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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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A teacher tells her​ students, "When you do your math homework​ assignments, you must use white lined paper.​ Please, no​ tear-o
Sindrei [870]

Answer:

The white lined paper

Explanation:

The teacher is most likely putting the while line paper in jeopardy because of the detail process involved in taking care of the paper prior to the submission of the home work.

The fact that a mistake must not be visible due to the instruction of every erasures being thorough and clean.  this can cause jeopardy to the paper.

8 0
3 years ago
Design a plate and frame heat exchanger for the following problem:
qwelly [4]

Answer:

See explaination and attachment.

Explanation:

Iteration method is a repetitive method applied until the desired result is achieved.

Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below

x = pi(x)

Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).

Similarly for second, thrid and so on. approximation

x2 = pi(x1)

x3 = pi(x2)

x4 = pi(x3)

xn = pi(xn-1).

please go to attachment for the step by step solution.

8 0
3 years ago
Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k =
Artyom0805 [142]

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of  weight of the piston and the force the spring exerts on the piston

Mathematically we can write

Force_{pressure}=Force_{spring}+Weight_{piston}

we know that

Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons

Weight_{piston}=mass\times g=100\times 9.81=981Newtons

Now the force exerted by an spring compressed by a distance 'x' is given by Force_{spring}=k\cdot x=5\times 10^{3}\times x

Using the above quatities in the above relation we get

5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters

5 0
3 years ago
A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t
bekas [8.4K]

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

a=450\ mm/s^2

Explanation:

Given that

s = 15t^3 - 3t\ mm

a)

When t= 2 s

s = 15t^3 - 3t\ mm

s = 15\times 2^3 - 3\times 2\ mm

s= 114 mm

At t= 4 s

s = 15t^3 - 3t\ mm

s = 15\times 4^3- 3\times 4\ mm

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

b)

We know that velocity V

V=\dfrac{ds}{dt}

\dfrac{ds}{dt}=45t^2-3

At t=  5 s

V=45t^2-3

V=45\times 5^2-3

V=1122 mm/s

We know that acceleration a

a=\dfrac{d^2s}{dt^2}

\dfrac{d^2s}{dt^2}=90t

a= 90 t

a = 90 x 5

a=450\ mm/s^2

4 0
3 years ago
If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

3 0
4 years ago
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