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Dvinal [7]
3 years ago
12

Air enters the combustor of a jet engine at Pressure=10 atm, T=1000R and M=0.2. Fuel is injected and burned with fuel-air ratio

(by mass) of 0.06. The heat released during the combustion is 4.50000000 ft-lb per slug of fuel. Assuming one-dimensional frictionless flow with r=1.4 for the fuel-air mixture, calculate M2, p2 and T2 at the exit of combustor?
Engineering
1 answer:
kkurt [141]3 years ago
5 0
Air enters the combustor of a jet engine at p1 = 10 atm, T1 = 1000 °R and M1 = 0.2. The
heat released during combustion is 4.5x108 ft-lb per slug of fuel. Assuming onedimensional
frictionless flow with γ = 1.4 for the fuel air mixture, calculate M2, p2, T2 at
the exit of the combustor. Hope this helps! Can I get brainly please? Plus safety first if you see any links don’t klick them please don’t.
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What is the difference between a refrigeration cycle and a heat pump cycle?
sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0
3 years ago
How will you maintain the orderliness of your storage area 2pts?​
fiasKO [112]

Explanation:

Label and group products. One would think that a general cleanup would be the first step, but no, it's not. ...Clean up the area. ...Put up demarcation lines. ...Stack properly. ...Keep the aisles, paths and ramps clear. ...Have all the safety signs in place.

4 0
2 years ago
A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t
maksim [4K]

Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

A) The flow velocity intensity for the input radial coordinate "r" is given by;

U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

Thus,

U(r) = (V_c)[1 - (2r/D)²]

From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m

But we are to find the velocity at the centre of the tube, thus;

We will use the radius across the horizontal distance which will be;

0.05 - 0.012 = 0.038m

Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

Thus;

U(r) = (V_c)[1 - (2r/D)²]

0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]

0.8 = (V_c)[1 - (0.76)²]

V_c = 0.8/0.4224 = 1.894 m/s

B) flow rate is given by;

ΔV = Average Velocity x Area

Now, average velocity = V_c/2

Thus, average velocity = 1.894/2 = 0.947 m/s

Area(A) = πr² = π x 0.05² = 0.007854 m²

So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s

4 0
3 years ago
What is the minimum number of crewmembers needed when using laser plane equipment?
Goshia [24]

Answer:

two

Explanation:

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8 0
3 years ago
A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.
Nataliya [291]

Answer:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

Explanation:

We can assume that the general formula for the drag force is given by:

D= C_D \frac{\rho}{2}V^2 A

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D1} \frac{\rho}{2}V^2 (wh) V

We can assume the same drag coeeficient C_{D1}=C_{D2}=C_{D} and we got:

\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D} \frac{\rho}{2}V^2 (wh) V

\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

\frac{l}{h}=\frac{5}{1.7}2.941

And we can estimate the calue of C_D = 1.2 from a figure.

And we can calculate the power difference like this:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

8 0
3 years ago
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