Ask question

Ask question

All categories

3 years ago

10

An electric power plant uses solid waste for fuel in the production of electricity. The cost Y in dollars per hour to produce el

ectricity is Y = 12+ .3x+.27x2, where x is in megawatts. Revenue in dollars per hour from the sale of electricity is 15x-.2x2. Find the value of X that gives maximum profit.

1 answer:

andrew-mc [135]3 years ago

8 0

Answer:

15.64 MW

Explanation:

The computation of value of X that gives maximum profit is shown below:-

Profit = Revenue - Cost

= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2

= 14.7x - .47x^2 - 12

After solving the above equation we will get maximum differentiate for profit that is

14.7 - 0.94x = 0

So,

x = 15.64 MW

Therefore for computing the value of X that gives maximum profit we simply solve the above equation.

You might be interested in

The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo

Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

$P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)$

consider the equation of bernoulli at the top of the pool

$P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)$

where $P_1=P_0$ atm pressure

At the top of the pool $v_1=0m/s$ , substitute in V_1 in equation (2)

$P_0+\rho gh_1 =constant ...(3)$

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

$P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)$

where $P_2=P_0$ atm pressure and $h_2=0m$

$P_0+\rho v_1^2 =constant ...(6)$

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

$P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)$

Hence velocity is $v_2=(\sqrt{2gh_1})m/s$

d.) consider (7)

$v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)$

This is the norminal value of velocity

e.) consider the equation of flow rate interval of v and t

flow(t)= $\frac{dv}{dt}(m^3/s)$ hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

$AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8$

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

$A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2$

6 0

3 years ago

2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?

mash [69]

D...................

3 0

2 years ago

A 0.50 m3 drum was filled with 0.49 m3 of liquid water at 25oC and the remaining volume was water vapor without any air. The dru

timurjin [86]

Answer:

There is not going to be pressure build up in the system,that is isobaric process.

Explanation:

Assumptions to be made

1. No mass is gained or lost during the heating process.

2. There are no friction losses,so work is transmitted efficiently.

3. It was started the water in the drum and its surrounding have same temperature.

4. This system is closed,so there is no mass transfer across its boundaries.

5 0

3 years ago

Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a

nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

$R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}$

Now by putting the values

$R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}$

$R_{th}=4.083/A\ K/W$

We know that

Q=ΔT/R

$Q=\dfrac{\Delta T}{R_{th}}$

$Q=A\times \dfrac{200-40}{4.086}$

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0

3 years ago

An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out

Elenna [48]

Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

Q1 - Q2 + Q + W = 0

b) rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

<u>Write the First law balance around the system and rate of work input to the system</u>

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw

Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw

Q = 18 Kw

Insert values into equation 2 above

W = 6.88 Kw

5 0

3 years ago