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allsm [11]
3 years ago
13

The displacement volume of an internal combustion engine is 2.2 liters. The processes within each cylinder of the engine are mod

eled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by p1 = 1.6 bar, T1 = 325 K and V1 = 2.36 liters.
1) If the cycle is executed 2100 times per min, determine:
a) the compression ratio.
b) the net work per cycle, in kJ.
c) the maximum temperature, in K.
d) the power developed by the engine, in kW.
e) and the thermal efficiency.

Engineering
1 answer:
soldier1979 [14.2K]3 years ago
5 0

Answer:

A) 14.75

B) 3.36Kj

C) 2384.2k

D) 117.6kW

E) 57.69%

Explanation:

Attached is the full solutions.

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Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in
Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time =   ∑  (\frac{Clock cyles}{Clock rate})

Clock cycles can be determined using following formula

Clock cycles = (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

Clock cycles = ( 50 x 10^{6} x 1) + (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

    = \frac{512(10^{6}) }{2(10^{9}) }

 Initial execution Time =  256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

(\frac{Clockcycles}{2} )=   (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

CPI_{FP improved} x No. FP instructions  =  (\frac{Clockcycles}{2} ) -[ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)]

CPI_{FP improved} x 50 x 10^{6}  = ( \frac{512(10)^{6} }{2} ) - [ (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)]

CPI_{FP improved} x 50 x 10^{6}  =  - 206 x 10^{6}

CPI_{FP improved}  = - 206 x 10^{6} / 50 x 10^{6}

CPI_{FP improved} = - 4.12 < 0

3 0
3 years ago
Calculate the reactions at 4 ends (supports) of this bookshelf. Assume that the weight of each book is approximately 1 lb. The w
hoa [83]

Answer:

R = 4.75 lb  (↑)

Explanation:

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

We can see the bookshelf in the pic.

8 0
3 years ago
A storm sewer is carrying snow melt containing 1.2 g/L of sodium chloride into a small stream. The stream has a naturally occurr
galina1969 [7]

Answer:

Given Data:

concentration of sewer Csewer = 1.2 g/L

converting into mg/L = Csewer = 1.2 g/L x 1000 mg/g = 1200 mg/L

flow rate of sewer Qsewer = 2000 L/min

concentration of sewer Cstream = 20 mg/L

flow rate of sewer Qstream = 2m3/s

converting Q into L/min = 2m3/s x 1000 x 60 = 120000 L/min

mass diagram is

6 0
3 years ago
At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut
professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ \frac{1}{\mu(1-\frac{\lambda}{\mu})}

thus,

on substituting the respective values, we get

Average time spent by the vehicle = \frac{1}{60(1-\frac{30}{60})}

or

Average time spent by the vehicle = \frac{1}{60(1-0.5)}

or

Average time spent by the vehicle = \frac{1}{60(0.5)}

or

Average time spent by the vehicle = \frac{1}{30} hr/veh

or

Average time spent by the vehicle = \frac{1}{30}\times60 min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0
3 years ago
Need help with both giving out brainlest for the people to help me
sammy [17]
The first one is d or the 4th answer choice and the second one is false. Hope this helps!
4 0
3 years ago
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