Fresh cut potato strips with a moisture content of 50% (w/w) on wet basis are fried in peanut oil to produce French fries.

Engineering

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

Percentage of oil uptake by the potato = 5.57% on a wet basis (including the moisture content)

Percentage of oil uptake by the potato = 7.24% on a dry basis (excluding the moisture content)

Explanation:

Starting with 1000kg of fresh potatoes,

It is given that this consists of 50% (w/w) of water.

Meaning that, 1000 kg of fresh potatoes consists of (1000 × 50%) of water = 500 kg of water.

If 1000 kg of potatoes consist of 500 kg of water, then the remaining 500 kg is pure potatoes (since it is stated that freshly cut potatoes have no oil content)

After frying, the weight of the French fries is 700 kg now.

But of this 700 kg, 23 % of its weight is moisture content, i.e. water,

23% of 700 kg = 161 kg.

This means that the amount of potatoes and oil in the French fries is 700 - 161 = 539 kg

But, recall, that the amount of potatoes in this process from the start is 500 kg. This amount doesn't change even on frying the fresh potatoes into French fries.

So, amount of oil in the French fries = 539 - 500 = 39 kg

Percentage of oil uptake by the potato = 100% × (amount of oil intake)/(total mass of potatoes) = 100% × 39/700 = 5.57% on a wet basis (including the moisture content)

And 100% × 39/539 = 7.24% on a dry basis (excluding the moisture content)

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P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si

klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V 0.184 W

2.499 mV 125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

$V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\$ *5

= 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:

V_o = A_voc*V_i*(R_o/R_l+R_o)

= 4.545*(100/100+50)

= 3.03 V

Now we can determine delivered power:

P_L = V_o^2/R_L

= 0.184 W

Apply voltage-divider principle to the circuit:

V_o = (R_o/R_o+R_s)*V_s

= 50/50+100*10^3*5

= 2.499 mV

Now we can determine delivered power:

P_l = V_o^2/R_l

= 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.