Can some one help me on this?

Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E

White raven [17]

Answer:

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

Explanation:

To solve this problem we use the expression for the temperature film

$T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5$

Then, we have to compute the Reynolds number

$Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}$

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

$Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77$

but we also now that

$Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\$

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0

3 years ago

Does somebody know how to do this?

maksim [4K]

No I don’t sorry, I hope you do well

4 0

3 years ago

Two wastewater treatment plant workers (one male and one female) are exposed to hydrogen sulfide in confined spaces in the treat

Vlada [557]

Answer:

Go to explaination for the details of the answer.

Explanation:

In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:

CDI = Chronic Daily Intake

C= Chemical concentration

CR= Contact Rate

EFD= Exposure Frequency and Distribution

BW= Body Weight

AT = Average Time.

Having names our variables lets create the equations that will be used to derive our answers.

Please kindly check attachment for details of the answer.

5 0

3 years ago

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0

3 years ago

If you were driving the blue Prius in the situation pictured above, explain why the red Mustang should be given right-of-way at

dexar [7]

Answer:

The blue Prius, because the Mustang arrived almost in the same time. And when you arrive in an intersection at the same time of other vehicle you need yield for the car on your right if the car is on your left you have the right of way.

Explanation:

4 0

3 years ago