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DerKrebs [107]
2 years ago
5

Answer ppeeeeeaaaalll

Engineering
1 answer:
Bad White [126]2 years ago
3 0

Answer:

what

Explanation:

is this an exam or an test or what is it

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A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

dQ = m \times c_p \times (T_2 -T_1)

For ideal gas, c_p = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0
3 years ago
What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A
ikadub [295]

Answer:

A.

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

8 0
2 years ago
A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s
sergiy2304 [10]

Answer:

D_o=11.9inch

Explanation:

From the question we are told that:

Thickness T=0.5

Internal PressureP=2.2Ksi

Shear stress \sigma=12ksi

Elastic modulus \gamma= 35000

Generally the equation for shear stress is mathematically given by

 \sigma=\frac{P*r_1}{2*t}

Where

r_i=internal Radius

Therefore

 12=\frac{2.2*r_1}{2*0.5}

 r_i=5.45

Generally

 r_o=r_1+t

 r_o=5.45+0.5

 r_o=5.95

Generally the equation for outer diameter is mathematically given by

 D_o=2r_o

 D_o=11.9inch

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

 D_o=11.9inch

7 0
3 years ago
soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new
timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0
3 years ago
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m ? K), and the wire/sheath interface
Svet_ta [14]

Question

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.

If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Answer:

a. 4.52W/m

b. 13mm

Explanation:

Given

Diameter of electrical wire = 2mm

Wire Thickness = 2-mm

Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)

Thermal contact resistance = 3E-4m².K/W

Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,

Temperature of the ambient air = 20°C.

Maximum Allowable Sheet Temperature = 50°C.

From the thermal circuit (See attachment), we my write

E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)

= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))

Where r in,i = D/2

= 2mm/2

= 1 mm

= 0.001m

r in,o = r in,i + t = 0.003m

T in, i = Tmax = 50°C

Hence

q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]

= 30/[(Ln3/0.26π) + 1/0.06π)]

= 30/[(1.34) + 5.30)]

= 30/6.64

= 4.52W/m

The critical radius is unaffected by the constant resistance.

Hence

Critical Radius = k/h

= 0.13/10

= 0.013m

= 13mm

5 0
3 years ago
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