Answer:
Part A:
Hourly production rate of acceptable parts=9.525 parts/hr
Part B:
Scrap rate=0.042713=4.2713%
Part C:
Proportion uptime=0.9774215=97.74215%
Explanation:
Part A:
Acceptable (non-defective) parts=381
Hours Per week=40
Hourly production rate of acceptable parts=Acceptable (non-defective) parts/Hours Per week
Hourly production rate of acceptable parts=![\frac{381}{40}](https://tex.z-dn.net/?f=%5Cfrac%7B381%7D%7B40%7D)
Hourly production rate of acceptable parts=9.525 parts/hr
Part B:
Total Parts Produced=Non-defective parts-Defective parts
Total Parts Produced=381+17
Total Parts Produced=398
Scrap rate=17/398
Scrap rate=0.042713=4.2713%
Part C:
Total Time for operation=5.23+0.58+(
)
Total Time for operation=5.894 min
Total up time=398(5.894)=2345.812 min=39.09686 hr
Proportion uptime=![\frac{39.09686}{40}](https://tex.z-dn.net/?f=%5Cfrac%7B39.09686%7D%7B40%7D)
Proportion uptime=0.9774215=97.74215%
To solve the exercise, it is necessary to apply the concepts related to the conservation of the energy flow given by Bernoulli and the equation of head loss in the pipe for laminar flow. Through them and the calculation of the flow we can identify the flow rate of oil.
Bernoulli is defined by,
![\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f](https://tex.z-dn.net/?f=%5Cfrac%7Bp_1%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D%2Bz_1%3D%5Cfrac%7Bp_2%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_2%5E2%7D%7B2g%7D%2Bz_2%2Bh_f)
Where,
P = Pressure
= Density
z= Datum height
V = Velocity
g = Gravitaty constant
= frictional head loss in the pipe
There is no change in the final height datum nor initial speed. So,
![\frac{p_{atm}}{850*9.81}+\frac{0^2}{2(9.81)}+1.45=\frac{p_{atm}}{850*9.81}+\frac{V_2^2}{2(9.81)}+0+h_f](https://tex.z-dn.net/?f=%5Cfrac%7Bp_%7Batm%7D%7D%7B850%2A9.81%7D%2B%5Cfrac%7B0%5E2%7D%7B2%289.81%29%7D%2B1.45%3D%5Cfrac%7Bp_%7Batm%7D%7D%7B850%2A9.81%7D%2B%5Cfrac%7BV_2%5E2%7D%7B2%289.81%29%7D%2B0%2Bh_f)
Re-arrange to find the head loss,
![h_f = 1.45-\frac{V^2}{19.62}](https://tex.z-dn.net/?f=h_f%20%3D%201.45-%5Cfrac%7BV%5E2%7D%7B19.62%7D)
In the case of the head loss in the pipe for laminar flow we have that
![h_f = \frac{32\mu VL}{\rho gd^2}](https://tex.z-dn.net/?f=h_f%20%3D%20%5Cfrac%7B32%5Cmu%20VL%7D%7B%5Crho%20gd%5E2%7D)
Equation we have,
![1.45-\frac{V^2}{19.62}=\frac{32\mu VL}{\rho gd^2}](https://tex.z-dn.net/?f=1.45-%5Cfrac%7BV%5E2%7D%7B19.62%7D%3D%5Cfrac%7B32%5Cmu%20VL%7D%7B%5Crho%20gd%5E2%7D)
![\frac{V^2}{19.62}+\frac{32\mu VL}{\rho gd^2}-1.45=0](https://tex.z-dn.net/?f=%5Cfrac%7BV%5E2%7D%7B19.62%7D%2B%5Cfrac%7B32%5Cmu%20VL%7D%7B%5Crho%20gd%5E2%7D-1.45%3D0)
At this point our values are given as,
![L=55*10^{-2}m](https://tex.z-dn.net/?f=L%3D55%2A10%5E%7B-2%7Dm)
![g = 9.81m/s^2](https://tex.z-dn.net/?f=g%20%3D%209.81m%2Fs%5E2)
![\rho = 850kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%20850kg%2Fm%5E3)
![\mu = 0.11 kg/m.s](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.11%20kg%2Fm.s)
![d = 4*10^{-3}m](https://tex.z-dn.net/?f=d%20%3D%204%2A10%5E%7B-3%7Dm)
Therefore,
![\frac{V^2}{19.62}+\frac{32(0.11)V(55*10^{-2})}{850(9.81)(4*10^{-3})^2}-1.45=0](https://tex.z-dn.net/?f=%5Cfrac%7BV%5E2%7D%7B19.62%7D%2B%5Cfrac%7B32%280.11%29V%2855%2A10%5E%7B-2%7D%29%7D%7B850%289.81%29%284%2A10%5E%7B-3%7D%29%5E2%7D-1.45%3D0)
![0.051V^2+14.51V-1.45=0](https://tex.z-dn.net/?f=0.051V%5E2%2B14.51V-1.45%3D0)
![V=0.1m/s](https://tex.z-dn.net/?f=V%3D0.1m%2Fs)
Finally the discharge is given as
![Q = AV](https://tex.z-dn.net/?f=Q%20%3D%20AV)
Where,
A= Area
V = Velocity
![Q = \frac{\pi}{4}d^2*V](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7Dd%5E2%2AV)
![Q=\frac{\pi}{4}*0.004^2*0.1](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%2A0.004%5E2%2A0.1)
![Q= 1.256*10^{-6}m^3s](https://tex.z-dn.net/?f=Q%3D%201.256%2A10%5E%7B-6%7Dm%5E3s)
That is equal in cm^3 per hour as,
![Q= 1.256*10^{-6}m^3s *(\frac{10^6cm^3}{1m^3})(\frac{3600s}{1h})](https://tex.z-dn.net/?f=Q%3D%201.256%2A10%5E%7B-6%7Dm%5E3s%20%2A%28%5Cfrac%7B10%5E6cm%5E3%7D%7B1m%5E3%7D%29%28%5Cfrac%7B3600s%7D%7B1h%7D%29)
![Q = 4521cm^3/h](https://tex.z-dn.net/?f=Q%20%3D%204521cm%5E3%2Fh)
Therefore the flow rate of oil is 4521cm^3/h
Answer:
Exceeding the rated load for the circuit wiring causes the circuit breaker to trip, shutting off the power to the entire circuit. If there were no breaker in the circuit, an overload would cause the circuit wiring to overheat, which could melt the wire insulation and lead to a fire
Explanation:
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Answer:
0.8 kilograms of fuel are consumed each second.
Explanation:
As turbines are steady-state devices, the thermal efficiency of a turbine is equal to the percentage of the ratio of the output power to fluid power, that is:
![\eta_{th} = \frac{\dot W}{\dot E} \times 100\,\%](https://tex.z-dn.net/?f=%5Ceta_%7Bth%7D%20%3D%20%5Cfrac%7B%5Cdot%20W%7D%7B%5Cdot%20E%7D%20%5Ctimes%20100%5C%2C%5C%25)
The fluid power is:
![\dot E = \frac{\dot W}{\eta} \times 100\,\%](https://tex.z-dn.net/?f=%5Cdot%20E%20%3D%20%5Cfrac%7B%5Cdot%20W%7D%7B%5Ceta%7D%20%5Ctimes%20100%5C%2C%5C%25)
![\dot E = \frac{8\,MW}{20\,\%}\times 100\,\%](https://tex.z-dn.net/?f=%5Cdot%20E%20%3D%20%5Cfrac%7B8%5C%2CMW%7D%7B20%5C%2C%5C%25%7D%5Ctimes%20100%5C%2C%5C%25)
![\dot E = 40\,MW](https://tex.z-dn.net/?f=%5Cdot%20E%20%3D%2040%5C%2CMW)
Which means that gas turbine consumes 40 megajoules of fluid energy each second, which is heated and pressurized with help of the fuel, whose amount of consumption per second is:
![50\,\frac{MJ}{kg} = \frac{40\,MJ}{m_{fuel}}](https://tex.z-dn.net/?f=50%5C%2C%5Cfrac%7BMJ%7D%7Bkg%7D%20%3D%20%5Cfrac%7B40%5C%2CMJ%7D%7Bm_%7Bfuel%7D%7D)
![m_{fuel} = 0.8\,kg](https://tex.z-dn.net/?f=m_%7Bfuel%7D%20%3D%200.8%5C%2Ckg)
0.8 kilograms of fuel are consumed each second.