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2 years ago

A resistance of 30Ω is placed in a circuit with a 90 volt battery. What current flows in the circuit?

Engineering

1 answer:

Svetlanka [38]2 years ago

6 0

Answer:

30 Amperes

Explanation:

if V ⇒ voltage

I ⇒ current

R ⇒ resistance

V = IR

90 = I x 30

90/30 = I

3 = I

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7–53 Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and leaves at

qwelly [4]

Answer:

a. The coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is 1.96368kW

Explanation:

Given

First we need to get the enthalpy of R-34a.

When T = 35°C and P = 800kPa;

h1 = 271.24kj/kg

When x2 = 0 and P = 800kPa;

h1 = 95.48kj/kg

To calculate the COP, first we need to calculate the energy balance.

This is given as

Q = m(h1 - h2)

Where m = 0.018kg

Q = 0.018(271.24 - 95.48)

Q = 3.16368Kw

COP is then calculated as Q/W

Where W = Power consumption of the compressor = 1.2kW

COP = 3.16368Kw/1.2Kw

COP = 2.6364

Hence, the coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is calculated as ∆Heat Rate

∆Heat Rate = Q - W

Where Q = Energy Balance = 3.16368Kw

W = Power consumption of the compressor = 1.2kW

∆Heat Rate = 3.16368Kw - 1.2kW

∆Heat Rate = 1.96368kW

Hence, The rate of heat absorption from the outside air is 1.96368kW

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3 years ago

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Nastasia [14]

C! hope this helps i got it right

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2 years ago

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Consider a pond that initially contains 10 million gallons of fresh water. Water containing a chemical pollutant flows into the

Ostrovityanka [42]

Answer / Explanation:

To solve this , we start by recalling the conversion principle:

Thus, dc/dt = Rate of C in - Rate of C out,

Therefore, for the number of chemical C(t) , i gram, in the pond at time (t), the rate of flow of C is modeled as:

Rate of Cin = Cin F,

Where Cin is the concentration of the chemical going in ( in this case, Cin = 2 grams per gallon), and F is the rate of flow of water into the pond (in this case, F = 5 million gallons per year). We then have that

Rate of C in = 10 million grams per year

Therefore the rate of Cout = c(t)F, where c(t) = C(t) / V is the concentration of the chemical in the pond ( here, we are assuming instant mixing). The volume, (V) of the water in the pond is 10 million gallons (we are assuming that the rate of flow of water into the pond is the same as the rate of flow out, so that the volume of water in the pond remains constant). Thus,

Rate of C out = 1/2 C(t) million grams per year.

Thus, the differential equation describing the evolution of C =

C(t) in time is: dc/dt = 10 - C/2 in millions of grams per year.

Now to plot the solution using Maple and to describe in words the effect of the variation in the incoming chemical,

we have: C(t) = 20 +ce ∧ -t/2, where c is an arbitrary constant since there was no chemical in the pond initially.

Therefore, C(t) = 0 , then C = -20

So that Q(t) = 20 ( 1 - e ∧ -t/2)

Kindly find the expression of the above statement C = C(t) below:

Note: In the diagram, Q is represented by C

Therefore, C = Q