Answer:
Amount reduction is 29.06 gal/year
Cost of reduction is $90.09 / year
Explanation:
Assume The effect of reduction of the frontal area on the drag coefficient is
negligible
Given that:
The drag coefficient (
) = 0.4 for a passenger car.
A = frontal area of car = 18 ft²
Velocity (V) = 55 mile per hour = (55 × 1.4667) ft/s = 80.6685 ft/s, density of air (ρ) = 0.075 lbm/ft³
The frontal drag force can be calculated by:
![F_D=C_DA\frac{\rho V^2}{2} =0.3 *18*\frac{0.075*80.6685^2}{2} =1317.75\\F_D=1317.75lbm.ft/s^2=(1317,75*\frac{1}{32.2})lbf=40.9lbf](https://tex.z-dn.net/?f=F_D%3DC_DA%5Cfrac%7B%5Crho%20V%5E2%7D%7B2%7D%20%3D0.3%20%2A18%2A%5Cfrac%7B0.075%2A80.6685%5E2%7D%7B2%7D%20%3D1317.75%5C%5CF_D%3D1317.75lbm.ft%2Fs%5E2%3D%281317%2C75%2A%5Cfrac%7B1%7D%7B32.2%7D%29lbf%3D40.9lbf)
The amount of work done to overcome this drag force is calculated by:
d = 12000 miles per year
![W_{DRAG}=F_D*d=(40.9lbf)*(12000miles/year)*\frac{5280ft}{1mile}*\frac{1lbf}{778.169 lbf ft} =3.330* 10^6 Btu/year](https://tex.z-dn.net/?f=W_%7BDRAG%7D%3DF_D%2Ad%3D%2840.9lbf%29%2A%2812000miles%2Fyear%29%2A%5Cfrac%7B5280ft%7D%7B1mile%7D%2A%5Cfrac%7B1lbf%7D%7B778.169%20lbf%20ft%7D%20%20%3D3.330%2A%2010%5E6%20Btu%2Fyear)
and the required energy input is:
![E_{in}=\frac{W_{DRAG}}{\eta_{car}} =\frac{3.330*10^6Btu/year}{0.30} =11.1*10^7Btu/year](https://tex.z-dn.net/?f=E_%7Bin%7D%3D%5Cfrac%7BW_%7BDRAG%7D%7D%7B%5Ceta_%7Bcar%7D%7D%20%3D%5Cfrac%7B3.330%2A10%5E6Btu%2Fyear%7D%7B0.30%7D%20%3D11.1%2A10%5E7Btu%2Fyear)
Heating value (HV) = 20000 Btu/lbm
![m_{fuel}=E_{in}/HV=1.11*10^7/20000\\](https://tex.z-dn.net/?f=m_%7Bfuel%7D%3DE_%7Bin%7D%2FHV%3D1.11%2A10%5E7%2F20000%5C%5C)
Amount of fuel = ![\frac{m_{fuel}}{\rho{fuel}} = \frac{1.11*10^7/20000}{50}=11.1ft^3/year *\frac{7.4804gal}{1ft^3}=83.03 gal/year](https://tex.z-dn.net/?f=%5Cfrac%7Bm_%7Bfuel%7D%7D%7B%5Crho%7Bfuel%7D%7D%20%3D%20%5Cfrac%7B1.11%2A10%5E7%2F20000%7D%7B50%7D%3D11.1ft%5E3%2Fyear%20%2A%5Cfrac%7B7.4804gal%7D%7B1ft%5E3%7D%3D83.03%20gal%2Fyear)
Cost of fuel = Amount of fuel × Price of fuel = 11.1 ft³/year × ($3.1/gal) × (7.4804 gal / 1 ft³) = $257.4 per year.
The percent reduction in the fuel consumption due to reducing frontal area (reduction ratio) is given by:
Reduction ratio = ![\frac{A-A{new}}{A}=\frac{20-13}{20} =0.35](https://tex.z-dn.net/?f=%5Cfrac%7BA-A%7Bnew%7D%7D%7BA%7D%3D%5Cfrac%7B20-13%7D%7B20%7D%20%20%3D0.35)
Amount reduction = Reduction ratio × Amount of fuel = 0.35 × 83.03 gal/year =29.06 gal/year
Cost of reduction = Reduction ratio × cost of fuel = 0.35 × $257.4 = $90.09 / year
Answer:
Explanation:
it is given that diameter = 8.6 cm
![radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m](https://tex.z-dn.net/?f=radius%20%3D%5Cfrac%7B8.6%7D%7B2%7D%3D4.3%5C%20cm%3D4.3%5Ctimes%2010%5E%7B-2%7D%5C%20m)
current =2.7 ampere
number of turns = 15
![area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}](https://tex.z-dn.net/?f=area%20%3D%5Cpi%20r%5E2%3D3.14%5Ctimes%20%5Cleft%20%28%204.3%5Ctimes%2010%5E%7B-2%7D%20%5Cright%20%29%5E%7B2%7D%3D0.005806%20m%5E%7B2%7D)
magnetic field =0.56 T
maximum torque= BINASINΘ for maximum torque sinΘ=1
so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm
Where are the master plans stored and order given?
Answer:
![COP_{HP} = \frac{1}{\eta_{th}}](https://tex.z-dn.net/?f=COP_%7BHP%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Ceta_%7Bth%7D%7D)
Explanation:
The coefficient of performance of the reversible heat pump is:
![COP_{HP} = \frac{T_{H}}{T_{H}-T_{C}}](https://tex.z-dn.net/?f=COP_%7BHP%7D%20%3D%20%5Cfrac%7BT_%7BH%7D%7D%7BT_%7BH%7D-T_%7BC%7D%7D)
![COP_{HP} = \frac{1}{1-\frac{T_{C}}{T_{H}} }](https://tex.z-dn.net/?f=COP_%7BHP%7D%20%3D%20%5Cfrac%7B1%7D%7B1-%5Cfrac%7BT_%7BC%7D%7D%7BT_%7BH%7D%7D%20%7D)
The thermal efficiency of the reversible power cycle is:
![\eta_{th} = 1 - \frac{T_{C}}{T_{H}}](https://tex.z-dn.net/?f=%5Ceta_%7Bth%7D%20%3D%201%20-%20%5Cfrac%7BT_%7BC%7D%7D%7BT_%7BH%7D%7D)
After a quick comparison between both expressions, the following relation is found:
![COP_{HP} = \frac{1}{\eta_{th}}](https://tex.z-dn.net/?f=COP_%7BHP%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Ceta_%7Bth%7D%7D)