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4 years ago

1. How many board feet in piece of Oak that is 1" thick 6" wide and 8' long?

Engineering

1 answer:

enyata [817]4 years ago

7 0

Answer:

Explanation:

A "board foot" is 1/12 of a cubic foot. It is the volume of a board 1" thick and 1 foot square.

Here, the board is 1" thick, so the number of board feet is numerically equal to the number of square feet of area it has.

(1/2 ft)(8 ft) = 4 ft²

The number of board feet is 4.

You might be interested in

What is the total inductance of a circuit that contains two 10 uh inductors connected in a parallel?

kolbaska11 [484]

Answer:

5 microhenries

Explanation:

The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.

10 uH ║ 10 uH = 5 uH

The effective inductance is 5 uH.

6 0

3 years ago

An undeformed specimen of some alloy has an average grain diameter of 0.050 mm. You are asked to reduce its average grain diamet

aleksandrvk [35]

Answer:

Yes it is possible

Explanation:

Procedures to be taken:

Step 1:

I will deform the specimen, that is, I will subject the specimen to plastic deformation at room temperature.

Step 2:

Also, I will anneal the deformed specimen at a high temperature.

Step 3:

Then, recrystallize the annealed specimen

Step 4:

Finally, I will facilitate the grain growth until the average grain diameter becomes 0.02mm.

5 0

3 years ago

Calculate the number of 12 V batteries (capacity 120 Ah) needed to run a 3 kW DC motor that operates in 240 V. How many hours th

Mrac [35]

Answer:

20 batteries
9.6 hours

Explanation:

To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...

240/12 = 20 . . . batteries needed

The current draw will be ...

(3000 W)/(240 V) = 12.5 A

Then the time available from the battery stack is ...

(120 Ah)/(12.5 A) = 9.6 h

The motor can run 9.6 hours from the series connection.

3 0

3 years ago

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)

vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0

4 years ago