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3 years ago

8

Before a rotameter can be used to measure an unknown flow rate, a calibration curve of flow rate versus rotameter reading must b

e prepared.
a. True
b. False

1 answer:

allochka39001 [22]3 years ago

6 0

This statement is b which is true: hope this helped

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How much power is needed to operate a Carnot heat pump if the pump receives heat 10°C and delivers 50 kW of heat at 40°C? at A)

mariarad [96]

Answer:

Power needed to pump=4.79 KW.

Explanation:

Given that: $T_{1}=283K,T_{2}=313K,Q_{H}=50KW$

We know that coefficient of performance of heat pump

COP= $\dfrac{T_{H}}{T_{H}-T_{L}}$

So COP= $\dfrac{313}{313-283}$

COP=10.43

COP= $\frac{Q_{H}}{W_{in}}$

10.43 = $\frac{50}{W_{in}}$

$W_{in}$ =4.79 KW

So power needed to pump=4.79 KW.

3 0

3 years ago

Using Pascal’s Law and a hydraulic jack, you want to lift a 4,000 lbm rock. The large cylinder has a diameter of 6 inches.

jolli1 [7]

Answer:

a diameter of D₂ = 0.183 inches would be required

Explanation:

appyling pascal's law

P applied to the hydraulic jack = P required to lift the rock

F₁*A₁ = F₂*A₂

since A₁= π*D₁²/4 , A₂= π*D₂²/4

F₁*π*D₁²/4 = F₂* π*D₂²/4

F₁*D₁²=F₂*D₂²

D₂ = D₁ *√(F₁/F₂)

replacing values

D₂ = D₁ *√(F₁/F₂) = 6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches

6 0

3 years ago

A company, studying the relation between job satisfaction and length of service of employees, classified employees into three le

Wewaii [24]

Answer:

Below see details

Explanation:

A) It is attached. Please see the picture

B) First to calculate the overall mean,

μ=65∗25/75+80∗25/75+95∗25/75

μ=65∗25/75+80∗25/75+95∗25/75 = 80

Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634

And E(MSE) = σ^2= 9

C) Yes, it is substantially large than E(MSE) in this case.

D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.

3 0

3 years ago

Giải giúp mình câu này được không ạ.em cảm ơn

sp2606 [1]

Answer:

çâdßèöñbvjsjaushdhshs

4 0

3 years ago

In a steam power plant, 1 MW is added in the boiler, 0.58 MW is taken out in the condenser and the pump work is 0.02 MW.

Free_Kalibri [48]

Answer:

a) $\eta = 42\,\%$ , b) $COP_{R} = 29$

Explanation:

a) The thermal efficiency is:

$\eta = \frac{\dot Q_{in} - \dot Q_{out}}{\dot Q_{in}}\times 100\,\%$

$\eta = \frac{1\,MW-0.58\,MW}{1\,MW} \,\times 100\,\%$

$\eta = 42\,\%$

b) The coefficient of performance is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{0.58\,MW}{0.02\,MW}$

$COP_{R} = 29$

3 0

3 years ago