Air enters a compressor operating at steady state at 176.4 lbf/in.^2, 260°F with a volumetric flow rate of 424 ft^3/min and exits at 15.4 lbf/in.^2, 80°F. Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in hp
Answer:
[tex]W_s =[/tex] 283.181 hp
Explanation:
Given that:
Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 and Temperature [tex]T_1[/tex] at 260°F
Volumetric flow rate V = 424 ft^3/min
Air exits at a pressure [tex]P_2[/tex] = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] at 80°F.
Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:
[tex]Q_{cv}[/tex] = -6800 Btu/h = - 1.9924 kW
Using the steady state energy in the process;
[tex]h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
where;
[tex]g(z_2-z_1) =0[/tex] and [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]
Then; we have :
[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]
[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)
Using the relation of Ideal gas equation;
P₁V₁ = mRT₁
Pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 = ( 176.4 × 6894.76 ) N/m² = 1216235.664 N/m²
Volumetric flow rate V = 424 ft^3/min = (424 × 0.0004719) m³ /sec
= 0.2000856 m³ /sec
Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K
Gas constant R=287 J/kg K
Then;
1216235.664 N/m² × 0.2000856 m³ /sec = m × 287 J/kg K × 399.817 K
[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]
m = 2.121 kg/sec
The change in enthalpy:
[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]
[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]
= 213.1605 kW
From (1)
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]
[tex]W_s =[/tex] - 1.9924 kW + 213.1605 kW
[tex]W_s =[/tex] 211.1681 kW
[tex]W_s =[/tex] 283.181 hp
The power input is [tex]W_s =[/tex] 283.181 hp
The liquid-phase reaction A + B → C follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of A and B feed streams are 2 M before mixing. The volumetric flow rate of each stream is 5 dm3 /min and the entering temperature is 300 K. The streams are mixed immediately before entering. Two reactors are available: One is a gray 200.0 dm3 CSTR that can be heated to 77°C or cooled to 0°C, and the other is a white 800.0 dm3 PFR operated at 300 K that cannot be heated or cooled but can be painted red or black. (Note: k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol.) How long would it take to achieve 90% conversion in a 200 dm3 batch reactor with CA ° = CB ° = 1 ???? after mixing at a temperature of 70°C?
Answer:
1.887 minutes
Explanation:
We are given k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol = 20000 cal/mol
To solve this, first of all let's calculate the rate constant(k);
For this question, The formula is;
K(t) = k(300K) × exp[(E/R)((1/300) - (1/T2))]
R is gas constant = 1.987 cal/mol.K
For temperature of 70°C which is = 70 + 273K = 343K, we have;
K(343) = 0.07 × exp[(20000/1.987)((1/300) - (1/343))]
K(343) = 4.7 dm³/mol.min
The design equation is;
dX/dt = -(rA/C_Ao) = K•(C_Ao)²•(1 - X)²/(C_Ao) = (KC_Ao)(1 - X)²
Since there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thus, integrating and solving for time for a 90% conversion we obtain;
(0.9,0)∫dX/(1 - X)².dX = (KC_Ao)((t, 0)∫dt
So, we'll get;
0.9/(1 - 0.9) = 4.77 × 1 × t
t = 9/4.77
t = 1.887 minutes
A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on a 30-kg carrier C until it impacts the end of the carrier.Knowing the impact between B and C is perfectly plastic determine (a) velocity of the bullet and B after the first impact, (b) the final velocity of the carrier
(Distance between C and B is 0.5 m)
Answer:
a.) 4.46 m/s
b.) 0.41 m/s
Explanation:
a) Given that the mass M of the bullet = 30g = 30/1000 = 0.03 kg
Velocity V = 450 m/s
From conservative of linear momentum,
Sum of momentum before impact = Sum of momentum after impact
0.03 × 450 = (0.03 + 3 ) × v₂
v₂ = 13.5/3.03 = 4.4554 m/s
Therefore the velocity of the bullet and B after the first impact = 4.46 m/s approximately
(b) To calculate the velocity of the carrier, you will consider the conservation of linear momentum again.
(m₁ + m₂)×v₂ = (m₁ + m₂ + m₃)×v₃
Where:
Mass of the carrier m₃ = 30 kg
Substitute all the parameters into the formula
3.03×4.4554 = (3.03 +30) × v₃
v₃ = 13.5 / 33.03 = 0.40872 m/s
Therefore the velocity of the carrier is 0.41 m/s approximately.
which of the following tells the computer wha to do
operating system
the ROM
the motherboard
the monitor
13- Convert the following numbers to the indicated bases. List all intermediate steps.
a- (36459080)10 to octal
b- (20960032010 to hexadecimal
c- (2423233303003040)s to base
25 36459080/8= 4557385 0/8 209600320/16=13100020 + 0/16 (2423233303003040)5 (36459080)10 =( 18 (209600320)10=( 1)16 (2423233303003040)5=( )125
Answer:
Following are the conversion to this question:
Explanation:
In point (a):
[tex]\to \frac{36459080}{8} = 4557385 + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673 + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\[/tex]
[tex]\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}[/tex]
In point (b):
[tex]\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}[/tex]
In point (c):
[tex]\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\[/tex]
[tex]\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}[/tex]
Symbols of Base 25 are as follows:
[tex]0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O[/tex]
Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabatically to an exit state of 1 bar, 160°C. Kinetic and potentialenergy effects are negligible. Determine for the turbine (a) the powerdeveloped, in kW, (b) the rate of entropy production, in kW/K, and (c)the isentropic turbine efficiency
Answer:
A) W' = 178.568 KW
B) ΔS = 2.6367 KW/k
C) η = 0.3
Explanation:
We are given;
Temperature at state 1;T1 = 360 °C
Temperature at state 2;T2 = 160 °C
Pressure at state 1;P1 = 10 bar
Pressure at State 2;P2 = 1 bar
Volumetric flow rate;V' = 0.8 m³/s
A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;
Specific volume;v1 = 0.287322 m³/kg
Mass flow rate of water vapour at turbine is defined by the formula;
m' = V'/v1
So; m' = 0.8/0.287322
m' = 2.784 kg/s
Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific enthalpy;h1 = 3179.46 KJ/kg
Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific enthalpy;h2 = 3115.32 KJ/kg
Now, since stray heat transfer is neglected at turbine, we have;
-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2 - h1)
Plugging in relevant values, the work of the turbine is;
W' = -2.784(3115.32 - 3179.46)
W' = 178.568 KW
B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific entropy: s1 = 7.3357 KJ/Kg.k
Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific entropy; s2 = 8.2828 KJ/kg.k
The amount of entropy produced is defined by;
ΔS = m'(s2 - s1)
ΔS = 2.784(8.2828 - 7.3357)
ΔS = 2.6367 KW/k
C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;
h2s = 2966.14 KJ/Kg
Energy equation for turbine at ideal process is defined as;
Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Again, Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2s - h1)
W' = -2.784(2966.14 - 3179.46)
W' = 593.88 KW
the isentropic turbine efficiency is defined as;
η = W_actual/W_ideal
η = 178.568/593.88 = 0.3
Air at 80 °F is to flow through a 72 ft diameter pipe at an average velocity of 34 ft/s . What diameter pipe should be used to move water at 60 °F and average velocity of 71 ft/s if Reynolds number similarity is enforced? The kinematic viscosity of air at 80 °F is 1.69E-4 ft^2/s and the kinematic viscosity of water at 60 °F is 1.21E-5 ft^2/s. Round your answer (in ft) to TWO decimal places.
Answer:
2.47 ft
Explanation:
Given that:
The initial temperature of air = 80°F
Diameter of the pipe = 72 ft
average velocity [tex]v_{air}[/tex] of the air flow through the pipe = 34 ft/s
The objective is to determine the diameter of the pipe to be used to move water at:
At a temperature = 60°F &
An average velocity [tex]v_{water}[/tex] of 71 ft/s
Assuming Reynolds number similarity is enforced;
where :
kinematic viscosity (V_air) of air at 80 °F (V_air) = 1.69 × 10⁻⁴ ft²/s
kinematic viscosity of water at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s
The diameter of the pipe can be calculated by using the expression:
[tex]D_{water} = \dfrac{V_{water}}{V_{air}}*\dfrac{v_{air}}{v_{water}}* D_{air}[/tex]
[tex]D_{water} = \dfrac{1.21*10^{-5} \ ft^2/s}{1.69*10^{-4} \ ft^2/s}*\dfrac{34 \ ft/s}{71 \ ft/s}* 72 \ ft[/tex]
[tex]D_{water} =[/tex] 2.4686 ft
[tex]D_{water} =[/tex] 2.47 ft ( to two decimal places)
Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft
Answer:
2.47 ft
Explanation:
Given that:
The initial temperature of air = 80°F
Diameter of the pipe = 72 ft
average velocity of the air flow through the pipe = 34 ft/s
The objective is to determine the diameter of the pipe to be used to move water at:
At a temperature = 60°F &
An average velocity of 71 ft/s
Assuming Reynolds number similarity is enforced;
where :
kinematic viscosity (V_air) of air at 80 °F (V_air) = 1.69 × 10⁻⁴ ft²/s
kinematic viscosity of water at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s
The diameter of the pipe can be calculated by using the expression:
2.4686 ft
2.47 ft ( to two decimal places)
Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft