2 years ago

Can anyone help me ?

Engineering

1 answer:

MrRa [10]2 years ago

3 0

That’s too hard for me lol oof

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Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 90-gallon hot

Rudik [331]

Answer:

Q=36444.11 Btu

Explanation:

Given that

Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

Volume of water = 90 gallon

Mass = Volume x density

$1\ gallon = 0.13ft^3$

Mass ,m= 90 x 0.13 x 62.36 lbm

m=729.62 lbm

We know that sensible heat given as

Q= m Cp ΔT

Now by putting the values

Q= 729.62 x 0.999 x (110-60) Btu

Q=36444.11 Btu

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2 years ago

A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c

Paraphin [41]

Answer:

$10.8\ \text{lb/ft^2}$

$101.96\ \text{lb/ft}^2$

Explanation:

$v_1$ = Velocity of car = 65 mph = $65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}$

$\rho$ = Density of air = $0.00237\ \text{slug/ft}^3$

$v_2=0$

$P_1=0$

$h_1=h_2$

From Bernoulli's law we have

$P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}$

The maximum pressure on the girl's hand is $10.8\ \text{lb/ft^2}$

Now $v_1$ = 200 mph = $200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}$

$P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2$

The maximum pressure on the girl's hand is $101.96\ \text{lb/ft}^2$

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2 years ago

The path taken to move items from their origin to their destination is known as which of the following?

mrs_skeptik [129]

Answer:d

Explanation:

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2 years ago

Match the given traits to their definitions.

damaskus [11]

Answer:

where the question

Explanation:

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2 years ago

Read 2 more answers

2. A trapezoidal channel has a bottom width of 4 m and side slopes of 2:1 (H:V). If the flow rate is 50 m3/s at a depth of 3.6 m

Daniel [21]

Answer:

Explanation:

Sum of the side slope = 2 + 1 = 3

Length of first slope = 2/3 X 3.6 = 2 X 1.2 = 2.4m

Lenght of second slope = 1/3 X 3.6 = 1.2m

Area of the trapezoidal channel = (2.4 + 1.2)/2 X 3.6 = 1.8 X 3.6 = 6.48m²

Alternate dept = 50m³/6.48m²= 7.716m

4 0

2 years ago