1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vikki [24]
3 years ago
6

Your ship has loaded 9,000 barrels of fuel oil at a cargo temperature of 35° C (95° F). API gravity is 44°. The volume correctio

n factor (VCF) is .0005. You are bound to New Jersey from Venezuela. How many barrels would you expect to unload if the cargo temperature is 55° F at the discharge port?
Engineering
1 answer:
evablogger [386]3 years ago
8 0

Answer:

8 820 barrels

Explanation:

Data:

The number of barrels loaded = 9 000

cargo temperature                    = 35 °C

API                                              = 44 ° C

Barrels when the cargo temperature is 55° F from the determination = 8 820 barrels.

You might be interested in
g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],
yKpoI14uk [10]

Explanation:

\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}

Therefore,

-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}

Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}

5 0
3 years ago
A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft
tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ;  f_{Lw}=6.6 mi/h

5' on right   ;    f_{Lc} = 0.4 mi/hr

3' on left   ;      no adjustment

3 lanes in each direction    f n = 3 mi/h

v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}

f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}

f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}

          = 0.877

v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}

       = 1,555 veh/hr/lane

FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}

      = (55 + 5) - 6.6 - 0.4 -3 -0

      = 50 mi/h

D = \dfrac{V_P}{s}

D = \dfrac{1555}{55} =28.27

level of service is D using speed flow curves and LOS for basic free moving of vehicle

5 0
3 years ago
Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes
loris [4]

Answer:

a). 8.67 x 10^{-3} m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L  = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X 10^{-5} Pa-s

We know density of air is ρ = 1.21 kg /m^{3}

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = \frac{\rho .U.x}{\mu }

Re = \frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = \frac{5.x}{\sqrt{Re}}

   = \frac{5\times 1}{\sqrt{332052.6}}

   = 8.67 x 10^{-3} m

a). Boundary layer thickness at x = 1 is δ = 8.67 X 10^{-3} m

b). Given Re = 100000

    Therefore the critical distance from the leading edge can be found by,

     Re = \frac{\rho .U.x}{\mu }

     100000 = \frac{1.21\times5\times x}{1.822 \times10^{-5}}

     x = 0.3011 m

c). Given x = 3 m from the leading edge

    The Reyonld no at x = 3 m from the leading edge

     Re = \frac{\rho .U.x}{\mu }

     Re = \frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }

     Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

    δ = \frac{0.38\times x}{Re^{\frac{1}{5}}}

       = \frac{0.38\times 3}{996158.06^{\frac{1}{5}}}

       = 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

  Re = \frac{\rho \times U\times x}{\mu }

5 X 10^{5} = \frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}

 x = 1.505 m

We know that the force acting on the plate is

F_{D} = \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

and C_{D} at x= 1.505 for a laminar flow is = \frac{1.328}{\sqrt{Re}}

                                                                         = \frac{1.328}{\sqrt{5\times10 ^{5}}}

                                                                       = 1.878 x 10^{-3}

Therefore, F_{D} =  \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

                                          = \frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}

                                         = 0.2137 N

e). The flow is turbulent at the end of the plate.

  Re = \frac{\rho \times U\times x}{\mu }

       = \frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }

       = 1992316

Therefore C_{D} = \frac{0.072}{Re^{\frac{1}{5}}}

                                           = \frac{0.072}{1992316^{\frac{1}{5}}}

                                           = 3.95 x 10^{-3}

Therefore F_{D} = \frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}

                                           = \frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}

                                          = 1.792 N

3 0
3 years ago
To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01
liq [111]

Answer:

138.9 °C

Explanation:

The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C

3 0
2 years ago
Provide an argument justifying the following claim: The average (as defined here) of two Java ints i and j is representable as a
ahrayia [7]

Answer:

public static int average(int j, int k) {

return (int)(( (long)(i) + (long)(j) ) /2 );

}

Explanation:

The above code returns the average of two integer variables

Line 1 of the code declares a method along with 2 variables

Method declared: average of integer data type

Variables: j and k of type integer, respectively

Line 2 calculates the average of the two variables and returns the value of the average.

The first of two integers to average is j

The second of two integers to average is k

The last parameter ensures average using (j+k)/2

3 0
2 years ago
Other questions:
  • A Scalar can only be a positive quantity that has a magnitude but no direction ? a)-True b)-False
    5·1 answer
  • Estimate the daily carbon utilization to remove chlorobenzene from 1.0 MGD of ground water saturated with chlorobenzene. Assume
    12·1 answer
  • Megan is an architect who is creating a building design. Which is a prominent technical aspect that she needs to keep in mind wh
    12·1 answer
  • WHO EVER COMMENTS FIRST GETS BRAINLIEST LOL
    15·2 answers
  • A train which is traveling at 70 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its
    12·1 answer
  • Why is the lubrication system of an internal combustion engine equipped with an oil filter?
    11·1 answer
  • Three possible career opportunities in embedded systems engineering
    11·1 answer
  • A partnership between a gaming company and moviemakers might happen in what two ways?
    6·1 answer
  • Define volume flow rate of air flowing in a duct of area A with average velocity V.
    13·1 answer
  • You’re going to write a program that models the Littleton City Lotto (not a real Lotto game). The program is going to allow to u
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!