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Svetlanka [38]
3 years ago
7

1 kg of oxygen is heated from 20 to 120°C. Determine the amount of heat transfer required when this is done during a (a) constan

t-volume process and (b) isobaric process.
Engineering
1 answer:
Ahat [919]3 years ago
4 0

(a) At constant-volume, Q = 66.7kJ

(b) At isobaric condition, Q = 92.7kJ

<u>Explanation:</u>

The specific heats of oxygen at the average temperature of (20+120)/2 = 70°C = 343 K are

Cp = 0.927  kJ/kg.K

Cv = 0.667 kJ/kgK

We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the  system. The energy balance for a constant-volume process can be expressed as

ΔE (system) = E(in) - E(out)

and

Q(in) = ΔH = mCp x (T₂ - T₁)

since ΔU + Wb = ΔH during a constant pressure quasi-equilibrium

process. Substituting for both cases,

(a)

Qin,V =const = mCv  (T2 - T1 )

                      = (1kg)(0.667 kJ/kg.K)(120 - 20)K

                      = 66.7kJ

(b)

Qin,P=const = mCp  (T2 - T1  )

                     = (1kg)(0.927 kJ/kg.K)(120 - 20)K

                     = 92.7kJ

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When the renewal period comes around (with respect to the license expiry date).

Explanation:

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2 years ago
A centimeter is Viooth of a meter, while a kilo-
tatuchka [14]

Answer:

b. False

Explanation:

A centimeter is a hundredth of a meter. This means 1/100 ,1 cm = 0.01 m

A kilo is 1000 grams.

The first answer is false because a hundredth isnot written well.

The second statement is false because the comparison given is of different units of measure. For distance is meters where as for weight is kilograms.

8 0
4 years ago
A farmer has 12 hectares of land on which he grows corn, wheat, and soybeans. It costs $4500 per hectare to grow corn, $6000 to
maw [93]

The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans

<h3>How to solve algebra word problem?</h3>

He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;

C + W + S = 12   ----(1)

It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;

45C + 60W + 50S = 63750  -----(2)

He will grow twice as many hectares of wheat as corn. Thus;

W = 2C    ------(3)

Put 2C for W in eq 1 and eq 2 to get;

C + 2C + S = 1200

3C + S = 1200     -----(4)

45C + 60(2C) + 50S = 63750

45C + 120C + 50S = 63750

165C + 50S = 63750    ------(5)

Solving eq 4 and 5 simultaneosly gives;

C = 250 and W = 500

Thus; S = 1200 - 3(250)

S = 450

Read more about algebra word problems at; brainly.com/question/13818690

5 0
2 years ago
Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

6 0
3 years ago
Discuss the difference between the observed and calculated values. Is this error? If yes, what is the source?
Scrat [10]

Answer and Explanation:

In any experiment, the observed values are the actual values obtained in any experiment.

The calculated values are the values that are measured by using the observed values in a formula.

The observed values are primary values whereas the calculated values are the secondary values as calaculations are made using observed values.

Yes, if the observed values are of low accuracy.

The values should be recorded with proper care and attention in order to avoid any error.

8 0
3 years ago
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