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kramer
3 years ago
9

A crude fermenter is set up in a shed in the backyard of a suburban house. Under anaerobic conditions with ammonia as the nitrog

en source, about 0.45 g ethanol are formed per g glucose consumed. At steady state, the ethanol production rate averages 0.4 kg h21. The owner of this enterprise decides to reduce her electricity bill by using the heat released during the fermentation to warm water as an adjunct to the household hot water system. Cold water at 10C is fed into a jacket surrounding the fermenter at a rate of 2.5 litres h21. To what temperature is the water heated? Heat losses from the system are negligible. Use a biomass composition of CH1.75 O0.58 N0.18 plus 8% ash.
Engineering
1 answer:
Aleks04 [339]3 years ago
5 0

Answer:

using calculations Heat losses will be 4512 J

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George and Ellen Rottweiler encourage their adult daughter Guinevere to break her engagement and continue living in their home,
Andrew [12]

Answer:

Negative feedback

Explanation:

In Biology, negative feedback refers to the counteraction of an effect by its own influence on the process producing it. For instance, the presence of   a high level of a particular hormone in the blood may inhibit further secretion of that hormone.

In other words, in negative feedback, the result of a certain action may inhibit further performance of that action

7 0
3 years ago
A body whose velocity is constant has a. positive acceleration b. negative acceleration g. zero acceleration d. all of the above
adoni [48]

Answer:

option (c) is the correct answer which is zero acceleration.

Explanation:

It is given in the question that the velocity is constant.

Now,

the options are provided in relation to the acceleration.

We know,

acceleration is rate of change of velocity per unit time i.e

acceleration = \frac{dV}{dt}

since, the change in velocity is given to be zero,

thus, dV/dt = 0

hence,  

acceleration = 0

therefore, option (c) is the correct answer which is zero acceleration.

4 0
3 years ago
A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a
OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS        A                B                    C

Benzene 1             4.72583     1660.652        -1.461

Toluene  2            4.07827     1343.943         -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5%    = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis     for all components

So for Benzene ; p × y1=x1 × p1s   ------let this be equation 1

for Toluene ; p × y2=x2 × p2s   ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0
2 years ago
Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.
lisov135 [29]

Answer:

specific energy  = 2.65 ft

y2 = 1.48 ft  

Explanation:

given data

average speed v = 6.5 ft/s

width = 5 ft

depth of the water y = 2 ft

solution

we get here specific energy that is express as

specific energy = y + \frac{v^2}{2g}     ...............1

put here value and we get

specific energy = 2 + \frac{6.5^2}{2\times 9.8\times 3.281}  

specific energy  = 2.65 ft

and

alternate depth is

y2 = \frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})  

and

here Fr² = \frac{v1}{\sqrt{gy}}  = \frac{6.5}{\sqrt{32.8\times 2}}  

Fr² = 0.8025

put here value and we get

y2 = \frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})

y2 = 1.48 ft  

7 0
3 years ago
A plastic molding machine produces a product whose annual demand is in the millions.
BabaBlast [244]

Answer:

a. 78.4 pieces/hr

b. $0.1806/min

c. $1.34/piece

Explanation:

(a) With a cycle time Tc = 45 sec = 0.75 min.

Production rate, Rp = 60/0.75 = 80 pieces/hr.

Let us factor in the 98% proportion uptime, so Production rate, Rp = 0.98*(80) = 78.4 pieces/hr

Quantity of product annualy = 6000 *(78.4) = 470,400 pieces/yr

(b) Equipment cost rate, Ceq = 500,000(1.30)/(60 x 10 x 6000) = $0.1806/min.

(c) Cost per piece of mould, Ct = 100,000/1,000,000 = $0.10/piece

Cost rate of labour ,CL = 18.00(0.20) = $3.60/hr = $0.06/min

Conclusively, cost per piece, Cpc = 1.20(0.88) + (0.06 + 0.1806)(0.75) + 0.10 = $1.34/piece

8 0
3 years ago
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