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skelet666 [1.2K]
2 years ago
8

R-134a vapor enters into a turbine at 250 psia and 175°F. The temperature of R-134a is reduced to 20°F in this turbine while its

specific entropy remains constant. Determine the change in the enthalpy of R-134a as it passes through the turbine.
Engineering
1 answer:
attashe74 [19]2 years ago
5 0

Answer:

Δ enthalpy = -23 Btu/Ibm

Explanation:

Given data:

Pressure ( P1 ) = 250 psi

Initial Temperature ( T1 ) = 175°F

Final temperature ( T2 ) = 20°F

<u>Calculate the change in the enthalpy of R-134a </u>

From R-134 table

h1 = 129.85 Btu/Ibm

s1 = 0.23281 Btu/Ibm.R

note : entropy is constant  

hence ; s1 = s2

by interpolation  ; h2 = 106.95

Δ enthalpy = h2 - h1

                  =  ( 106.95 - 129.85 ) = -23 Btu/Ibm

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A cantilever beam AB of length L has fixed support at A and spring support at B.
atroni [7]

The force in the spring will be F =\dfrac{KPl^3}{3EI}.

The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

<h3>What is a cantilever beam?</h3>

A rigid, horizontally extending structural member known as a cantilever is supported at only one end. Typically, it extends from a solidly affixed flat vertical surface, such as a wall.

Given that:-

  • A cantilever beam AB of length L has fixed support at A and spring support at B.
  • The spring behaves in a linearly elastic manner with stiffness k. If a concentrated load P is applied at B.

The spring force will be calculated as:-

F = kx

Deflection will be given by:-

x = \dfrac{PL^3}{3EI}

The spring force will be calculated by:-

F = \dfrac{KPL^3}{3EI}

The deflection of the beam will be given as:-

\rho = \dfrac{0.15KPL^3}{3EI}

Therefore the force in the spring will be F =\dfrac{KPl^3}{3EI}..The deflection of the beam will be \rho = 0.15(\dfrac{KPL^3}{3EI})

To know more about Cantilever beam follow

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7 0
2 years ago
An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,
blsea [12.9K]

Answer:

<em>the % recovery of aluminum product is 80.5%</em>

<em>the % purity of the aluminum product is 54.7%</em>

<em></em>

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the  machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg  is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = <em>80.5%</em>

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = <em>54.7%</em>

8 0
3 years ago
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str
Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 mm^4

So the shear stress at point  A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0
2 years ago
An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

3 0
3 years ago
Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.
likoan [24]

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

 print("Largest: " + string(number2))

else:

 print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))  

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))        

# END #

4 0
3 years ago
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