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3 years ago

Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters

Engineering

1 answer:

KonstantinChe [14]3 years ago

4 0

Answer:

Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37

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What is considered the greatest engineering achievement of the 20th century?

kvasek [131]

Answer:

There were a lot of great engineering achievements presented in the 20th century. To name some, we have the electricity, airplane, radio and television, water supply and distribution, computers, television, X-ray imaging, nuclear technologies, and of course the Internet.

6 0

2 years ago

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

Given the inherent costs of regulation it is safe to say that there is always a negative economic impact associated with regulat

Alexus [3.1K]

the answer is true. <u> </u>

6 0

2 years ago

The correct statement about the lift and drag on an object is:_______

Lisa [10]

Answer:

(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift

Explanation:

When a fluid flows around the surface of an object, it exerts a force on it. This force has two components, namely lift and drag.

The component of this force that is perpendicular (normal) to the freestream velocity is known as lift, while the component of this force that is parallel or in the direction of the fluid freestream flow is known as drag.

Lift is as a result of pressure differences, while drag results from forces due to pressure distributions over the object surface, and forces due to skin friction or viscous force.

Thus, drag results from the combination of pressure and viscous forces while lift results only from the<em> pressure differences</em> (not pressure forces as was used in option D).

The only correct option left is "A"

(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift

8 0

4 years ago

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A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at

BartSMP [9]

Answer:

a) C= 1/120

b) P(X>=5) = 0.333

Explanation:

The attached file contains the explanation for the answers

7 0

4 years ago

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