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:
heater wattage = 4.32 kW
Explanation:
given data
volume flow rate of air = 0.1 m³/s
temperature t1 = 15°C
temperature t2 = 50°C
relative humidity RH = 80%
pressure = 101.325 kPa
to find out
heater wattage
solution
we will get here h value from psychometric chart with the help of temperature
that is
h1 = 37 kJ/kg
and
h2 = 73 kJ/kg
and v1 = 0.829 m³/kg
so here heater wattage formula is
heater wattage = m ( h2 -h1) ......1
here m is
m = V/v1 = 0.1 / 0.819 = 0.12 kg/s
so from equation 1
heater wattage = 0.12 ( 75 -39 )
heater wattage = 4.32 kW
In the United States, fire codes are developed primarily by two model code organizations, the International Code Council (ICC) and the National Fire Protection Association (NFPA).
Answer:
a)
The crack and connecting rod is used in the design of car.This mechanism is known as slider -crank mechanism.
Components:
1.Inlet tube
2. Wheel
3. Exhaust
4. Engine
5.Air tank
6.Pressure gauge
7.Stand
8. Gate valve
b)
The efficiency of air engine is less as compare to efficiency of electric engine and this is not ecofriendly because it produce green house gases.These gases affect the environment.
c)
it can run around 722 km when it is full charge.