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3 years ago

Calculate the energy in kJ added to 1 ft3 of water by heating it from 70° to 200°F.

Engineering

1 answer:

Serga [27]3 years ago

3 0

Answer:

The energy in kJ is 8558.16 kJ.

Explanation:

Data presented in the problem:

Water is heated from 70 (T1) to 200 °F (T2).

Volume (V) of the water is 1 ft3.

It is required for the specific heat of water(HW), which is 1 BTU/lb°F.

First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.

M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.

.After that, we can calculate the heat required (Q).

Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)

Q = 62.4 * 130 BTU = 8112 BTU.

Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ

Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.

Finally, the energy required is 8558.16 kJ.

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Discuss four (4) advantages of direct and indirect water supply system.

Reil [10]

Answer and Explanation:

Some of the advantages of direct water supply system are:

Smaller size of the cistern
No risk of polluted water
Ease of installation
Lower pipe work

Some of the advantages of indirect water supply system are:

It can be pumped completely
Lower demand on the main source
Minimization of wear and tear on taps
Larger size of the cistern.

7 0

3 years ago

Read 2 more answers

A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p

Stella [2.4K]

Answer:

Cut-off ratio $\dfrac{V_3}{V_2}=6.15$

Cxpansion ratio $\dfrac{V_4}{V_3}=3.25$

The exhaust temperature $T_4=1997.5R$

Explanation:

Compression ratio CR(r)=20

$\dfrac{V_1}{V_2}=20$

$P_2=P_3=920 psia$

$T_1=520 R ,T_{max}=T_3,T_3=3200 R$

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

$\dfrac{T_2}{T_1}=r^{\gamma -1}$

$\dfrac{T_2}{520}=20^{1.4 -1}$

$T_2=1723.28R$

$\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}$

$\dfrac{V_3}{V_2}=\dfrac{3200}{520}$

So cut-off ratio $\dfrac{V_3}{V_2}=6.15$

$\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}$

Now putting the values in above equation

$\dfrac20=\dfrac{V_4}{V_3}\times 6.15$

$\dfrac{V_4}{V_3}=3.25$

So expansion ratio $\dfrac{V_4}{V_3}=3.25$ .

$\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}$

$\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}$

$T_4=1997.5R$

So the exhaust temperature $T_4=1997.5R$

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4 years ago

According to the eNotes, a program that eliminates sales and promotions in an effort to minimize the bullwhip effect would be ca

Ilia_Sergeevich [38]

Options: True or false

Answer:True, it will be called CRMWRONGEDLP.

Explanation:eNotes was founded in 1998 by Brad Satoris and Alexander Bloomingdale, to enable and enhance the ability of students to solve assignments as given and other home works,eNote is also used by students to prepare effectively and efficiently for examinations.

Since its launch eNote has been a great source of educational materials for students who have gained a lot in the use of its materials.

bullwhip effect is a concept in supply chain, where forcasts causes the supply chain to be inefficient, it usually starts from retailers who raise concern of high demands.

CRMWRONGEDLP, is the program that minimizes the bullwhip effects according to eNote.

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3 years ago

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

Answer:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

Explanation:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

(i) convert to polar form

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

(i) convert to polar form

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

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3 years ago

What disadvantages does traditional high-speed internet have compared with space-based internet?( choose all that apply)

N76 [4]

A,B I’m assuming I’m not sure

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3 years ago