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PtichkaEL [24]
3 years ago
11

Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The co

mpressor pressure ratio is 7, and the turbine inlet temperature is 1200 K. For constant specific heats with k = 1.4 and Cp = 1.005 kJ/kg, calculate the percent thermal efficiency (enter a number only)
Engineering
1 answer:
ICE Princess25 [194]3 years ago
7 0

Answer:

The thermal efficiency of cycle is 42.6%.

Explanation:

Given that

T_1=300 K

P_1=100KPa

mass flow rate = 6 kg/s

Compression ratio = 7

Turbine inlet temperature = 1200 K

γ=1.4

We know that thermal efficiency of Brayton cycle given as

\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}

Now by putting the values

\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}

\eta=1-\dfrac{1}{7^{\frac{1.4-1}{1.4}}}

η=0.426

So the thermal efficiency of cycle is 42.6%.

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Answer:

Explanation:

Step by step solved solution is given in the attached document.

8 0
3 years ago
Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha
Romashka-Z-Leto [24]

Answer:

Technician A is wrong

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Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

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3 years ago
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4 0
2 years ago
A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum
adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress \sigma = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder  there exist both the circumferential stress and the longitudinal stress.

Circumferential stress \sigma = \dfrac{pd}{2t}

Making thickness t the subject; we have

t = \dfrac{pd}{2* \sigma}

t = \dfrac{560000*3}{2*150000000}

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

\sigma = \dfrac{pd}{4t}

t= \dfrac{pd}{4*\sigma }

t = \dfrac{560000*3}{4*150000000}

t = 0.0028  mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value  with the maximum thickness = 5.6 mm

8 0
3 years ago
soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new
timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0
3 years ago
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