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Zigmanuir [339]
4 years ago
9

Hot and cold streams of a fluid are mixed in a rigid mixing chamber. The hot fluid flows into the chamber at a mass flow rate of

5 kg/s with an energy in the amount of 150 kJ/kg. The cold fluid flows into the chamber with a mass flow rate of 15 kg/s and carries energy in the amount of 50 kJ/kg. There is heat transfer to the surroundings from the mixing chamber in the amount of 5.5 kW. The mixing cham- ber operates in a steady-flow manner and does not gain or lose energy or mass with time. Determine the energy carried from the mixing chamber by the fluid mixture per unit mass of fluid, in kJ/kg.
Engineering
1 answer:
FromTheMoon [43]4 years ago
7 0

Answer:74.75 kJ/kg

Explanation:

Given

mass of hot fluid(m_h)=5 kg/s

Energy associated with it=150 kJ/kg

mass of cold fluid(m_l)=15 kg/s

Energy associated with it=50 kJ/kg

Energy lost=5.5 kW

Let the Enthalpy of outlet fluid be h

mass at outlet

m=m_l+m_h=20 kg/s

Using Energy Conservation

m_h\times E_h+m_l\times E_l+Q=m\times h

5\times 150+15\times 50-5.5=20\times h

h=74.75 kJ/kg

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Air is drawn from the atmosphere into a turbomachine. At the exit, conditions are 500 kPa (gage) and 130oC. The exit speed is 10
finlep [7]

Answer:

P=- 88.41 KW

Negative sign indicates that power is given to the system.

Explanation:

Given that

P₂=500 KPa

T₂=130°C

V₂=100 m/s

mass flow rate ,m= 0.8 kg/s

Lets take inlet condition for air

T₁=25°C

P₁=100 KPa

V₁=0 m/s

We know that

Heat capacity for air Cp=1.005 KJ/kg.k

We know that for air change in enthalpy only depends only on temperature

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}+W

1.005\times 298+\dfrac{0^2}{2000}=1.005\times 403+\dfrac{100^2}{2000}+W

W=1.005\times 298-1.005\times 403-\dfrac{100^2}{2000}

W=-110.52 KJ/kg

Shaft power P = m .W

P = -110.52 x 0.8

P=- 88.41 KW

Negative sign indicates that power is given to the system.

 

3 0
4 years ago
Rewrite the following loops, using the enhanced for an array of floating-point numbers. Write your code within comments
dmitriy555 [2]

Answer:

Explanation:poop

4 0
3 years ago
Consider two different versions of algorithm for finding gcd of two numbers (as given below), Estimate how many times faster it
juin [17]

Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

     gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

    gcd(31,415,14,142) =gcd(14,142,3,131)

                                  =gcd=(3,131, 1,618)

                                   =gcd(1,618, 1,513)

                                   =gcd(1,513, 105)

                                   =gcd(105, 43)

                                    =gcd(43, 19)

                                     =gcd(19, 5)

                                      =gcd(5, 4)

                                      =gcd(4, 1)

                                      =gcd(1, 0)

                                      =1

STEP 2

b)  The number of comparison of given input with the algorithm based on  checking consecutive integers and Euclid's algorithm is

     The number of division using Euclid's algorithm =10 from part (a)

      The consecutive integer checking algorithm:

      The number of iterations =14,142 and 1 or 2 division of iteration.

        14,142 ∠= number of division∠ = 2*14,142

         Euclid's algorithm is faster by at least 14,142/10 =1400 times

          At most 2*14,142/10 =2800 times.

5 0
3 years ago
In fuel oil piping systems, all exterior above-grade fill piping shall be __________ when tanks are abandoned or removed.
castortr0y [4]

Answer:

Removed

Hope this Helps!

6 0
3 years ago
ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in
Mama L [17]

Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

d_{p} =mN_{p}

Where

m = module = 5

Np = number of teeth of pinion = 26

d_{p} =5*26=130mm = 0.13 m

The pitch line velocity is equal to:

V_{t} =\frac{d_{p}*2*\pi  *w_{p} }{120}

Where

wp = speed of the pinion = 1800 rpm

V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s

The factor B is equal to:

B=\frac{(12-Q_{v})^{2/3}  }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}

The bending stress is equal to:

\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}}  \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa

4 0
3 years ago
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