Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

12

A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g

eneration in the wire per unit volume, in Btu/hrft3 , and the heat flux on the outer surface of the wire, in Btu/hrft2 , as a result of this generation.

1 answer:

SSSSS [86.1K]3 years ago

3 0

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

You might be interested in

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

3 years ago

For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho

ad-work [718]

Answer:d

Explanation:

Given

Temperature $=200^{\circ}\approc 473 K$

Also $\gamma for air=1.4$

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no. $=\frac{V}{\sqrt{\gamma RT}}$

(a) $1000 km/h\approx 277.78 m/s$

Mach no. $=\frac{277.78}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.63

(b) $500 km/h\approx 138.89 m/s$

Mach no. $=\frac{138.89}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.31

(c) $2000 km/h\approx 555.55 m/s$

Mach no. $=\frac{555.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=1.27

(d) $200 km/h\approx 55.55 m/s$

Mach no. $=\frac{55.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0

3 years ago

How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = $\frac{q^2Nd^2*Xn^2}{6Eo} * f$

also note that ; Pactive ∝ Nd2 (

tD = K . $\frac{Vdd}{(Vdd - Vt )^2}$ since K = constant

Hence : Nd ∝ rt

5 0

3 years ago

How do you build a engine best anwser will get marked as brainlist

marysya [2.9K]

Answer:

It is best to soak the new lifters in engine oil overnight in order to build a car engine from scratch, regardless of its length. Ensure all rings of the pistons are installed correctly. You should flip the block over and carefully install the top half of the main rod and bearing parts. Get the block out of your way and flip it over again. By flipping over the block, you’ll see the full effect. Ensure that the lifters are installed correctly.

Explanation:

5 0

2 years ago

Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss

klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

Δ $U = Q + W$ ⇒ $Q =$ Δ $U -W$

$Q = 79 - (-63) = 142 kJ$

Therefore, the gas gained heat by an amount of 142 kJ.

3 0

3 years ago