If you are unsure about holding a piece of wood to be drilled, then you should always use a

A 132mm diameter solid circular section

Ganezh [65]

Answer:

not sure if this helps but

5 0

3 years ago

You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210

Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity = 1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

= 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0

3 years ago

Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz

butalik [34]

Answer:

Exit velocity $V_2=1472.2$ m/s.

Explanation:

Given:

At inlet:

$P_1=100 bar,T_=600°C,V_1=35m/s$

Properties of steam at 100 bar and 600°C

$h_1=3624.7\frac{KJ}{Kg}$

At exit:Lets take exit velocity $V_2$

We know that if we know only one property inside the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

$h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}$

So the enthalpy of steam at the exit of turbine

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}$

$h_2=2541.62\frac{KJ}{Kg}$

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

In the case of adiabatic nozzle Q=0,W=0

$3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0$

$V_2=1472.2$ m/s

So Exit velocity $V_2=1472.2$ m/s.

4 0

3 years ago

Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t

almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

$h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}$

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

$n = \frac{h_1 - h_2*}{h_1 - h_2}$

$n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}$

solving for h2, we have

$h_2 = 1197.77 Btu/Ibm$

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

$(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}$

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0

3 years ago

Both model building codes and NFPA __________ can be used to determine the type of construction used in a building.

Alex777 [14]

Answer:

Both model building codes and NFPA 220 can be used to determine the type of construction used in a building.

8 0

2 years ago