Which of the following is a advantage of a chain and sprocket over a pulley and belt system?

Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour

Debora [2.8K]

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by $\frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}$

So availability of system $=\frac{900}{900+100}=\frac{900}{1000}=0.9$

So the availability of system will be 0.9

7 0

3 years ago

You are a designer of a new processor. You have to choose between two possible implementations (called M1 and M2) of the same ar

Kaylis [27]

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

CPI for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2. number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

CPI for M2 = 2.65

B ) Calculate the native MIPS ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

each clock cycle = 0.3846 * 10^-9. frequency = 2.6 Ghz

first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6 = 1083

For M2

Assumptions : number of instructions executed = 10

each clock cycle = 0.3846 * 10^-9. frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

8 0

3 years ago

A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl

bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1. Steady operating conditions exist.

2. Kinetic and potential energy changes are negligible.

Properties: The specific heat of geothermal water ( $c_{geo}$ [) is taken to be 4.18 kJ/kg.ºC.

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

P $P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }$

= $=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788$

b. Pump

$h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\$

$W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\$