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Slav-nsk [51]
3 years ago
14

A large truck drives down the highway at 10 m/s hauling a rectangular trailer that is 6 m long, 2 m wide, and 2 m tall. The trai

ler contains frozen food and is therefore temperature-controlled. Its external surface can be approximated to be a consistent 15°C, while the outside air is at 20°C. Assume the heat transfer on the front, back, and bottom of the trailer is negligible.
a) How much cooling power must be provided to maintain the temperature controlled trailer? (i.e. What is the total heat transfer rate from the air to the trailer)?b) What is the minimum local heat transfer coefficient on the surface of the trailer? Where does the minimum occur?c) What percentage of the total heat transfer is occuring over a laminar boundary layer?d) If the cooling system can only provide 5 KW of cooling, what is the fastest speed that this truck can drive while still adequately maintaining the temperature within the trailer?

Engineering
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

3w/m²k

Explanation:

Base on the scenario been described in the question, the solution to the given problem solve in the file attached below

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Txt you. U u u yxigzextctvyb
4 0
3 years ago
7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 v
Naddik [55]

Answer:

23.34 seconds

Explanation:

Flow rate = 1500

Arrival = 800 vehicle per hour

Cycle c = 60 seconds

Dissipation time = 10 seconds

Arrival time = 800/3600 = 0.2222

Rate of departure = 1500/3600 = 0.4167

Traffic density p = 0.2222/0.4167 = 0.5332

Real time = r

r + to + 10 = c

to = c-r-10 ----1

t0 = p*r/1-p ----2

Equate both 1 and 2

C-r-10 = p*r/1-p

60-r-10 = 0.5332r/1-0.5332

50-r = 0.5332r/0.4668

50-r = 1.1422r

50 = 1.1422r + r

50 = 2.1422r

r = 50/2.1422

r = 23.34 seconds

7 0
3 years ago
A room in the lower level of a cruise ship has a 30 cm diameter circular window If the midpoint of the window is 4 m below the w
AleksAgata [21]

Answer:

F = 2840.3 N

Explanation:

Given:

- Diameter of window D = 0.3 m

- Midpoint of window from sea level h = 4 m

- Specific gravity of sea water S.G = 1.024

- Density of water p = 1000 kg/m^3

Find:

The hydro-static force F_r acting on the mid-point of the window.

Solution:

- The average pressure P acting on the midpoint of the window:

                               P = S.G p*g*h

                               P = 1.024*1000*9.81*4

                               P = 40181.76

- The hydro-static force F_r acting on the mid-point of the window:

                               F = P*A = P*pi*D^2 / 4

                               F = 40181.76*pi*0.3^2 / 4

                               F = 2840.3 N

7 0
3 years ago
Use differentials to estimate the amount of metal in a closed cylindrical can that is 30 cm high and 6 cm in diameter if the met
PilotLPTM [1.2K]

Answer:

dV=113.55 \ cm^3

Explanation:

<u>The Differential of Multivariable Functions</u>

Given a multivariable function V(r,h), the total differential of V is computed by

dV=\displaystyle \frac{\partial V}{\partial  r} dr+\frac{\partial V}{\partial  h} dh

The volume of a cylinder of radius r and height h is

V=\pi\cdot r^2\cdot h

Let's compute the partial derivatives

\frac{\partial V}{\partial  r}=2\pi rh

\displaystyle \frac{\partial V}{\partial  h}=\pi r^2

The total differential is

dV=(2\pi rh)dr+(\pi r^2)dh

The differential dr is approximated to \Delta r and the differential dh is approximated to \Delta h. We can see the increment of radius is the thick of the metal in the sides, and the increment of the height is the thick of the metal in the top and bottom. Thus dr=0.05 cm, dr=0.2 cm

dV=(2\pi \cdot 3\ cm\cdot 30\ cm)0.2\ cm+(\pi\ (3\ cm)^2)0.05\ cm

dV=113.55 \ cm^3

5 0
3 years ago
(a) For the solidification of iron, calculate the critical radius r* and the activation free energy ΔG* if nucleation is homogen
CaHeK987 [17]

Answer:

r = 1.46 *10^{-9} m

\Delta G^* = 1.828 *10^{-18} j

Explanation:

given data:

latent heat of fusion \Delta H_f = -1.85*10^9 j/m2

surface free energy = 0.204 j/m2

meltinf point = 1538 degree celcius

\Delta T_c = 273 K

critical radius is given as

r^* = \frac{-2rT_m}{\Delta H_f} *\frac{1}{\DeltaT_c}

= \frac{-2*0.204*(1538+273)}{-1.85*10^9} *\frac{1}{273}

r = 1.46 *10^{-9} m

activation free energy is given as

\Delta G^* = \frac{16\pi r^3 t_m^2}{3\Delta H^2_f} * \frac{1}{\Delta T^2_C}

       = \frac{16\pi 0.204^3*(1538+273)^2}{3*(-1.85*10^9)^2} * \frac{1}{273^2}

\Delta G^* = 1.828 *10^{-18} j

5 0
3 years ago
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