Ask question

Ask question

All categories

3 years ago

10

The fan pressure differential gage on an air handler reads 12 cm H2O. What is this pressure differential in kiloPascals

1 answer:

Deffense [45]3 years ago

4 0

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 $1.18\ \text{kPa}$ .

You might be interested in

An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tens

Natalka [10]

Answer:

It will results in mechanical hardening.

5 0

3 years ago

Read 2 more answers

Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.

Nutka1998 [239]

Answer:

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = $\sqrt{\frac{k}{m} }$ = $\sqrt{2}$

next we determine the damping condition using the damping formula

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

6 0

2 years ago

Steam enters an adiabatic turbine at 400◦C, 2 MPa pressure. The turbine has an isentropic efficiency of 0.9. The exit pressure i

pychu [463]

Answer:

Explanation:

Find attached the solution

8 0

2 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

2 years ago

Can some help me with this !!! Is 26 points!!

Aleonysh [2.5K]

Third one
15,000,000 ohms because M=10^6

8 0

2 years ago