Hi plz delete this question i had to edit it cuz it was wrong question

A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen

Lerok [7]

Answer:

Q = 424523.22 kw

Explanation:

$\rho =7833 kg/m$

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

$\alpha = 3.91*10^{-6} m^2/s$

$T_s = 285 degree celcius$

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate $\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15$

$\dot m = 14764.85 kg/s$

$A_s = \pi DL = \pi 0.4*2 = 2.513 m^2$

$Q = \dot m C \Delta T = h A_s \Delta T$

$\dot m C \Delta T = h A_s \Delta T$

solving for h

$h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}$

h = 675.6 kw/m^2K

$Q = h A_s\Delta T$

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

7 0

3 years ago

What is A roofed structure that is similar to a porch, but is detached from the house.

agasfer [191]

Answer:

a gazebo

Explanation:

6 0

3 years ago

Read 2 more answers

A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.

AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass $m_{in}+m_{out}=\bigtriangleup m_{system}$

so, $m_e=m_1-m_2$

$Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1$

#Also, given the properties of water as;

$(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg$

#We assume constant properties for the steam at average temperatures: $h_e=\approx(h_1+h_2)/2$

#Replace known values in the equation above; $h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg$

#Using the mass and energy balance relations;

$m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg$

#We have $Q_i_n+m_eh_e=m_2u_2-m_1u_1$ : we replace the known values in the equation as;

$Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ$

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0

3 years ago

A person is planning a bungee jump from a 40 meter high bridge. Under the bridge is a river with crocodiles, so the person does

Nonamiya [84]

Answer:

a. l = 19.7m, b. 18.55m, c. Impact Force = 3889.84 N

Explanation:

The total energy of the system when the person is at top of the bridge is

Potential energy = mgh, Kinetic energy = 0

The total energy of the system when the person reaches just above the surface

Potential energy = 0, Kinetic energy = 0, Spring energy = ½ K X2, where k is the spring constant and X is the deflection

Applying conservation of energy

mgh = 0 + 0 + ½ K X²

80 x 9.81 x 40 = ½ (3600/l) X²

31392 = ½ (3600/l) X²

We can also conclude that

l+ X + 1.75 = 40

l + X = 38.25

a. Substitute the value of x from above into the energy conversion expression

31392 = ½ (3600/l)(38.25 - l)²

31392 x 2/3600 = (38.25 + l² – 2l(38.25))/l

17.44l = l2 – 76.5l + 38.25²

l² – 76.5l – 17.44l +1463.0625 = 0

Solving for l we get

L = 19.7

Hence, length of the rope is 19.7m

b. The deflection is calculated by using the relation between l and X

L + X = 38.25

X = 38.25 – 19.7 = 18.55m

c. The impact force is calculated using the impact force formula which relates the impact force with the deflection

F = KX

F = (3600/l) . X

F = (3600/19.7) . (18.55) = 3889.84 N

Thus, the impact force is 3889.84 N

3 0

3 years ago

What are the three main sections of a nuclear power plant?

strojnjashka [21]

The reactor, generator, and the cooling towers

7 0

3 years ago