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Darya [45]
3 years ago
11

Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)

? b. What would be the equivalent reading of a Mercury barometer (mm of Hg)?
Engineering
1 answer:
Fofino [41]3 years ago
6 0

Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi     ( 1 psi= 6894.76 \frac{N}{m^2} )

we know that   P=\rho\times g\timesh

where \rho =Density,g=9.81\frac{m}{s^2}

     h=height.

Given that P=14.769 psi ⇒P= 101828.6 7\dfrac{N}{m^2}

(a) P=\rho_{w}\times g\times h_{w}  

     \rho_{w}=1000\frac{Kg}{m^3}

⇒101828.67=1000\times 9.81\times h_{w}

h_{w}=10.38 m

So water barometer will read 408.66 in.            (1 m=39.37 in)

(b)  P=\rho_{hg}\times g\times h_{hg}

     \rho_{hg}=13600

So 101828.67=13600\times 9.81\times h_{hg}

h_{hg}=0.763 m

So mercury barometer will read 30.04 in.

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Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b
svetlana [45]

complete question:

attached

Answer:

2356.11 W/m^2

6100.11 W/m^2

Explanation:

Assumptions:

1. Steady-state conditions.

2. The cake is placed in a large surrounding.

3. Heat flux delivered to the cake is due to convection and radiation.  

Case 1

Since convection feature is disabled the mode of heat transfer associated with this situation is through free convection and radiation.  

q''(free) = [q''(free convection+q''(radiation) ]W/m^2

            = h_free(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

            = 3 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -  

               (24+273K)^4 ]

            = 468 +1881.11

            = 2356.11 W/m^2

Case 2

Since convection feature is enabled or activated the mode of heat transfer associated with this situation is through forced convection and radiation.  

q''(free) = [q''(forced convection+q''(radiation) ]W/m^2

            = h_forced(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

            = 27 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -  

               (24+273K)^4 ]

            = 4212 +1881.11

            = 6100.11 W/m^2

1. The total heat flux is is 2.58 times higher when the convection feature is activated. Therefore the cake will bake faster during this condition.  

2. The contribution of convection heat flux under natural(free) convection is very low as compared to the contribution during forced convection.  

3. The heat transfer due to radiation is same in both the cases.  

4. Only 19.9 % of the total heat flux is contributed by free convection in the first case.  

5. In the second case 69 % of the total heat flux is contributed by forced convection.  

5 0
3 years ago
What could happen in the aviation
Alina [70]

Answer:

Yes

Explanation:

7 0
2 years ago
A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when
lianna [129]

Answer:

1. 3.4^{o}

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

                                              = (50)(1000/3600) m/s

                                               = 13.89 m/s

The acceleration due to gravity, g = 9.81 m/s^{2}

The frictional force, f = μsN     ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

        ΣF = mg sinθ-f = 0      ...... (2)

Substitute the equation (1) in equation (2), we get

 ΣF = mgsinθ-μsN = 0

 mgsinθ-μsmgcosθ =0

 μs = sinθ/cosθ

   tanθ = μs

    θ = tan-1( μs) = tan-1(0.06) = 3.4^{o}

(b)The vertical component of the force is

N cosθ = fsinθ+mg

 N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg     ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma  (Centripetal acceleration, a = \frac {v^{2}}{r}

  Nsinθ+fcosθ = m(\frac {v^{2}}{r})

   Nsinθ+μsNcosθ = m(\frac {v^{2}}{r})

N[sinθ+μs cosθ] = m(\frac {v^{2}}{r})     ...... (4)    

Dividing the equation (4) with equation (3),

 [sinθ+μscosθ]/[cosθ- μs sinθ] = \frac {v^{2}}{rg}

 cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =\frac {v^{2}}{rg}

[tanθ+μs]/[1-μs tanθ] = \frac {v^{2}}{rg}      

 From part (1), tanθ = μs

 Then the above equation becomes

 \frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

Therefore, the minimum radius of the curvature of the curve is

               r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g} 

                   = \frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}

                   = \frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}

                 = 163.3 m

5 0
3 years ago
Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.
Mekhanik [1.2K]

Answer:

98,614.82 W/m²

Explanation:

Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

Where;

Q = the amount of heat loss from the pipe

h =  the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂  = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

r₂ = outer radius of the pipe = 25mm

To determine the amount of heat loss from the pipe per unit length

From the equation above

\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})

\frac{Q}{L} = 314.159(\frac{70}{0.223})

\frac{Q}{L} = 314.159(313.901) = 98,614.82 W/m²

3 0
3 years ago
Select three functions of catalysts.
Korolek [52]

Answer: speed up food processing

speed up plant growth

Increase fuel efficiency

Explanation:

A catalyst simply refers to a substance that leads to an increase in the reaction rate when it's added to a substance. When the activation energy is reduced by catalysts, this.hwlpa on the speeding up of a reaction.

Therefore,the functions of catalysts include speed up food processing, speeding up plant growth and increase fuel efficiency

5 0
2 years ago
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