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pantera1 [17]
3 years ago
8

A plumbed eyewash station is portable.

Engineering
1 answer:
Hitman42 [59]3 years ago
8 0
Plumbed stations are permanently connected to a source of potable water, whereas portable stations are self-contained gravity-fed units with their own flushing fluid that must be replaced after each use. ... Eyewash fluid must irrigate and flush both eyes simultaneously.
Hopefully this helped.
You might be interested in
For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta
Crank

Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

c=\sqrt(\gamma*R*T)

Where R is the gas's particular constant defined in terms of Cp and Cv:

R=Cp-Cv

For particular values given:

R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}

R=188.9 \frac{J}{Kg-K}

The gamma undimensional constant is also expressed as a function of Cv and Cp:

\gamma=Cp/Cv

\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}

\gamma=1.29

And the variable T is the temperature in Kelvin. Thus for the known temperature:

c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)

c=\sqrt(91867.73 \frac{J}{Kg})

The Jules unit can expressing by:

J=N.m=\frac{Kg.m}{s^2}* m

J=\frac{Kg.m^2}{s^2}

Replacing the new units for the speed of the sound:

c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})

c=\sqrt(91867.73 \frac{m^2}{s^2})

c=313 m/s

3 0
3 years ago
Read 2 more answers
A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 10
bagirrra123 [75]

Answer:

Explanation:

Step by step solved solution is given in the attached document.

8 0
3 years ago
A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig
jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

             Section: 9-ft section of pipe.

                     Sum of forces perpendicular to member AC = 0

                     F_d - 0.8*W*L = 0

                     F_d = 0.8*10*9 = 72 lb

                     Sum of forces perpendicular to member BC = 0

                     F_e - 0.6*W*L = 0

                     F_e = 0.6*10*9 = 54 lb

              F_d = 72 lb ,  F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

              Section: Entire Frame.

                    Sum of moments about point B = 0

                    -A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

                    -A_y*(1.5625) + 15 + 39.375 = 0

                    A_y = 34.8 lb  

                   Sum of forces in vertical direction = 0

                     A_y + B_y - 0.8*F_d - 0.6*F_e = 0

                     B_y = 0.8*(72) + 0.6*(54) - 34.8

                     B_y = 55.2 lb  

                   Sum of forces in horizontal direction = 0

                     A_x + B_x - 0.6*F_d + 0.8*F_e = 0

                     A_x + B_x = 0

               Section: Member AC

                    Sum of moments about point C = 0

                     F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

                     72*2.5 - 34.8*12 - 9*A_x = 0

                     A_x = -237.6 / 9 = - 26.4 lb

                     B_x = - A_x = 26.4 lb

                     A_x = -26.4 lb  ,  B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

               Member AC:

                    From point A till just before point D

                     -0.6*A_x*x - A_y*0.8*x + M = 0

                     15.84*x - 27.84*x + M = 0

                      M = 12*x   ..... max value at D, x = 12.25 in

                      M_max = 12*12.25/12 = 12.25 lb-ft

               Member BC:

                    From point B till just before point E

                     -0.8*B_x*x + B_y*0.6*x + M = 0

                     -21.12*x + 33.12*x + M = 0

                      M = -12*x   ..... max value at E, x = 11.25 - 2.5 = 8.75 in

                      M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

                      AC: at D , M_max = 12.25 lb-ft

                      BC: at E , M_max = 8.75 lb-ft

5 0
2 years ago
: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the
Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}

V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s

\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}

taking f = 0.0185

at Z₁ = Z₂

\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P= 1.13\ psi

c) When flow is downhill  z₂-z₁ = -2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P=-0.601\ psi

7 0
3 years ago
The fan pressure differential gage on an air handler reads 12 cm H2O. What is this pressure differential in kiloPascals
Deffense [45]

Answer:

1.18\ \text{kPa}

Explanation:

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

h = Height of reading = 12 cm

\rho = Density of water = 1000\ \text{kg/m}^3

Pressure due to height difference is given by

P=\rho gh\\\Rightarrow P=1000\times 9.81\times 12\times 10^{-2}\\\Rightarrow P=1177.2\ \text{Pa}=1.1772\ \text{kPa}\approx 1.18\ \text{kPa}

The pressure differential is 1.18\ \text{kPa}.

4 0
3 years ago
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