Answer:
5.833
Explanation:
Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.
where RE is refrigeration effect and P is power input
Here, the power input is given as 30 kW
We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW
Now the ![COP=\frac {175}{30}=5.833](https://tex.z-dn.net/?f=COP%3D%5Cfrac%20%7B175%7D%7B30%7D%3D5.833)
Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:
Answer:
LOL where is the question, that u need help with?
Explanation:
Answer:
8 mm
Explanation:
Given:
Diameter, D = 800 mm
Pressure, P = 2 N/mm²
Permissible tensile stress, σ = 100 N/mm²
Now,
for the pipes, we have the relation as:
where, t is the thickness
on substituting the respective values, we get
or
t = 8 mm
Hence, the minimum thickness of pipe is 8 mm
Answer:
The answer is below
Explanation:
a) The weight of the combined system is the sum of the weight of the water and the weight of the tank
![m_{water}=V_{tank}.\rho_{wtaer}\\\\m_{water}=0.2m^3*1000kg/m^3\\\\m_{water}=200 \ kg\\\\m_{total} = m_{water}+m_{tank}\\\\But\ m_{tank}=3kg,therefore:\\\\m_{total} =200kg+3kg\\\\m_{total} =203\ kg\\\\weight_{total}=m_{total}g\\\\weight_{total}=203kg*9.81m/s^2\\\\weight_{total}=1991.43\ N](https://tex.z-dn.net/?f=m_%7Bwater%7D%3DV_%7Btank%7D.%5Crho_%7Bwtaer%7D%5C%5C%5C%5Cm_%7Bwater%7D%3D0.2m%5E3%2A1000kg%2Fm%5E3%5C%5C%5C%5Cm_%7Bwater%7D%3D200%20%5C%20kg%5C%5C%5C%5Cm_%7Btotal%7D%20%3D%20m_%7Bwater%7D%2Bm_%7Btank%7D%5C%5C%5C%5CBut%5C%20m_%7Btank%7D%3D3kg%2Ctherefore%3A%5C%5C%5C%5Cm_%7Btotal%7D%20%3D200kg%2B3kg%5C%5C%5C%5Cm_%7Btotal%7D%20%3D203%5C%20kg%5C%5C%5C%5Cweight_%7Btotal%7D%3Dm_%7Btotal%7Dg%5C%5C%5C%5Cweight_%7Btotal%7D%3D203kg%2A9.81m%2Fs%5E2%5C%5C%5C%5Cweight_%7Btotal%7D%3D1991.43%5C%20N)
b) Since the weight of a system can be divided into smaller portions, hence weight is an extensive property.
c) When analyzing the acceleration of gases as they flow through a nozzle, the geometry of the nozzle which is an open system can be chosen as our system.
d) Given that:
![\rho_{water}=1000kg/m^3\\\\1kg/m^3=0.062428lb/ft^3\\\\1000kg/m^3=1000kg/m^3*\frac{0.062428lb/ft^3}{kg/m^3}=62.43lb/ft^3\\ \\\rho=SG*\rho_{water}=1.03*62.43=64.272lb/ft^3\\\\P=P_{atm}+\rho g H\\\\P=14.7\ psia+64.272\ lb/ft^3*32.2\ ft/s^2*175\ ft*\frac{1\ ft^2}{12^2\ in^2}*\frac{1\ lbf}{32.2\ lbm.ft/s^2} \\\\P=92.8\ psia](https://tex.z-dn.net/?f=%5Crho_%7Bwater%7D%3D1000kg%2Fm%5E3%5C%5C%5C%5C1kg%2Fm%5E3%3D0.062428lb%2Fft%5E3%5C%5C%5C%5C1000kg%2Fm%5E3%3D1000kg%2Fm%5E3%2A%5Cfrac%7B0.062428lb%2Fft%5E3%7D%7Bkg%2Fm%5E3%7D%3D62.43lb%2Fft%5E3%5C%5C%20%5C%5C%5Crho%3DSG%2A%5Crho_%7Bwater%7D%3D1.03%2A62.43%3D64.272lb%2Fft%5E3%5C%5C%5C%5CP%3DP_%7Batm%7D%2B%5Crho%20g%20H%5C%5C%5C%5CP%3D14.7%5C%20psia%2B64.272%5C%20lb%2Fft%5E3%2A32.2%5C%20ft%2Fs%5E2%2A175%5C%20ft%2A%5Cfrac%7B1%5C%20ft%5E2%7D%7B12%5E2%5C%20in%5E2%7D%2A%5Cfrac%7B1%5C%20lbf%7D%7B32.2%5C%20lbm.ft%2Fs%5E2%7D%20%20%5C%5C%5C%5CP%3D92.8%5C%20psia)