Answer:
Any circuit that goes directly from the fixture to the breaker panel. For dedicated outlets, isolated fixtures and high-load items, you would want them to be on their own breakers. Those would get a home run, a line that goes all the way back without stopping.
Consider a turbofan engine installed on an aircraft flying at an altitude of 5500m. The CPR is 12 and the inlet diameter of this engine is 2.0m The bypass ratio of this engine 8. The bypass ratio (BPR) of a turbofan engine is the ratio between the mass flow rate of the bypass stream to the mass flow rate entering the core. The inlet temperature is 253K and the outlet temperature is 233K. Determine the thrust of this engine in order to fly at the velocity of 250 m/s. Assume cold air approach. The engine is ideal.
Answer:
The thrust of the engine calculated using the cold air is 34227.35 N
Explanation:
For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as
[tex]\dot{m}=\rho AV_a[/tex]
Here
ρ is the density which is given as [tex]\dfrac{P}{RT}[/tex]P is the pressure of air at 5500 m from the ISA whose value is 50506.80 PaR is the gas constant whose value is 286.9 J/kg.KT is the temperature of the inlet which is given as 253 KA is the cross-sectional area of the inlet which is given by using the diameter of 2.0 mV_a is the velocity of the aircraft which is given as 250 m/sSo the equation becomes
[tex]\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}[/tex]
Now in order to find the flow from the fan, the Bypass ratio is used.
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}[/tex]
Here BPR is given as 8 so the equation becomes
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}[/tex]
Now the exit velocity is calculated using the total energy balance which is given as below:
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2[/tex]
Here
h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as [tex]c_pT_4[/tex] and [tex]c_pT_5[/tex] respectively.The value of T_4 is the inlet temperature which is 253 KThe value of T_5 is the outlet temperature which is 233KThe value of c_p is constant which is 1005 J/kgKV_a is the inlet velocity which is 250 m/sV_e is the outlet velocity that is to be calculated.So the equation becomes
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2[/tex]
Rearranging the equation gives
[tex]\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s[/tex]
Now using the cold air approach, the thrust is given as follows
[tex]T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N[/tex]
So the thrust of the engine calculated using the cold air is 34227.35 N
Could anyone answer this, please? It's about solid mechanics. I will give you 100 points!!! It's due at midnight.
Answer:
sorry i don't know
Explanation:
A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected
to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an
elongation of 0.50 mm?
Answer:
"7.654 mm" is the correct solution.
Explanation:
According to the question,
[tex]E=110\times 10^3 \ N/mm^2[/tex][tex]\sigma_y = 240 \ mPa[/tex][tex]P = 6660 \ N[/tex][tex]L = 380 \ mm[/tex][tex]\delta = 0.5 \ mm[/tex]Now,
As we know,
The Elongation,
⇒ [tex]E=\frac{\sigma}{e}[/tex]
[tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]
or,
⇒ [tex]\delta=\frac{PL}{AE}[/tex]
By substituting the values, we get
[tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]
then,
⇒ [tex]D^2=58.587[/tex]
[tex]D=\sqrt{58.587}[/tex]
[tex]=7.654 \ mm[/tex]
A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that rises 0.25 m above the channel bottom. Will a hydraulic jump occur?
Answer:
The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )
Explanation:
width of channel = 3.0 m
Flow rate = 5 m^3/s
Normal depth = 0.50 m
Flow encounters a dam rise of 0.25 m
To know if the hydraulic jump will occur
we will Determine the new normal depth
Y2 = 3.77m
Yc ( critical depth )= 0.66m
Attached below is the detailed solution
Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.
[10 marks] What is the cost to install a 5kW PV system for a residence?
[10 marks] If the solar irradiance in Ibra is on average 800W/m2 and the installed panels have efficiency of 18%. How many panels are required if the panel’s area is 2m2?
[15 marks] Assume Ibra has an average of 10 day-hours, dusty environment which causes the efficiency of the solar system to drop by 10% on average, and 30 cloudy days/year which cause the efficiency of the solar panels drops by 50%. If electrical power cost per kWh is 0.05 OMR determine the break-even time for the 5kW PV system.
[10 marks] If the system to be off-grid, what would be the backup time if three 12-V batteries were selected each with a capacity of 200Ah. Assume that you can discharge the batteries up to 80% of their capacities.
Answer:
so hard it is
Explanation:
I don't know about this
please mark as brainleast
byýyy