Identify the characteristics of parallel flow heat exchangers.

The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.

liberstina [14]

Answer:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

i (t) = 6*e^(-2*t)

v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

- The amount of charge Q delivered can be determined by:

dQ = i(t) . dt

$Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt$

- Integrate and evaluate the on the interval:

$= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C$

- The power can be calculated by using v(t) and i(t) as follows:

v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

$W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt$

- Integrate and evaluate the on the interval:

$W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ$

6 0

4 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

3 years ago

Read 2 more answers

Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

nikitadnepr [17]

Answer:

a) $m_e= 3.05 Kg$

b) $\rho=1072.3kg/m^3$

c) $m_e= 3.05 Kg$

Explanation:

From the question we are told that:

Beaker Mass $m_b=1.20$

Liquid Mass $m_l=1.85$

Balance D:

Mass $m_d=3.10$

Balance E:

Mass $m_e=7.50$

Volume $v=4.15*10^{-3}m^3$

a)

Generally the equation for Liquid's density is mathematically given by

$m_e=m_b+m_l+(\rho*v)$

$\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}$

$\rho=1072.3kg/m^3$

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

$m_d=m- (\rho *v )$

$m=3.10+ (1072.30 *4.15*10^{-3}m^3 )$

$m=18.10kg$

c)

Generally the equation for E's Reading at A pulled is mathematically given by

$m_e=m_b+m_l$

$m_e = 1.20 + 1.85$

$m_e= 3.05 Kg$

6 0

3 years ago

Disadvantage of metal

ivolga24 [154]

You can get hurt if u don’t use it properly.

7 0

3 years ago

A keyboard, monitor and mouse are all considered as computer hardware and a program is known as a software.

Iteru [2.4K]

Answer:

This should be true because program is a software and mouse and keyboard are specific computer hardware to make your exsperience with a computer more efficent and faster.

5 0

3 years ago