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kramer
4 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]4 years ago
5 0

Answer:

using calculations Heat losses will be 4512 J

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HELP FAST WILL MARK BRAINLIEST (for a real answer)
Nata [24]

Answer:

B. 180 million joules

Explanation:

Apply the formula for heat transfer given as;

Q=m*c*Δt  where

Q = electrical energy consumed by the heater in joules

m= mass of air in the chamber in kg

c= specific heat of air in joules per kg degrees Celsius

Δt= change in temperatures in degrees Celsius

Given in the question;

m= 1200 kg

c= 1000 J/°C /kg

Δt = 180°-30°= 150° C

Substitute values in the equation to get Q as;

Q=m*c*Δt

Q= 1200 * 1000* 150

Q= 180000000 joules

Q = 180 million joules

<u>The correct answer option is B : 180 million joules.</u>

7 0
3 years ago
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kirza4 [7]

Answer:

203.0160

Explanation:

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7 0
2 years ago
Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0)
Gre4nikov [31]

Answer:

Explanation:

Given that:

y = \int^t_og'(t-s) f(s) ds \  \text{is  solution to } \ my"ky= f(t)

where;

g'(0) = \dfrac{1}{m}     and mg"+kg = 0

\text{Using Leibniz Formula to prove the above equation:}

\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt

So, y = \int ^t_0  g' (t-s) f(s) \ ds

\text{By differentiation with respect to t;}

y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\  y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds

y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)

Since \ \ mg" (t) +kg (t) = 0  \\ \\  \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\  put \  t \  =0 \  we  \ get;\\g" (0) = - \dfrac{k}{m } g(0)  \\ \\  g"(0) = 0 \ \ \ \   ( because \  g(0) =0) \\ \\

Now \ differentiating \ equation (111) \ with \ respect \ to \ t  \\ \\  g"'(t) = -\dfrac{k}{m}g(t)  \\ \\  replacing  \ it \ into  \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o  \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\  y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\  my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}

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3 years ago
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UNO [17]

Answer:

I believe its emotional because it says satisfaction and most of the time that has to deal with emotions running through the body.

Explanation:

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1 Define Engineering <br><br> 2 Engineering use a (n) _______ process to solve problems
OverLord2011 [107]
Engineering use a (n) Engineering design process to solve problems.

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