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gayaneshka [121]
2 years ago
9

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rat

e of heat addition to this cycle when it produces 150 hp of power, the cycle is repeated 1200 times per minute, and the state of the air at the beginning of the compression is 95 kPa and 17°C. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

Engineering
1 answer:
defon2 years ago
8 0

Answer:

A) Rate of heat addition = 228.53 Hp

B) Maximum Air temperature = 1109 °C

Explanation:

We are given;

Cut off ratio; r_c = 1.5

Compression ratio; r = 18

Power produced; W` = 150 Hp

cp = 1.005 kJ/kg·K

cv = 0.718 kJ/kg·K

R = 0.287 kJ/kg·K

k = 1.4

Temperature;T1 = 17°C = 17 + 273 K = 290K

I've attached the rest of the explanation below.

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4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and
hodyreva [135]
Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
6 0
2 years ago
An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th
makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)    

v=total volume i.e. volume of container"80L"   80L=0.08 cubic meter

vf=vol of liquid phase  vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg          hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

<u>Explanation</u>:        

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                                                      

                                                                                   

                                                                                                                                                   

                                                                                                                                                             

5 0
2 years ago
A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35
lukranit [14]

Answer:

P_O = 0.989 watt = 19.9 dBm

Explanation:

Given data:

P_1 power = 20 dBm  = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor = 10 log \frac{P_1}{P_f}

solving for  final power

20 = 10 log\frac{P_1}{P_f}

2 = log \frac{P_1}{P_f}

100 = \frac{0.1}{P_f}

P_f = 0.001 watt = 0 dBm

Directivity D =  10 \frac{P_f}{P_b}

35 = 10 \frac{0.001}{P_b}

P_b = 3.162 \times 10^{-7} wattt

output Power  = P_1 -P_f - P_b

                       = 0.1 - 0.001 - 3.162 \times 10^{-7}

P_O = 0.989 watt = 19.9 dBm

6 0
2 years ago
Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III,
Rom4ik [11]

Answer:

See explanations for step by step procedures to get answer.

Explanation:

Given that;

Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III, and M0M0M_0. Enter positive value if the deflection is upward and negative value if the deflection is downward.

4 0
2 years ago
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